aieee physics practical material - doubtion · volume of a cylinder r2l and volume of a sphere 3 4...
TRANSCRIPT
Page No. 1
AIEEE PHYSICS PRACTICAL MATERIAL VERNIER CALLIPERS
SYNOPSIS
The maximum precision upto which a scale can measure is 1mm or 0.1 cm. To measure lengths less than 1mm, vernier calipers is used. Vernier calipers is also called as slide calipers.
Volume of a cylinder 2r l and volume of a sphere 34 r3
can be determined by using vernier
calipers by finding radius and length of the cylinder and by finding the radius of the sphere. Vernier calipers consists of two scales 1) Main Scale 2) Vernier Scale Main Scale: It is caliberated in cm on one side and in inches on the other side. The value of one
main scale division (MSD) is 1mm or 0.1 cm. Vernier Scale: It is the movable scale that can slide on the main scale. Generally 10 vernier scale
divisions is equal to 9 main scale divisions. Therefore the value of one vernier scale division
(VSD) is equal to 910
mm.
The upper two jaws are used to find the internal diameter of a pipe, test tube and that of hollow cylinder.
The lower two jaws are used to measure the length and diameter of a cylinder or the diameter of a sphere by placing the object between the lower jaws.
A thin metallic strip projects at one end of the calipers which is used to measure the depth of the vessel.
The smallest length upto which the vernier calipers can measure is called least count or vernier constant.
Least count of vernier calipers is defined as the ratio of the value of one MSD to the number of divisions on the vernier scale.
Least Count = S Valueof 1MSDN Number of divisionson the vernier scale .
Least count of vernier calipers is also defined as the difference between the value of one MSD and the value of one VSD i.e. Least Count = 1MSD – 1 VSD
N divisions of vernier scale coincide with (N-1) divisions of the main scale N division of V.S. = (N-1) divisions of MS
1 div of V.S. = N 1
N
div. of M.S.
Least Count = 1MSD – 1VSD
=1MSD - N 1
N
MSD
1 MSDN
When the lower two jaws touch each other, the zero of the vernier coincides with the zero of the
main scale. Then there is no error. If the zero of the vernier doesnot coincide with zero of the main scale when the jaws are in
contact, then error exists and necessary correction has to be applied. If the zero of the vernier is to the right of the zero of the main scale, the extent of shift multiplied
by the least count gives zero error or zero reading. In this case, the final reading is greater than the actual reading of the object by an amount equal to zero error. Hence the zero error is substracted from the observed length to get correct length i.e. zero correction is negative.
Page No. 2
If the zero of the vernier is to the left of the zero of the main scale, then the zero error is added to the observed reading to get correct length i.e. zero correction is positive.
Zero correction is opposite to zero error. The jaws (lower) of the calipers should not be pressed too hard on the object placed between
them. At any position, the diameter should be measured in two directions at right angles to each other. Same units should be used while calculating the result. Vernier coincidence must be noted without any parallax error by repeating the observations five
times at different positions of the ball. Slide calipers is called vernier calipers since it was first designed by French mathematician named
Vernier. Rules for taking readings with vernier calipers. Least count is determined. For taking the main scale reading, the division next before the zero of the vernier should be
considered. The division of the vernier scale which coincides with any main scale division should be noted. (
Vernier coincidence ) Correct reading or length or diameter
= Main scale reading + (Least count Vernier Coincidence) A lens must be used to note vernier coincidence. Vernier principle is also used for measurement of fraction of angles as in extant.
PHYSICS – QUESTIONS ON VERNIER CALLIPERS
1. One centimeter on the main scale of Vernier calipers is divided into 10 equal parts.. If 10 divisions
of vernier coincide with 8 small divisions of main scale, the least count of the calipers is 1) 0.01cm 2) 0.02 cm 3) 0.05 cm 4) 0.005 cm 2. The error in the measurement of radius of the sphere by using vernier calipers is 0.3%. The
permissible error in the measurement of surface area is 1) 0.6% 2) 1.2 % 3) 1.8% 4) 0.9 %
3. A person performs an experiment with vernier calipers and takes 100 readings. He repeats the
same experiment but now takes 400 readings. Then probable error is
1) Remain the same 2) Is halved 3) Doubled 4) reduced by a factor of ¼
4. Least count of vernier callipers is 110N
cm. The value of one division on the main scale is 1mm.
then the no of divisions of the main scale that coincide with ‘N’ divisions of vernier scale is
1) 10 N 2) 10N 3) ( N -1 ) 4) N – 10
5. Choose the wrong statement
1) If the zero of vernier scale doesnot coincide with the zero of the main scale, then the vernier
callipers is said to be having zero error
2) Zero correction has a magnitude equal to zero error but sign is opposite to that of zero error.
3) Zero error is positive when the zero of vernier scale lies to the left of the zero of the main scale
4) Options 2, 3 are wrong
Page No. 3
6. The reading of the vernier callipers as shown in the fig. is _______________ mm
mm scaleo
5object
50 6o
vernier scale
1) 53.7 2) 53.2 3) 54.6 4) 54.7
7. Vernier constant is
1) The value of one MSD divided by total no of divisions on the main scale
2) The value of one VSD divided by total number of divisions on the vernier scale
3) Total number of divisions on the main scale divided by total no of divisions on the venier scale
4) The difference between value of one main scale division and one vernier scale division.
8. If the error in the measurement of the diameter of the sphere is 1 %, then the error in the
measurement of the volume is
1) 3 % 2) 5 % 3) 10 % 4) 12 %
9. The volume of the sphere measured by vernier callipers is 36 . Then dVdR
is
1) 9 2) 18 3) 27 4) 36
10. The least count of the vernier callipers is 0.01 cm. The length and breadth of the rectangular
lamina are 0.93cm and 0.45 cm respectively. The percentage error in the measurement is
1) 1 % 2) 2 % 3) 3 % 4) 4 %
11. The vernier of a circular scale is divided into 30 divisions which coincide against 29 main scale
divisions. Each main division is 01
2. The least count of the instrument is
1) 110 2) 10 1. 3) 11 4) 130
12. The least count of vernier callipers is 0.01 cm. The percentage error in the measurement of length
and breadth of a rectangular lamina by using vernier callipers is 0.125 % and 0.2 % respectively.
Then length of the lamina is
1) 8 cm 2) 12cm 3)15cm 4) 23 cm
13. In the above question, the breadth of the lamina is
1) 2 cm 2) 5 cm 3) 9 cm 4) 14 cm
14. The length of the cylinder measured by vernier callipers is 10.3 cm, 10.2 cm, 10.1 cm, 9.7 cm, 10
cm, 9.8 cm and 9.9 cm. The mean deviation of the observations is
1) 0.241 2) 0.452 3) 0.171 4) 0.425
Page No. 4
15. In the above problem, the variance is
1) 0.047 2) 0.032 3) 0.052 4) 0.027
16. Which of the following is the most accurate
1) Vernier callipers having 20 divisions on the vernier scale which coincide with 19 divisions on
the main millimeter scale
2) Vernier callipers having 50 VSD coinciding with 40 MSD
3) A vernier scale of least count 0.01 mm
4) A vernier callipers of least count 0.01 cm
17. The least count of vernier callipers is 0.01 cm. Then the error in the measurement is
1) > 0.01 2) 0.01cm 3) <0.01 4) 0.01
18. The length of the cylinder is measured as 4.5cm by a vernier callipers of least count
0.01cm. The radius is measured as 0.125cm with a screwgauge of least count 0.001cm
The percentage error in the determination of the volume of the cylinder is
1) 1% 2) 2% 3) 3% 4) 4%
19. The length of the strip measured with metre rod is 10.0cm. Its width is measured with
Vernier callipers is 1.00cm. The least count of metre rod is 0.1cm and that of Vernier
Callipers is 0.01cm. The error in the area is
(1) 0.01cm2 2) 20 1. cm 3 ) 20 11. cm 4) 20 2. cm
20. The side of a cubical block when measured with a vernier calipers is 3.5cm .The vernier constant
is 0.01. The maximum possible error in the area of the side of the block is
1) 20 01. cm 2) 20 02. cm 3) 20 1. cm 4) 20 07. cm
21. The length and breadth of the block are 12.00cm and 13.50cm. They are measured with vernier
calipers having least count 0.01cm. The perimeter of the block is
1) 51 00 0 00. . cm 2) 51 00 0 01. . cm 3) 51 00 0 02. . cm 4) 51 00 0 04. . cm
22. 1 cm of main scale of a vernier calipers is divided into 10 divisions. The least count of the calipers
is 0.005 cm, then the vernier scale must have
1) 10 divisions 2) 20 divisions 3) 25 divisions 4) 50 divisions
23. The atmospheric pressure is measured with a fortin’s barometer having a vernier scale whose 20
divisions coincide with 19 divisions of main scale. Each division of the main scale is equal to 0.5
mm. The correct atmospheric pressure is
1) 75.0050 cm of Hg 2) 75.005 cm of Hg 3) 75.00 cm of Hg 4) 71.05 cm of Hg
Page No. 5
24. The accurate value of the length of the pencil is 6.04cm Two students A and B measure this
length, one by a metre scale and another by a vernier calipers respectively. ‘A’ measures the
length as 5.9 cm while that of B is 6.37 cm. Then accuracy is more in the measurement of
1) A 2) B 3) A and B equally 4) Neither A nor B
25. Vernier calipers measures the dimensions of a rectangular block as 5 10 5mm mm mm. The least
count of vernier calipers is 0.01 cm. The maximum percentage error in the measurement of
volume of the block is
1) 20 % 2) 15 % 3) 10 % 4) 5 %
26. The least count of vernier calipers is 0.1mm. The main scale reading before the zero of the vernier
scale is 10 and zeroth division of the vernier scale coincides with MSD. Each MSD is 1mm, the
measured value should be expressed as
1) 0.01 m 2) 1 cm 3) 1.0 cm 4) 1.00 cm
27. The diameter of the cylinder is measured with vernier calipers having least count of 0.01cm. The
diameter is 1.95 cm. The radius (w.r.t. significant figures) is
1) 0.975 cm 2) 0.98 cm 3) 1.0 cm 4) 1 cm
28. Each division on the main scale is 1mm. Which of the following vernier scales give vernier
constant equal to 0.01 mm?
1) 9mm divided into 10 divisions 2) 90 mm divided into 100 divisions
3) 99 mm divided into 100 divisions 4) 9 mm divided into 100 divisions
KEY
1 – 10 : 2 1 2 3 3 2 4 1 4 3 11 – 20 : 3 1 2 3 1 3 4 2 4 4 21 – 30 : 3 2 1 1 4 4 2 3
HINTS
1. Least count = 1MSD 1VSD
0.1 0.08 0.02cm .
2. A r% 2 100A r
.
3. 1errorNo.of obervations
.
4. 1VSD 1MSD L.C.
Page No. 6
8. DV Dd% 3 100V d
.
9. Volume = 3
34 R
10. .100100 b
DbL
DL
11. 30VSD = 29MSD
1 VSD = 00
6029
21
3029
3029
MSD
12. D 100 0.125ll , where D 0.01cml .
13. Db 100 0.2b
, where Db 0.01cm .
14. nx xd
n
, where nx = each value, and x =arithmetic mean.
15. 2 2 2
1 2 nd d ........ dn
.
16. Less least count implies more accuracy.
18. 1001002100 l
Dlr
DrV
DV
20. 1002100 L
DLA
DA
1002100
L
DLADA
21 Perimeter 2 l b 4 (least count).
23 Least count of barometer 0.5mm 0.0025cm2D
so measurement is up to 4 decimal places in cm .
25 Error % 310100
5.
10.
5.
CLCLCL .
26 Length measured =Reading MSR L.C Vernier cerncidence 10mm 0 0.1m 1.00cm .
28 L.C 991MSD 1VSD 1mm mm100
Page No 13
SCREW GAUGE 1. Which instrument is used for measuring the thickness of very thin plates or wires. Ans. Screw Gauge 2. What is the working principle of screw gauge? Ans. It works on the principle of screw. In this rotational motion is converted into translational motion. 3. Is the screw gauge a directly measuring instrument? Ans. Yes, the reading directly gives the measurement. 4. Of screw gauge and vernier callipers, which is more sensitive? Ans. Screw gauge is more sensitive. 5. What is pitch of the screw? Ans. It is the distance between two consecutive threads of screw, measured along axis of screw.
(or) The distance between two consecutive threads of screw
(or) The distance moved by the screw tip for one complete rotation of the screw is called pitch of the
screw. Distance movedPitch of the screw=No.of rotations
6. What is least count of screw gauge?
Ans. Pitch of the screw 1L.C= 0.01No.of divisions on the head scale 100
mm mm
( If pitch of the screw is 1mm and No.of head scale divisions are 100) 7. What are the errors present in the measurement made by a screw gauge Ans. The systematic errors like least count error, zero error and the random error like backlash error are
present in the measurement. 8. Can we make the zero error cancelled? Ans. Yes, by rotating the screw in the fixed stud with a screw driver 9. How do you increase the accuracy of the measurement? Ans. By decreasing the least count, the least count error is decreased. Zero error is estimated and
corrected. The backlash error is avoided by taking the average of number of readings. Thus the accuracy is increased by decreasing the errors.
10. What are different zero errors? How they can be corrected? Ans. The head of the screw is moved until the two shafts are in touch with each other, if zero division of
the head scale coincides with the reference line, the zero error is nil, so correction is also nil. If zero division does not coincide with reference line then the instrument has zero error.
(or) If zero of the main (pitch) scale doesn’t coincide with zero of circular scale on bringing the two
shafts in contact, the screw gauge is said to have zero error. Positive error:- If the head scale zero is below the index line, zero error is positive, and the zero correction is negative
Corrected head scale reading = observed reading – (error) Negative error: - If the head Scale zero is above the index line, zero error is negative and zero correction is positive.
Corrected head scale reading = observed reading + (error)
11. The head scale of a screw gauge has 100 divisions engraved on it. If its spindle moves by 2mm on sleeve, when four complete rotations are given, calculate (i) pitch (ii) least count
Page No 14
Ans. i) Distances moved by the screw on sleeve 2Pitch = 0.5 0.05No.of complete rotations 4
mm mm cm
ii) Pitch 0.05L.C = 0.0005No.of divisions on the head scale 100
cm cm
12. Figure 1.15 shows a screw gauge in which head scale has 100 divisions. Calculate (i)Least count (ii) diameter of wire(Assume that zero error is nil)
Ans. (i) Pitch L.C =No.of head scale divisions
=1 0.01 0.001100mm mm cm
(ii) Pitch scale reading (PSR) = 4mm = 0.4cm Head scale reading (HSR) = 34 divisions Observed diameter (d) = P.S.R + (L.C x H.S.R) = 0.4 + 0.001 x 34 = 0.4 + 0.034 = 0.434 cm 13. Pitch (Main) scale of a screw gauge has 10 divisions to a centimeter and its circular scale (head
scale) has 100 divisions, such that the spindle advances by one division on one complete rotation. Calculate the (i) Pitch (ii) least count. If this instrument has a positive zero error of 4 divisions and the reading on the pitch scale is 4 divisions and that on the head scale is 74 divisions, find the diameter of wire.
Ans. (i) 1/10Pitch scale division Pitch = 0.1No.of rotations 1
cmcm
(ii) Pitch 0.1L.C = 0.001Head scale divisions 100
cm
Pitch scale reading (PSR) = 4 divisions x 0.1 cm = 0.4cm HSR (observed ) = 74 Correct H.S.R = 74 – error = 74- 4 = 70 Correct diameter = P.S.R + (L.C x corrected H.S.R) = 0.4 + (0.001 x 70) = 0.4 + 0.07 = 0.470 cm
Page No 7
MICROMETER-SCREW GAUGE SYNOPSIS
1) It is an instrument used for measuring the diameter of very thin wires or similar objects. Its accuracy
is upto 0.001 cm and is commonly called micrometer screw gauge.
2) Construction : The screw gauge consists of the following parts:
a. U-Frame : It is a steel frame, shaped in the form of U. On one end of U-frame a screw is fixed
permanently. It is commonly called stud and forms the fixed jaw of the screw gauge. On the other
end of U-frame is fixed a nut, through which slides a screw. The end B of the screw forms the
movable jaw of screw gauge.
b. Nut and screw : The nut is threaded from inside and the screw from outside. The screw can move in
and out of nut by circular motion.
c. Thimble or circular cylinder : The screw is connected to a hollow circular cylinder(S), which
rotates along with nut on turning.
d. Sleeve cylinder :To the nut is attached a hollow cylinder, commonly called sleeve cylinder. The
spindle of the screw passes through sleeve cylinder.
e. Base line : A reference line or base line, graduated in mm, is drawn on the sleeve cylinder, parallel
to the axis of nut. It is commonly called main scale or sleeve scale.
f. Circular scale or Thimble scale: The hollow cylinder moving over the sleeve cylinder is tapered
from one end. On the tapered end are made graduations, which are either 50 or 100 in number. The
scale marked on sleeve is called circular scale or thimble scale or head scale.
g. Ratchet: The ratchet is attached to screw by means of a spring. When the flattened end B of the
screw comes in contact with stud A, the ratchet becomes free and makes a rattling noise. Thus, end
B of the screw is not further pushed towards the stud A.
3) Pitch of Screw : The pitch of screw is defined as, the distance between two consecutive threads of
screw, measured along the axis of screw. (OR)
Pitch of screw can also be defined as the distance traveled by the tip of screw (end B) when head of
screw is given one complete rotations.
Page No 8
Working – Principle of Screw Gauge : It works on screw principle that rotatory motion can be
converted into translatory motion.
4) Determination of Pitch of screw : In order to determine pitch, the screw is given five complete
rotations. The distance moved by the thimble on the main scale is recorded. The pitch can be found
by the formula.
Distance moved bythimbleon M.SPitch
Number of rotations of thimble [M.S stands for main scale]
Ex :- If 5 mm is the distance moved by the thimble on the main scale for 5 rotations then :
5 15mm mm Pitch
5) Least count of screw : Least count of the screw is the smallest distance moved by its tip when the
screw turns through 1 division marked on it.
6) Determination of Least Count : Determine the pitch and count the number of divisions on circular
scale. The least count is determined by the formula given below.
. .LC PitchNumber of C.S divisions
[C.S. stands for circular scale]
It pitch of screw is 1 mm and number of divisions marked on its thimble are 100, then:
1. . 0.01 0.001100mmLC mm cm .
7) Determination on Zero Error : If the zero of the main scale does not coincide with zero
of circular scale on bringing the screw end B in contact with stud A, the micrometer is said to have
zero error.
a) Positive Zero Error : If the zero line, marked on circular scale, is below the reference line of the
main scale, then there is a positive zero error and the correction is negative. In the diagram (Fig. 1),
5th circular scale division is coinciding with reference line.
Ex :- Correction = - coinciding division of . . 5 0.001 0.005C S L C cm cm .
If the observed diameter is 0.557 cm, then:
Corrected diameter = Observed diameter Correction
= 0.557 cm – 0.005 cm. = 0.552 cm
Fig - 1
b) Negative Zero Error : If the zero line marked on circular scale, is above the reference line of the
main scale, then there is a negative error and the correction is positive.
Page No. 8
Page No 9
Ex :- Correction = +[n – coinciding division of C.S. x L.C], where n is the total number of
circular scale divisions.
Correction = + [100 – 96 ] x 0.001 cm= 0.004 c,
If observed diameter is 0.557 cm then:
Corrected diameter- Observed diameter Correction = 0.557 cm + 0.004 cm = 0.561 cm.
Fig - 2
NUMERICAL PROBLEMS ON SCREW GAUGE Numerical problem 1: The thimble of a screw gauge has 100 divisions engraved on it. If the thimble
advances 2 mm, when four complete rotations are given, calculate : (i) Pitch (ii) least count.
Solution:
(a) Distance moved byscrew on sleevePitch
Number of complete rotations = 2 0.5 0.05
4mm mm cm
(b) L.C. 0.05 0.0005100
cm cm Pitch
No.of circular scale divisions
Numerical problem 2 : Figure shows a screw gauge in which thimble has 100 divisions. Calculate (i) Least
count (ii) Diameter of wire.
Least count = 1 0.01 0.001100mm mm cm
PitchNo.of circular scale divisions
Main scale reading = 4 mm = 0.4 cm
Circular scaled reading = 34 div
Observed diameter = M.S reading + L.C x C.S.D = 0.4 cm + 0.001 cm x 34 = 0.434 cm
Page No 10
EXERCISE 1. The diagram shows a screw gauge in which circular scale has 200 divisions. Calculate the least count
and radius of wire.
[(i) 0.0005 cm (ii) 0.2585 cm]
2. The diagram shows a screw gauge in which circular scale has 100 divisions. Calculate the least count
and the diameter of a wire.
[(i) 0.0005 cm (ii) 0.4865 cm]
3. A atomic screw gauge has a negative error of 8 divisions. While measuring the diameter of a wire the
reading on main scale is 3 divisions and 24th circular scale division coincides with base line.
If the number of divisions on the main scale are 20 to a centimeter and circular scale has 50
divisions, calculate
(i) pitch (ii) least count (iii) observed diameter (iv) corrected diameter
4. A micrometer screw gauge has a negative error of 7 divisions. While measuring the diameter of a
wire the reading on main scale is 2 divisions and 79th circular scale division coincides with base line.
If the number of divisions on main scale is 10 to a centimeter and circular scale has 100 divisions,
calculate
(i) pitch (ii) least count (iii) observed diameter (iv) corrected diameter
Multiple Choice Questions 1. The least count of Screw guage is?
a) 0.01 cm *b) 0.01 mm c) 0.1 mm d) 0.001 mm
2. The pitch of screw gauge is
*a) 1 mm b) 0.1 mm c) 1 cm d) 0.1 cm
3. Screw gauge can measure the diameter of thin wires or similar objects with accuracy upto
a) 1cm b) 0.1 cm c) 0.01 cm *d) 0.001 cm
4. Pitch of screw of a screw gauge is
*a) distance moved by thimble on main scalenumber of rotation of thimble
b) pitchnumber of circular scale divisions
Page No 11
c) number of rotation of thimblenumber of circular scale divisions
d) None of the above
5. Least count of screw gauge is defined as
a) distance moved by thimble on main scalenumber of rotation of thimble
*b) scaleheadordivisionsscalecircularofnumber
screwtheofpitch
c) number of rotation of thimblenumber of circular scale divisions
d) None of the above
6. Screw gauge can be used to determine
a) Thickness of glass piece b) diameter of a wire
*c) to measure thickness of glass piece and to measure diameter of wire d) None of the above
7. Screw gauge contains following scales
a) main scale, vernier scale b) main scale, ordinary scale
*c) main scale, head scale d) only main scale
8. Screw gauge is said to have negative error
a) when head scale zeroth division coincides with base line of main scale
*b) when head scale zeroth division is above with base line of main scale
c) when head scale zeroth division is below with base line of main scale
d) None of the above
9. For negative error correction is
*a) positive b) negative c) no correction d) none of the above
10. Screw gauge is said to have positive error when
a) when head scale zeroth division coincides with base line of main scale
b) when head scale zeroth division is above with base line of main scale
*c) when head scale zeroth division is below with base line of main scale
d) None of the above
11. For positive error correction is
a) positive *b) negative c) no correction d) none of the above
12. If the observed diameter of a wire with screw gauge is 0.431 cm and the device negative error is 19
then corrected diameter is
*a) 0.450 cm b) 0.419 cm c) 0.431 cm d) None of the above
13. If the observed diameter of a wire with screw gauge is 0.456 cm and the device positive error is 16
then corrected diameter is
a) 0.472 cm *b) 0.440 cm c) 0.4 cm d) 0.441 cm
14. The diameter of a wire is measured using screw gauge of zero error as
a) main scale reading + circular scale reading L.C.
b) circular scale reading + main scale reading L.C.
Page No 12
c) main scale reading + vernier scale reading L.C. d) None of the above
15. In a Screw gauge value of each main scale division and circular scale division is
a) 1 mm & 0.01 cm b) 0.01 cm & 1 mm *c) 1 mm & 0.001 cm d) 0.001 cm & 1 mm
16. The number of circular scale or head scale divisions of a screw gauge is
*a) 100 b) 150 c) 200 d) 50
17. Screw gauge is preferable to measure thickness of a wire instead of using a ordinary scale because
*a) least count is more b) least count is less c) no difference in reading d) None of the above
18. For more accuracy which of vernier and screw gauge is preferable to measure thickness of glass plate
a) vernier *b) screw gauge c) both vernier & screw gauge d) none
Page No. 15
SIMPLE PENDULUM A point mass suspended by means of a torsionless string is called simple pendulum.
Oscillations of a simple pendulum are simple Harmonic provided the amplitude should be very small (< 05 )
The restoring force mgsin (Component of gravitational force) is responsible for the S.H.M.
Tension in the string of the pendulum T mg cos
As increases restoring force increases and tension decreases.
Simple pendulum will be stopped after sometime because of friction of thread and air resistance
Simple pendulum can oscillate in vacuum
Period of a simple pendulum T 2 / g
LAWS OF SIMPLE PENDULUM
Period of simple pendulum is independent of the mass, material size & shape of the bob.
Period of simple pendulum is independent of amplitude.
T when g is constant.
1Tg
when is constant.
The length of a simple pendulum is halved its period decreases by 2 times
The length of a simple pendulum is doubled its period increases by 2 times
The length of a simple pendulum is increased by 21% its period increases by 10%
If the period of a simple pendulum increases by 20% its length increases by 44%
If the period of a simple pendulum decreases by 20% its length decreases by 36%
The length of a simple pendulum is decreases by 1 % its period decreases by 0.5 %
The length of a simple pendulum is increased by 1 % the time its lose per a day 432 sec.
A hollow bob of a simple pendulum is filled with a liquid. If the liquid leaks through a hole at the bottom.
Its time period increases first then decreases and finally reaches the original value.
If the hole at the middle on one side of the bob then the time period increases.
In summer the pendulum of a clock expands time period increases and it goes slow or lose time
In winter the pendulum of a clock contract its period decreases so it gains time and goes fast.
A simple pendulum is suspended in a lift moving up with an acceleration “a” then its period decreases &
frequency increases.
T 2g a
, g an 1/ 2
Page No. 16
A simple pendulum is suspended in a lift moving down with an acceleration “a” its period increases &
frequency decreases.
T 2g a
, g an 1/ 2
When a simple pendulum suspended in a freely falling lift. Its time period is infinity & frequency is zero.
Simple pendulum is does not oscillate at places where g = 0.
Example : at the center of earth, in an artificial satellite & freely falling lift
A pendulum whose period of oscillation is 2 sec is called “Second Pendulum”.
Length of a seconds pendulum 2g / about 100 cm
L - T graph of a simple pendulum is a parabola
2L T graph of a simple pendulum straight line passing through the origin
L - T & 2L T graph intersection at T = 1 & L = 25 cm
The time period of simple pendulum of length comparable of earth’s radius.
1T = 21 1g
R
CASE 1 : The time period of simple pendulum of infinite length
RT = 2g
= 84 min.
CASE 2 : The time period of simple pendulum of length comparable of earth radius R
RT = 22g
= 59.4 min.
ENERGY OF A PARTICLE IN S.H.M.
POTENTIAL ENERGY
The potential energy at a displacement y is the work done against the restoring force in moving the
body from the mean position to this position.
Potential Energy = y
2
0
1kydy ky2
Potential Energy 2 2 2 2 21 1m y m A sin wt2 2
The maximum value of P.E. is 2 21 mw A2
.
P.E. is maximum at the extreme position, i.e. when y = A and zero (minimum) when y = 0.
Page No. 17
KINETIC ENERGY
Kinetic Energy = 2 2 2 21 1mv m A cos wt2 2
2 2 21 m A y2
The maximum value of K.E. is = 2 21 m A2
.
K.E. is maximum at the mean position, y = 0, and zero when y = A
TOTAL MECHANICAL ENERGY
Total mechanical energy
E = kinetic Energy PotentialEnergy
2 21E mw A cons tan t2
So, we observe that total mechanical energy is a constant.
Page No. 18
MULTIPLE OPTIONS
1. If A is amplitude of a particle in SHM, its displacement from the mean position when its kinetic energy is
thrice that of its potential energy
1) A 2) A/4 3) A/2 4) 3A/4
2. A particle of mass 0.1kg is executing SHM of amplitude 0.1m. When the particle passes through the mean
position, its kinetic energy is 38 10 J. Then the equation of its motion when the intial phase is 045 is
1) 0.2sin 4t / 4 2) 0.1sin 4t / 4 3) 0.1sin 2t / 4 4) 0.1sin t / 4
3. Two simple pendulums of lengths 100cm and 1600 cm are in phase at the mean position at a certain instant
of time. If T is the first pendulum, then the minimum time after which they will be in phage again is
1) 3T4
2) 4T3
3) 3T 4) 4T
4. A pendulum clock of length is moving up with deceleration g/3. Then the number of oscillations made by
that pendulum in 2 seconds is
1) 1 2g3
2) 1 2g2 3
3) 1 3g2
4) 1 3g2 2
5. Two simple pendulums oscillate simultaneously from mean positions. The first pendulum made 20
oscillations in certain time in which the second pendulum made 25 oscillations. Then the ratio of the lengths
of the pendulums is
1) 5 : 4 2) 4 : 5 3) 25 : 16 4) 16 : 25
6. A seconds pendulum is taken to an altitude above the earth, which is equal to radius of the earth, there its
new time period would be
1) 2 sec 2) 4 sec 3) 2 sec 4) 2 2 sec
7. If a simple pendulum of length ' ' has maximum angular displacement , then the maximum kinetic
energy of its bob of mass m is
1) mg2 2) mg
2 3) mg 1 sin 4) mg 1 cos
8. A simple pendulum is suspended from the roof of a cart that moves down an inclined plane without friction.
It ' ' is the inclination of the inclined plane to the horizontal, time period of oscillation of the pendulum is
1) cos2g
2) sin2
g
3) 2
gsin
4) 2
g cos
9. A pendulum which gives correct time beats seconds on the ground. It is moved to an altitude 640 m above
the earth. If radius of the earth is 6400 km, how much time will the pendulum lose or gain in one day ?.
Page No. 19
1) Loses 4.32s 2) Gains 4.32s 3) Loses 8.64s 4) Gains 8.64s
10. A simple pendulum has a bob which is a hollow sphere full of sand and oscillates with certain period. If all
that sand is drained out through a hole at its bottom, then its period.
1) Increases 2) Decreases 3) Remains same 4) Is zero
11. A simple pendulum is suspended inside a trolly which is sliding down on an inclined plane, which is
frictionless. It oscillates, mean position of that pendulum will be
1) Exactly vertical 2) Exactly horizontal
3) Normal to the inclined surface 4) Parallel to the inclined surface
12. If a pendulum clock is shifted from Guntur to Thirumala it will (altitude of latter is more)
1) Gain time 2) loss time 3) Neither gain nor lose time 4) Not work
13. A simple pendulum oscillates with certain frequency in a stationary lift. If the lift is in motion, its frequency
would decrease when.
1) The lift moves up with uniform acceleration
2) The lift moves down with uniform acceleration
3) The lift moves with uniform velocity
4) Any of these
14. The period of a simple pendulum depends on
1) Length of the pendulum 2) Acceleration due to gravity
3) Both 1 and 2 4) None of these
15. If a hole is drilled along the diameter of the earth and you leave a coin in the hole, then
1) If falls off and leaves the earth
2) It falls off and finally stops at the center of the earth
3) It falls off but does not leave the earth
4) It falls and comes back to you
16. A pendulum clock shows correct time at 00 C. On a summer day
1) It runs slow and gains time 2) It runs fast and loses time
3) It runs slow and loses time 4) It runs fast and gains
17. The seconds pendulum is taken from earth to moon, to keep the time period constant
1) length of the seconds pendulum should be decreased
2) the length of the seconds pendulum should be raised
3) the amplitude should increase
4) the amplitude should decrease
18. A simple pendulum hanging freely and at rest is vertical because in that position
Page No. 20
1) K.E. is zero 2) K.E. is minimum 3) P.E. is zero 4) P. E is minimum
19. The percentage change in the time period of a seconds pendulum when its amplitude is reduced by 50% is
1) 75% 2) 25% 3) 0% 4) 50%
20. The length of a seconds pendulum at the surface of earth is 1m. the length of second pendulum at the
surface of moon where g is 1/6th that at earth’s surface is
1) 1/6m 2) 6m 3) 1/36m 4) 36m
21. The time period of a seconds pendulum is 2sec. The spherical bob which is empty from inside has a mass
50g. This is now replaced by another solid bob of same radius but having a different mass of 100g. the new
time period will be
1) 4sec 2) 1sec 3) 2sec 4) 8sec
22. A simple pendulum is made of a body which is a hollow sphere containing mercury suspended by means of
a wire. If a little mercury is drained off, the period of pendulum will
1) Remain unchanged 2) Increase 3) Decrease 4) Become erratic
23. The period of oscillation of a simple pendulum of constant length at earth’s surface is T. Its period inside a
mine is
1) Greater than T 2) Less than T 3) Equal to T 4) Can not be compared
24. To total energy of a particle executing SHM is proportional to
1) Displacement from equilibrium position 2) Frequency of oscillation
3) Velocity in equilibrium position 4) Square of amplitude of motion
25. The time period of a simple pendulum is doubled, when
1) Its length is doubled 2) Mass of bob is doubled
3) Length is made four times 4) The mass of bob and length of pendulum are doubled
Key
1. 3 2. 2 3. 2 4. 1 5. 3 6. 2 7. 4 8. 4 9. 3 10. 3
11. 3 12. 2 13. 2 14. 3 15. 4
16. 3 17. 1 18. 4 19. 3 20. 1
21. 3 22. 2 23. 1 24. 4 25. 3
Page No. 21
Moments and Equilibrium Turning Effects of a Force If we have hinged or pivoted body, any force applied changes the rotation of that body about the pivot. The turning effect is called a moment.
The equation is:
Moment = force × perpendicular distance
In Physics code:
= Fs
This strange looking symbol, , is “gamma”, a Greek capital letter ‘G’. Units are newton metre (Nm).
Moments have a direction. As they are turning effects, we can talk of clockwise and anti-clockwise moments. By convention, clockwise is positive.
Question 1 The spanner in the picture is 30 cm long and the nut in question has to be tightened to a torque (moment) of 85 Nm. What force must the fitter apply?
Page No. 22
Ans :
The spanner in the picture is 30 cm long and the nut in question has to be tightened to a torque (moment) of 85 Nm. What force must the fitter apply?
Moment = Force x distance
85 Nm = Force x 0.30 m
F = 85 ÷ 0.30 = 283 N
Confess! Did you forget to convert the distance to metres? A very common bear trap.
Consider a trap door held by a piece of string, BC. P and Q are forces. The trap door is hinged about point O.
The perpendicular distance of the line of action of force Q is the length of the line OC.
Moment of P about O = P × OA
Moment of Q about O = Q × OC
Note that the Moment of Q about O is [Q ×OC]. This is because OC is the perpendicular distance of the force Q from the hinge O.
If the trap door remains in equilibrium:
Anticlockwise moment (Q × OC) = clockwise moment (P ×OA)
This is the Principle of Moments.
Page No. 23
Since OC = OB sin we can say that Q × (OB sin ) = P × OA
Question 2 The trap door in the diagram has a mass of 12 kg. The centre of mass is at A. The weight is force P. What is the value of force P?
Ans :
The force P is the weight.
Weight = mass x acceleration due to gravity
Weight = 12 kg x 10 m/s2 = 120 N
Watch out for this very common bear trap.
Question 3 The angle is 25 degrees. The trap door is 100 cm wide. What is the value of force Q?
Ans :
Distance OA = 50 cm = 0.5 m
Clockwise moment = 0.5 m x 120 N = 60 Nm
Anticlockwise moment = 60 Nm = Q x 1.0 m x sin 25 = 0.423 x Q
Q = 60 Nm ÷ 0.423 = 142 N
Centre of Gravity We treat objects as point masses referring to a single point called the centre of gravity.
In regular object like a cube or a sphere, the centre of gravity is in the middle. In some objects the centre of mass is outside the object.
Page No. 24
The centre of gravity is the point through which the entire weight is said to act. Objects with a very low centre of gravity tend to be very stable. Some objects are so stable that they never fall over. Objects with a high centre of gravity are unstable.
Balancing
Many questions involve the balancing of see-saw around pivots. Let us look at some situations:
This is the simplest case. The pivot is in the middle of a uniform bar. It means that the object is totally regular and the centre of mass is in the middle. Therefore we can ignore the mass of the bar. If the bar is balanced, we can say that:
anticlockwise moments = clockwise moments
Ax = By
Let us look at a case where we move the pivot P.
Page No. 25
In this case, the line of action of the weight, W, is z metres from the pivot P. Applying the principle of moments we can say:
Ax = By + Wz
Question
4
Some children are playing on a see-saw as shown in the diagram. The see-saw is a plank of wood 3.0 m long, with a pivot exactly in the middle. The centre of mass is directly above the pivot, so we can ignore it.
(a) What are the weights of the children?
(b) Child B is sitting 0.4 m from the pivot. Where should child C sit so that the see-saw remains level?
(c) Child C misses its footing and falls off the end. What will happen to the others?
(a) What are the weights of the children?
A: 35 kg = 350 N B: 25 kg = 250 N C: 40 kg = 400 N
(b) Child B is sitting 0.4 m from the pivot. Where should child C sit so that the see-saw remains level?
Anticlockwise moment caused by child A = 350 N x 1.5 m = 525 Nm
Total clockwise moment = 525 Nm Clockwise moment caused by child B = 250 N x 0.4 m = 100 Nm Clockwise moment caused by child C = 525 Nm - 100 Nm = 425
Nm 425 Nm = d x 400 N
Page No. 26
d = 425 Nm ÷ 400 N = 1.06 m
C should sit 1.06 m from the pivot.
(c) Child C misses its footing and falls off the end. What will happen to the others?
Child A would end up on the floor, because its anti-clockwise moment of 525 Nm is much larger than the clockwise moment from child B (100 Nm)
Child B would go up in the air.
Couples
If two forces act about a hinge in opposite directions, there is an obvious turning effect called a couple. The resulting linear force from a couple is zero.
The couple is given by the simple formula:
= 2 Fs
This strange looking symbol, , is “gamma”, a Greek capital letter ‘G’. Couples are measured in Newton metre (Nm)
The turning effect is often called the torque. It is a common measurement made on motors and engines, alongside the power. Racing engines may be quite powerful but not have a large amount of torque. This is why it would not make sense for a racing car to be hitched to a caravan, any more so than trying to win a Formula 1 race in a 4 x 4.
Principle of Moments redirects here. For the Robert Plant album, see The Principle of Moments. See also Moment (mathematics) for a more abstract concept of moments that evolved from this concept of physics.
In physics, the moment of force (often just moment, though there are other quantities of that name such as moment of inertia) is a pseudovector quantity that represents the magnitude of
Page No. 27
force applied to a rotational system at a distance from the axis of rotation. The concept of the moment arm, this characteristic distance, is key to the operation of the lever, pulley, gear, and most other simple machines capable of generating mechanical advantage. The SI unit for moment is the newton meter (Nm).
Moment = Magnitude of Force × Perpendicular distance to the pivot (Fd)
Overview
In general, the (first) moment M of a vector B is represented as M r B
where
r is the position where quantity B is applied. × represents the cross product of the vectors.
If r is a vector relative to point A, then the moment is the "moment M with respect to the axis that goes through the point A", or simply "moment M around A". If A is the origin, one often omits A and says simply moment.
н== Parallel axis theorem == Since the moment is dependent on the given axis, the moment expression possess a common y,
where
or alternatively,
Related quantities Some notable physical quantities arise from the application of moments:
Angular momentum (L = Iω ), the rotational analog of momentum. Moment of inertia (I), which is analogous to mass in discussions of rotational motion. Torque ( or ), the rotational analog of force. Magnetic moment (M), a dipole moment measuring the strength and direction of a
magnetic source.
Page No. 28
PHYSICS Determination of Young’s modulus Young’s modulus of the material of a given wire can be determined by using Searle’s apparatus. It consists of two wires in which both the wires taut with dead load. One wire will be used as reference
wire. Other wire is called as experimental wire. With the increase in load attached to the experimental wire it elongates. The elongation in the wire for a fixed increase in load in stops can be measured with micrometer screw. Micrometer reading will be taken after adjusting the micrometer screw so that air bubble in the spirit level lies at the middle of the spirit level. Spirit leve will be placed on the bar connceting the two wires. Air bubble will be disturbed from the middle of spirit level when the experimental wire elongates due to increase in the load attached to it, Average elongation e for a load M is determined. If is the length and r is radius of the wire then Young’s modulus Y the material of the given
wire is 2
g MYr e
Slope = load Melongation e
g is acceleration due to gravity. If elongations are measured for different values of load M then graph drawn by taking load M on y-axis
and elongation e on x-axis will give Me
value.
To avoid the backlash error micrometer screw is to be moved in upward direction only. Young’s modulus is the ratio of longitudinal stress to longitudinal strain.
F / AY
el
AMgYor
le
AMgY )(/
Extension or elongation increases with the load, length of the wire or Extension is more in a thin wire than thick wire. Extension is less if the substance is more elastic.
Young’s modulus of a substance increases with the addition of material of high young’s modulus and decreases with the addition of material of low Young’s modulus.
Young’s modulus of a substance decreases with increase of temperature. Young’s modulus of a substance increases with hammering and rolling because of breaking up of the
crystal grain. Young’s modulus of a substance decreases due to annealing. If K is inter atomic force constant, r0 is inter atomic distance then Young’s modulus of a wire Y and 0r are
related as 0K Yr . Y is independent of stress and strain. Elasticity of steel is more than that of copper. Elasticity of copper is more than that of gold.
Energy stored in a wire of length L and area of crossection A when a force F is applied is 2YAeE
2L
Where e is the extension in the wire and Y is young’s modulus of the wire.
Load
Elongation
Page No. 29
Young’s modulus (Y) of a solid related to its rigidity modulus (n) and bulk modulus (K) as 9 3 1Y n K
If is poisson’s ratio then Y 3K 1 2 and Y 2n 1 Young’s modulus value of liquids and gases is zero.
Slope of load (M) and elongation (e) graph is YALg
if L is the length, A is area and Y is youngs modulus of the wire when load (M) is taken on y-axis elongation (e) is taken on x-axis
Material Young’s modulus in N.m–2 Aluminium 0.7 1011
Brass 0.91 1011 Copper 1.1 1011 Iron 1.9 1011 Steel 2.0 1011 Tungsten 3.6 1011 Determination of Young’s modulus 1. A metal wire of length 30 cm and diameter 1 mm has an elongation 0.01 mm. When it is loaded with 500
gram. Young’s modulus of the material is ...... N.m–2 1) 1.87 1011 2) 18.7 1011 3) 3.74 1011 4) 1.87 10–11 2. Which of the following statement is wrong regarding Searle’s apparatus. 1) Reference wire will not be elongated with the increase of load in experiment. 2) Micrometer reading is to be noted only when the air bubble is at the centre of the spirit level. 3) Elongation will be propotional to mass attached to the wire 4) Average elongation of two wires will be taken as the elongation of wire. 3. Radius of the experiment wire is to be measured with .......... in Searle’s experiment to determine Y. 1) Ordinary scale 2) Vernier caliperse 3) Spherometer 4) Screw gauge 4. Which of the following wires A, B and C will have more elongation if they are identical but of difference
materials with same load if YA = 20 1011 N.m–2 YB = 2 1011 N.m–2 YC = 0.2 1011 N.m–2 1) A 2) B 3) C 4)Same elongation for all the sires. 5. Which of the following statements is wrong 1) The area under the stress-strain curve is a measure of the energy stored by the material during
stretching. 2) At the yield point, the material deforms under constant stress 3) At the breaking point the actual stress is greater than that at the ultimate strength point 4) Normal forces can change the shape of the body 6. Which of the following statements is true. 1) Plastic deformations are permanent 2) Stress is defined as external force applied upon area of the surface 3) The stiffness of a body is measured in terms of its modulus of rigidity. 4) Rubber is more elastic than steel. 7. A wire of a substance which elongates more due to stress will have 1) high Young’s modulus 2) low Young’s modulus 3) more initial length 4) both
(2) and (3) 8. A wire of a substance which elongates less will have more 1) Area of crossection 2) Less length 3) more length 4) both (1) and (2) 9. Inter atomic distance r0 force constant K Youngs modulus Y are related as ..........
Page No. 30
1) K = Yr0 2) Y = Kr0 3) KY = r0 4) Y + K = r0 10. To increase the elastic nature of a substance .......... is to be added. 1) Substance of high elasticity 2) Subtance of low elasticity 3) Plastic substance 4) Same substance 11. Approximation to perfectly elastic body 1) Quartz fibre 2) optical fibre 3) human bone 4) copper rod 12. If the stress in a wire is 1/200th of Young’s modulus, the strain produced is .......... 1) 1/100 2) 1/200 3) 1/10 4) 1/20 13. A force of 100 N increases the length of a given wire by 0.1 mm then the force required to increase the
length by 0.25 mm is .......... 1) 50 N 2) 150 N 3) 250 N 4) 500 N 14. The dimensions of the wires A and B are the same but materials are different if the graphs for load versus
elongation for those two wires as shown, Young’s modulus is less for
1) A 2) B 3) Same for both 4) None 15. Load-elongation graph for two wires of same material as shown in the graph. If their lengths are same
which of the two wires A and B is thin.
1) B 2) A 3) both are of same radius 4) None 16. Which of the following is wrong 1) glass is more elastic than rubber 2) steel is more elastic than rubber 3) copper is more elastic than gold 4) copper is more elastic than steel. 17. A : Hammering on a solid increases elasticity of the solid R : Due to hammering crystals will be broken and hence elasticity will be increased. 1) A and R true R is correct explanation of A. 2) A and R true R is not correct explanation of A 3) A is False R is true 4) A is true R is False 18. Which of the following is wrong. 1) A material with high young’s modulus will have less elongation. 2) A thin wire elongate more when compareed with thick wire of same material and length with same load 3) A long wire will have more elongation when compared with short wire of same material and radius for
same load. 4) Gold is more elastic than copper. 19. Which of the following is wrong regarding Searle’s apparatus method in finding Young’s modulus of a
given wire. 1) Average elongation of wire will be determined with a particular load while increasing the load and
decreasing the load. 2) Reference wire will be just taut and experimental wire will undergo for elongation. 3) Air bubble in the spirit level will be disturbed from the central position due to relative displacement
between the wires due to elongation
Load
Elongation
A
B
Load
Elongation
A
B
Page No. 31
4) Average elongation of the wires is to be determined by increasing the load attached to both the wires. 20. Fig. shows the stress, strain lines for brass, steel and rubber the lines A, B and C are for ..........respectively
1) steel, brass, rubber 2) rubber, steeel, brass 3) brass, rubber and steel 4) brass, steel, rubber 21. Which is true for a metal 1) Y > n 2) n Y 3) Y = n 4) Y < 1/n 22. In steel, the young’s modulus and the strain at the breaking point are 2 1011 N.m–2 and 0.15 respectively.
The stress at the breaking point for steel is .......... 1) 1.33 1011 N.m–2 2) 1.33 10–12 N.m–2 3) 7.5 10–3 N.m–2 4) 3 1010 N.m–2 23. On increasing the length by 0.5 mm in a steel wire of length 2 m and area of crossection 2mm2 the force
required is .......... N (Y for steel = 22 1011 N.m–2) 1) 1.1 104 2) 1.1 103 3) 1.1 102 4) 11 24. The tension on a wire is removed at once then 1) It will break 2) Its temperature will reduce 3) there will be no change in temperature 4) Its temperature increases 25. A wire of area of crossection 10–6 m2 is increased in length by 0.1%. The tension produced is 1000 N. The
Young’s modulus of wire is .......... 1) 1012 N.m–2 2) 1011 N.m–2 3) 1010 N.m–2 4) 109 N.m–2 26. The following data were obtained when a wire was stretched with in the elastic region. Force applied to
wire 100 N. Area of crossection of wire is 10–6 m2. Extenstion of wire is 2 10–3 m. Original length of wire 2 m. Which of the following deductions can be correctly made from this data? a) The value of Young’s modulus is 1011 N.m–2 b) The strain is 10–3 c) The energy stored in the wire when the load applied is 10 J
1) a, b, c are correct 2) a, b are correct 3) a only correct 4) b only correct 27. A steel wire A and a copper wire B have same length. The ratio of radii of A and B is 2 : 1. The ratio
YA/YB is 2. Both the wires are extended by the same force. The ratio of the increase of length of wire A and B is ..........
1) 1 : 4 2) 4 : 1 3) 1 : 8 4) 8 : 1 28. The energy is spent in stretching a wire of length L m and radius r m by m is ...... if Y is the Young’s
modulus of the material of the wire.
1) 2Y r L
2) 2Y r
2L 3)
2 2Y r L2
4) LrY l
222
29. A cable is replaced by another one of the same length and material but twice the diameter. The elongation would now become
1) double 2) one-half 3) one-fourth 4) one-sixth 30. A 100 N force stretches the length of a hanging wire by 0.5 mm. The force required to stretch a wire of the
same material and length but having four times the diameter by 0.5 mm is .......... 1) 100 N 2) 400 N 3) 1200 N 4) 1600 N 31. A 2 m long iron rod of crossectional area 50 mm2 stretches by 0.5 mm when a mass of 250 kg hung from
its lower end then Young’s modulus of elasticity of iron is .......... g = 10 m.s–2
A B
C Stress
Strain
Page No. 32
1) 2 107 N.m–2 2) 2 108 N.m–2 3) 2 1010 N.m–2 4) 2 1011 N.m–2 32. A 1 m long wire of crossectional area 10–6 m2 is used to suspend a load of 840 N. Which of the following is
incorrect? If Y = 12 1010 N.m–2 and P = 7000 Kg.m–3 S = 420 J.Kg–1K–1 1) The stress developed in the wire is 840 106 N.m–2 2) The strain is 70 10–4 3) Energy stored is 3.94 J 4) If the wire snaps temperature raise by 10C 33. The extension produced in a copper wire of length 2 m and diameter 3 mm when a force of 30 N is
applied. Young’s modulus for copper is 1.1 1011 N.m–2 and g = 10 N. kg–1 1) 0.02 mm 2) 0.04 mm 3) 0.08 mm 4) 0.68 mm
Key
1-10 1. 1 2.1 3. 4 4. 1 5. 4 6. 3 7. 4 8. 4 9. 2 10.1 11-20 11. 1 12. 2 13. 3 14. 1 15. 1 16. 2 17. 1 18. 4 19. 4 20.1 21-30 21. 1 22. 4 23. 2 24. 4 25. 1 26. 2 27. 3 28. 4 29. 2 30.4 31-33 31. 4 32. 3 33. 3
Page No. 33
SURFACE TENSION OF WATER BY CAPILLARY RISE AND EFFECT OF DETERGENTS (EXPERIMENTAL SKILLS)
1) The surface tension of water at room temperature by capillary rise method can be determined by
rh rdg3T dynes / cm
2
here angle of contact of water is 0° ‘h’ is height of the liquid in the capillary tube which is measured upto the lower miniscus of the liquid.
‘r’ is radius of the capillary tube ‘d’ is density of water ‘g’ is accelaration due to gravity
the term r3
is added to ‘h’ by considering the weight of the liquid in the miniscus
by neglecting r3
we can write hrdgT2
tap water is prefered over distilled water because due to the greasy nature of distilled water its surface tension is lowered
2) The shape of the miniscus is concave for water 3) The shape of the miniscus is convex for mercury 4) Pressure on the concave side of the liquid surface is more than the convex side of the liquid surface 5) In the experiment to measure ‘h’ value the microscope is to be shifted in vertical direction.
To measrue ‘r’ value the microscope can be moved both in horizontal and vertical directions. 6) The capillary tube should have a uniform bore throughout the whole length. 7) The capillary tube, beaker and surface of water should be clean and free from oil and grease 8) The capillary tube should be clamped vertically 9) The radius of the capillary tube is measured in the mutually perpendicular directions to minimise the
error if the bore is not circular 10) The tip of the bent wire should just touch the free surface of water in the beaker and not dipped in it. 11) Beaker should be raised or lowered before taking the reading in order to allow the water meniscus to
settle down at the correct position. 12) While taking observations, the microscope should be moved in one direction only to avoid back-lash
error 13) There should not be any air bubble in the capillary tube.If there is an air bubble the height of the
liquid column cannot be measured accurately. 14) The readings are noted without parallox error
Angle of contact: the angle made by the tangent at the point of contact of a liquid surface with tangent plane the solid surface inside the liquid is called the angle of contact.
15) The angle of contact depends upon the nature of the liquid, nature of the material of the container and temperature of liquid.
16) The angle of contact for pure water is 0o 17) The angle of contact for mercury is 140o 18) As temperature of liquid increases angle of contact also increases but surface tension decreases 19) Surface tension is a molecular phenomenon. Different liquids will have different value of S.T at
same temperature. 20. This method is not applicable for all types of liquids. It should be applicable only to those liquids for
which the angle of contact is zero and wet the walls of the tube. 21. Cohesive force is not same for all substances. Cohesive force is large for solids, because of this the
solid has a definite shape it is less in the case of liquids and gases. 22. The capillary tube should kept exactly vertical, other wise the liquid meniscus will become elliptical
and present formulae will not be applied.
SURFACE TENSION OF WATER AT DIFFERENT TEMPERATURES
Page No. 34
Temperature
oC surface tension
dynes/cm 0 75.7
10 74.2 15 73.5 20 72.75 25 72 30 71.2 40 69.6 50 67.9 60 66.6 70 64.4 80 62.9 100 58.8
20) The rise of water inside a capillary tube depends on i) radius of the tube ii) nature of the liquid iii) angle of contact 21) At different places as ‘g’ value is different the rise of water inside a capillary tube is also different 22) When a partially solvable impurity is added to the solution then its S.T decreases 23) When a completely solvable impurity is added to the solution then its S.T increases 24) When a capillary tube of insufficient length is used dipped in a beaker of water (i) radius of curvature(R) of the miniscus changes as the height(h) changes but
Rh = 2T cons tan tdg
25) The angle of contact changes as 1 2
2 1
cos hcos h
26) When detergents are added to the water, the detergent molecule weaker the strength of hydrogen bonding because the polar end of the detergent molecule is also attracted to the water therefore on adding detergents the S.T of liquid decreases
Problems 1. Least count of traveling microscope is 1) 0.01 cm 2) 0.01 mm 3) 0.001 cm 4) 0.001 mm 2. Surface Tension of water is 72 dynes/cm. How much force is required to break a water film of length
10 cm 1) 72 N 2) 720 N 3) 572 10 N 4) 5720 10 N 3. By what nature of liquid surface tension is arised 1) polar 2) nonpolar 3) viscous 4) non volatile 4. which of the following is a better cleaning agent 1) cold water 2) hot water 3) soap water 4) hot soap water 5. The vernier scale of travelling microscope can be moved 1) horizontally 2) vertically 3) both horizontally and vertically 4) circularly 6. A capillary tube is dipped in water and the top is closed then the rise of water 1) increases 2) decreases 3) remains same 4) water does not rise 7. If a capillary tube of insufficient length is used the liquid 1) rises and over flows slowly 2) does not overflow 3) rises and comes out like fountain 4) does not rise into the tube even 8. When a capillary tube of insufficient length is taken what will change along with the height of the
liquid in the capillary tube 1) radius of the capillary tube 2) radius of curvature of the meniscus 3) surface tension of liquid 4) density of liquid
liquid S.T at 20Oc dynes/cm
Alcohol 22.0 Benzene 29.0
Ether 16.5 Glycerine 64.0 Kerosene 30.0 Mercury 465.5
Turpentine 27.3 Olive oil 32.0
Paraffin oil 25.5 Paraffin oil 25.5
Soap solution 20.0 Water 72.75
Page No. 35
9. What happens to the length of liquid column in the capillary tube if it is inclined with respect to vertical
1) decreases 2) increases 3) remains same 4) overflow 10. Which force is responsible for the shape of a body 1) cohesive forces 2) adhesive force 3) gravitational force 4) electric force 11. The capillary rise method is suitable to 1) soap water 2) tap water 3) distilled water 4) cold water
12. In the calculation of S.T on what purpose a term r3
is added to the ‘h’
1) because ‘r’ is very small 2) by considering the weight of the liquid in the miniscus 3) it is an error in the formula 4) considering the weight of the capillary tube 13. ‘h’ is height of the liquid column in the capillary tube when it is clamped vertically. If the tube is
making an angle ( ) with vertical. Then length of the liquid column (l) is
1) l h cos 2) l = h cos2 3) l = 2h cos 4) l h
cos
14. What error does get when the microscope is in to and fro motion 1) parallax error 2) gross error 3) backlash error 4) percentage error 15. What is ascent formula
1) hrdgT2cos
2) rh dg3T
2cos
3) hdgr
2T cos
4) 2Tcosh
rdg
16. Two students A & B determining the surface tension in summer and winter as SA& SB respectively(other conditions are same in the experiment)
1) A BS S 2) A BS S 3) A BS S 4) BA
SS2
17. A student ‘A’ does the capillary rise experiment at equator and gets the S.T value SA.Another student does the same experiment at same conditions at poles and get the S.T value SB.(both have taken g = 980 cms-2 ) Then
1) A BS S 2) A BS S 3) A BS S 4) cannot be find 18. To measure the radius of the capillary tube it is clamped 1) vertically 2) horizontally 3) inclined to the surface of water 4) inclined to the microscope 19. To measure the height of the liquid column in the capillary tube it is clamped 1) vertically 2) horizontally 3) inclined to the surface of water 4) inclined to the microscope 20. The rise of liquid in a capillary tube depends on 1) radius of the tube 2) nature of the liquid 3) angle of contact 4) all of the above 21. The height upto which water will rise in a capillary tube will be 1) maximum when temperature is 4oC 2) minimum when temperature is 4oC 3) minimum when temperature is 0oC 4) same at all temperature 22. When a clean capillary tube is dipped vertically in a beaker containing water, the water rises about 6
cm. What will happen if another capillary tube of same radius and length 3 cm is dipped vertically in the same beaker containing water. Angle of contact of water is 0oC
1) water will flow out like a fountain 2) water will rise to a hight of 3 cm only and angle of contact is 0o 3) water will rise to a hight of 3 cm only and angle of contact is 60o
4) water will not rise at all 23. Olive oil drops are poured into a container which has the mixture of water and kerosene then oil drops
reaches 1) bottom of the container 2) top of the container 3) middle of the container 4) cannot say 24. What is the shape of the miniscus of water in the capiller tube
Page No. 36
1) convex 2) concave 3) spherical 4) elliptical 25. What is the shape of the miniscus of mercury when a capillary tube is dipped vertically 1) convex 2) concave 3) spherical 4) elliptical 26. On which side of a liquid surface pressure is more 1) convex 2) concave 3) plane 4) all are equal 27. What happens to the S.T of a liquid when a partially solvable impurity is added 1) increases 2) decreases 3) remains same 4) cannot be measured 28. What happens to the S.T of a liquid when completely solvable impurity is added 1) increases 2) decreases 3) remains same 4) cannot be measured 29. What happens to the S.T if temperature of liquid is increased 1) increases 2) decreases 3) remains same 4) becomes zero 30. What happens to the angle of contact if temperature is increases 1) decreases 2) increases 3) becomes zero 4) remains same 31. How the height of the liquied column changes if radius of the tube is decreased 1) increases 2) decreases 3) unchanged 4) over flows 32. What happens to the height of the liquid column if the capillary tube contains air bubble 1) not affented 2) decreases 3) increases 4) does not rise 33. Teflon is coated on the surface of non-sticking pans because 1) it makes the angle of contact abtuse 2) it makes the angle of contact accute 3) it makes the angle of contact zero 4) it gives smoothness to the surface 34. Oil is poured to calm sea wally because 1) to decrease the S.T 2) to increase the S.T 3) to get colours on the surface of water 4) to increase the breed of fishes 35. If the angle of contact is 900 then the liquid level in a capillary tube 1) increases 2) decreases 3) does not rise(or) fall 4) does not enter the tube even 36. What happens to the S.T if detergents are added to the water 1) increases 2) decreases 3) remains same 4) becomes zero 37. What happens to the angle of contact if detergents are added to the water 1) decreases 2) increases 3) becomes zero 4) remains same
38. How many number of scales in traveling microscope 1) none 2) one 3) two 4) three
39. The length of 1 division on main scale in traveling microscope 1) 0.05 m 2) 0.05 cm 3) 0.05 mm 4) 0.5 cm
40. Total number of divisions in vesnier scale in traveling microscope 1) 50 2) 100 3) 75 4) 10
41. This method is applicable for all types of liquids ? 1) Yes 2) No 3) Both 4) None
42. This method is applicable for the liquids 1) Angle of contact is 0° 2) Angle of contact is140°
3) Angle of contact is 90° 4) None 43. Which of the following having large cohesive force
1) Liquids 2) Gases 3) Solids 4) Liquids & Gases 44. In this experiment the capillary tube should not be kept exactly vertically then the shape of the
meniscus is 1) Concave 2) Convex 3) Elliptical 4) None
Page No. 42
Determination of Coefficient of Viscosity of a liquid 1. Define coefficient of viscosity Ans. The viscosity force acting tangentially on unit area of the liquid when there is a unit velocity gradient
in the direction perpendicular to the flow is called the coefficient of viscosity.
/ Tangentialstress/ Velocitygradient
F Adv dx
C.G.S Unit : Poise S.I. Unit : Pa – sec Dimensional formula : 1 1 1M L T 2. What is terminal velocity of a body? Ans. When a body is falling through a viscous medium it soon acquires a constant velocity due to the
viscous forces balancing the gravitational force. This velocity is called its terminal velocity. 3. Define velocity gradient. Ans. The change in velocity per unit length in a direction perpendicular to the direction of velocity is
called velocity gradient. Velocity gradient = dvdx
S.I Unit: S-1 4. What is viscosity? Ans. The property of a fluid which opposes the relative motion between its adjacent layers is called
viscosity. This is called as fluid friction. 5. Why rain drops are falling with less velocity? Ans. Due to viscosity of air, rain drops acquire a less but constant velocity known as terminal velocity 6. Write the formula for coefficient of viscosity of a liquid by measuring terminal velocity of a given
spherical body of radius ‘r’.
Ans. 229 t
grv
= density of solid body
229 /
grd t
= density of liquid
tdistance fallen dv =terminal velocity= =
time t
7. Which method is useful for measuring coefficient of viscosity of a less viscous liquid like water. Ans. Poiseuille’s method. (By measuring volume rate of flow of liquid through a capillary tube) 8. Which method is useful for measuring coefficient of viscosity of high viscous liquids. Ans. Stokes method. By measuring terminal velocity of a small spherical body falling through the given
viscous liquid. 9. How the viscosity varies with temperature. Ans. The viscosity of a liquid decreases as its temperature increases (because of decrease in cohesive
forces) The viscosity of gases increases with temperature due to increase in transfer of momenta during
collision between the molecules. 10. A solid sphere falls with a terminal velocity of 20m/s in air. If it is allowed to fall in vacuum. . a) terminal velocity will be 20m/s
b) terminal velocity will be less than 20m/s c) terminal velocity will be more than 20 m/s d) There will be no terminal velocity Ans. d Explanation: Because there is no viscous opposing force in vacuum.
Page No. 43
11. A spherical ball is dropped in a long column of a viscous liquid. The speed of the ball as a function
of time may be best represented by the graph. 1) A 2) B 3) C . 4) D
Ans. B
12. A solid sphere moves at a terminal velocity of 20m/s in air at a place where g = 9.8 m/s2. The sphere
is taken to a gravity free hall having air at the same pressure and pushed down at a speed of 20m/s. . 1) Its initial acceleration will be 9.8 m/s2 downwards 2) Its initial acceleration will be 9.8 m/s2 upwards 3) The magnitude of acceleration will decrease as the time passes 4) It will eventually stops. The answer is 2, 3, 4 13. What is the effect of pressure on viscosity? Ans. For liquids the viscosity increases with increase of pressure. For gases at normal temperature,
viscosity is independent of pressure or density. 14. Among water and honey, viscosity is more for . 1) Water 2) Honey 3) Same for both 4) None Ans. 2 15. For a body falling with terminal velocity, the net force on it is 1) buoyant free
2) weight of the body 3) difference of viscous drag and weight of the body 4) zero
Ans. 4 16. To reduce friction between the moving parts of a machine, the liquid used must have 1) High viscosity 2) Low viscosity 3) Low density 4) Low boiling point Ans. 1 17. The clouds float in air due to 1) Viscosity of air 2) Clouds are heavier
3) At higher altitudes density of air is more 4) None Ans. 1 18. For liquids, the graph between temperature and coefficient of viscosity is Ans. Rectangular hyperbola 19. The viscosity of water is 1) more in deep waters 2) more in shallow water 3) same throughout 4) none Ans. 1
Page No. 44
20. What is stokes formula for the terminal velocity
Ans. 2t
2v = a g9
= density of body = density of air or liquid If , body will rise instead of falling (Example air bubble in water) 21. Which liquids are less viscous ? Ans. The liquids like alcohol, water and spirit Etc. 22. Which liquids are more viscous ? Ans. The liquids like coal tar, castor oil, glycerin Etc. 23. Why should the capillary tube kept horizontal. Ans. To avoid the effect of gravity on the flow of water through the tube. 24. Is the velocity of the liquid flowing through the capillary tube same everywhere ? Ans. No. the velocity is maximum along the axis of the tube and gradually decreases towards the wall of
the tube. 25. Can poisculle’s method used for very high viscous liquids ? Ans. No, this method is suitable for only mobile liquids.
Page No. 37
Determination of coefficient of viscosity of a liquid By Measuring Terminal Velocity
Viscosity of a liquid can be determined by measuring the terminal velocity of
a solid sphere in it. The figure shows the apparatus A test tube ‘A’ contains the experimental liquid and is fitted into a water bath
‘B’ A thermometer ‘T’ measures the temperature of the bath A tube ‘C’ is fitted in the cork of the test tube ‘A’ There are three equidistant marks ‘P’, ‘Q’ and ‘R’ on the test tube well below
the tube ‘C’ A spherical metal ball is dropped in the tube ‘C’ The time interval taken by the ball to pass through the length PQ and through
the length QR are noted with the help of a stop watch. If these two intervals are not equal the same process will be repeated with
balls of small size until two time intervals are equal. If two time intervals are equal the velocity achieved by the metal ball before
passing through the mark ‘P’ is called terminal velocity which is constant. The radius of the ball is determined by a screwgauge Mass of the ball is calculated by weighing it. The length PQ=QR is measured by using a scale The coefficient of viscosity can be determined by using the formula
vgr 2)(
92
Where = density of metal ball =ballmetaltheofvolume
ballmetaltheofmass 3
34 r
m
= density of liquid in which the ball is moving g = acceleration due to gravity
Page No. 38
r= radius of metal ball
v = terminal velocity of metal ball = ervalTime
PQoflengthint
This method is useful for highly viscous liquids
M.C.Q.S
1. A rain drop falls near the surface of the earth with almost uniform velocity because
1) Its weight is negligible 2) The force of surface tension balances its weight 3) The force of viscosity of air balances its weight 4) The drops are charged and atmospheric electric field balances its weight 2. A piece of wood is taken deep inside a long column of water and released. It will move up 1) With a constant upward acceleration 2) With a decreasing upward acceleration 3) With a increasing upward acceleration 4) With a uniform velocity 3. A solid sphere falls with a terminal velocity of 20ms-1 in air. If it is allowed to fall in vacuum 1) Terminal velocity will be 20ms-1 2) Terminal velocity will be less than 20ms-1 3) Terminal velocity will be greater than 20ms-1 4) There will be no terminal velocity. 4. A spherical ball is dropped in a long column of a viscous liquid. The speed of the ball as a function of time may be the best represented by the graph 1) A 2) B 3) C D) D 5. A solid sphere moves at a terminal velocity of 20ms-1 in air at a place where g=9.8ms-2. The sphere is taken in a gravity free hall having air at the same pressure and pushed down at a speed of 20ms-1
1) Its initial acceleration will be 9.8ms-2 downward. 2) Its initial acceleration will be 9.8ms-2 upward 3) The magnitude of acceleration will decrease as the time passes. 4) It will eventually stop 1) ‘a’ is wrong 2) ‘b’ is wrong 3) ‘c’ is wrong 4) ‘d’ is wrong 6. Eight spherical rain drops of the same mass and radius are falling down with a terminal speed with a terminal speed v. If they coalesce to form one big drop its terminal speed will be (neglect buoyancy due to air)
1) 4v 2) v 3) 4v 4) 8v
7. A spherical steel ball released at the top of a long column of gleycerine of
length L, falls through a distance2L with accelerated motion and remaining
Page No. 39
distance 2L with a uniform velocity. If t1 and t2 denote the times taken to
cover the first and second half and w1 and w2 are the works done against gravity in the two halves then 1) 2121 , wwtt 2) 2121 , wwtt 3) 2121 , wwtt 4) 2121 , wwtt 8. Two spheres of equal masses but radii R and 2R are allowed to fall in a liquid. The ratio of their terminal velocities is 1) 1: 4 2) 1 : 2 3) 1 : 32 4) 2 : 1 9. A tiny sphere of mass ‘m’ and density ‘x’ is dropped in a tall jar of glycerine
of density ‘y’. When the sphere acquires terminal velocity, the magnitude of the viscous force acting on it is
1) y
mgx 2) x
mgy 3)
xymg 1 4)
yxmg 1
10. The viscous force on a spherical body moving through a fluid depends upon a) the mass of the body b) the radius of the body c) the velocity of the body d) the viscosity of the fluid 1) a, b, c, are true 2) a,c,d are true 3) b,c,d are true 4) ‘b’ and ‘d’ are true 11. Choose the correct statement. When a small body falls freely in a viscous fluid
a) Its speed increases indefinitely b) Its speed first increases due to gravity and then decreases due to viscosity
of the fluid, eventually the body comes to rest in the fluid. c) It accelerates first and then experiences deceleration until it acquires a
constant velocity called terminal velocity d) The speed of the body remains constant through out the motion 1) d 2) c 3) b 4) a
12. If the upthrust on a body is negligible compared to its weight, the terminal velocity of a small spherical body falling through a viscous liquid depends upon a) The mass of body 2) The diameter of body c) The viscosity of liquid 4) The acceleration due to gravity 1) a & b are true 2) b & C are true 3) b, c, d are true 4) All are true 13. A spherical ball of radius 10-4 m and of density 10 4 kgm-3 falls freely under gravity through a distance ‘h’ before entering a tank of water. If after entering the water, velocity of the ball does not change, the value of ‘h’ is nearly ( the coefficient of viscosity of water is 9.8x10-6 N-S/m2 ) 1) 25.2m 2) 10.4m 3) 15.6m 4) 29.3m 14. A small metal sphere of radius ‘r’ and density ‘ ’ falls from rest in a viscous liquid of density and coefficient of viscosity . Due to friction heat is produced. The expression for the rate of production of heat when the sphere has acquired the terminal velocity is
Page No. 40
1) 52)(278 rg
2) 52
2
)(278 rg
3) 52
)(278 rg
4) 5
2
2
)(278 rg
15. A steel ball of density ‘ 1 and radius ‘r’ falls vertically through a liquid of density 2 . Assume that viscous force acting on the ball is F=Krv, Where ‘k’ is a constant and ‘v’ is velocity. The terminal velocity of the ball is
1) K
gr3
4 212 2)
gkgr
34 21
3) kgr
g2
21
34 4) 213
4
kgr
16. A small sphere of mass ‘m’ is dropped from a great height. After it has fallen 100 metres, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 metres of fall is 1) Greater than the work done by air friction in the second 100 metres 2) Less than the work done by air friction in the second 100 metres 3) Equal to 100 mg 4) Greater than 100 mg 17. A ball of mass ‘m’ and radius ‘r’ is released in viscous liquid. The value of its terminal velocity is proportional to
1) onlyr1 2)
rm 3)
21
rm 4) ‘m’ only
18. Two equal drops are falling through air with a steady velocity of 5cms-1. If the drops coalesce, the new terminal velocity will be 1) 125 cms 2) 125 cms
3) 131
)4(5 cms 4) 1
25 cms
19. A solid ball of volume ‘V’ is dropped in a viscous liquid. It experiences a viscous force F. If the solid ball of volume 2V of the same material is dropped in the same fluid, then the viscous force acting on it will be
1) 2F 2) F 3) 2F 4) 4F
20. A small spherical ball falls through a viscous medium of negligible density with terminal velocity v. If another ball of twice the radius of the first one but having the same mass falls through the same medium, its terminal velocity will be
a) 2v b) 2
v c) 2v d)
4v
KEY
Page No. 41
1-10 3 2 4 2 1 3 4 4 3 3 11-20 2 4 1 2 1 2 2 3 3 3
HINTS 1) Conceptual 2) Conceptual 3) Conceptual 4) Conceptual 5) Conceptual
6) rnRrv 312 ;
7) For first half vLttvL
1120
2 For second half
vLttvL22 22
21 tt The work done against both halves = 2Lmg
8)
16 mgrvt = density of liquid
= density of sphere vt r= Constant
r
vt1
9) Viscous force ygrxgrF 33
34
34
yxgrF 3
34
xyxgr 1
34 3
xymg 1
10) Conceptual 11) Conceptual 12) Conceptual
13)
92 2 grv and use ghuv 222
14) tvfP 15) Weight – upthrust = viscous force. 16) Conceptual
17) trvmg 61
18) 2rvt 19) vF 20) 2rvt
NEWTON’S LAW OF COOLING
Page No: 45
PHYSICS
01. The rate of cooling of a body is directlyproportional to the excess of temperatureof the body over that of surroundingswhen the temperature difference is smalland the area and temperature of the bodyremain constant.
i.e. 0dtdQR
0KdtdQ
Where K = cooling constant whose valueis 3
0A4 . = temperature of body, , 0= temperature of surroundings
dtdQ
Rate of cooling.
The negative sign in dictates that the amountof heat is decreasing with time.
02. If the temperature of body decreasesfrom 1 to 2 in time ‘t’, then rate of fall oftemperature
0
2121
2k
tR and rate of
cooling
0
21
2KR Here
221
03 (a) The heat loss by a body at tempera-ture higher than that of surroundings isknown as cooling(b) Rate of cooling - The rate of loss of heat
is known as the rate of cooling dtdQR
(c) Its unit is calorie / second and its dimen-sions are 32TML
04. (a) Rate of fall of temperature R - Therate of decrease of temperature with respect
to time is known as R . i.e. dtdR
(b) Its unit is K/second and its dimensions are1T
(c) Rate of cooling
dtdJmSK
dtdQ
0
Rate of fall of temperature
0Kddt JmS
But 30eA4k = cooling constant.
JmSeA4
dtd 0
30
(d) Because for a spherical body. 4A r² and m = Volume x density
dr34 3
dr
34Js
r4e4dtd
3
20
30
rdJs
e12dtd 0
30
rds1
dtd
05. Cooling curve(a) The curve drawn between the temperature
of body and time is known as cooling curve
- - - - - -- - -
- - -
- --
- - -
- --
- - -
1
2
2t1t TimeTem
pera
ture
>
>
Synopsis
NEWTON’S LAW OF COOLING
NEWTON’S LAW OF COOLING
Page No: 46
PHYSICS
(b) Because 0Kdtdms
on integrating this equation, Kt0 Ae ,
which indicates that the temperature of bodydecreases exponentially with increasing time.
06. (a) The curve between rate of cooling andtemperature difference[ 0R ]This is a straight line passing throughorigin(b) Curve between the rate of cooling and
body temperature 0 0R K K K
(c) Because AlogKtlog e0e
This is the equation of a straight line, hence thecurve between 0log and ‘t’ will be astraight line. also(d) Curve between Rlog e and 0elog
(e) ktlog
02
01e
If the mass and spectic heat of the body are mand s respectively
ms
ktlog02
01e
[ kdtd
0
]
t
00
dtkd2
1
2
10loge kt
ktlog
01
02e
or
1 0
2 0
log e kt
07. If the rates of cooling of two bodies aresame then the rate of fall of temperatureof the body with highest heat capacity willbe the least
i.e. ms1
dtd
if R is constant
08. If two liquids are cooled under identicalcondition (i.e surface area, temperaturedifference, time difference same) then theirrates of cooling will be samei.e. if 2121 tt,AA and 21 RR or
II0I0 09. If two liquids are cooled under identical
circumstances then their rates of fall oftemperature will not be same. The rate offall of temperature R of that liquid will beminimum whose specific heat is maximum and
>
R
>
0
- - -
->
R >
t
>
>
>
0elog >
NEWTON’S LAW OF COOLING
Page No: 47
PHYSICSvice versa
i.e. ms1R
10. Limitations of law of cooling(a) The surface of the body must be black(b) The loss of heat must take place onlyvia radiation process(c) The excess of temperature 0
must not be large i.e. 35300
(d) The surface area (A), its nature and its
mean temperature
2
21 must
remain constant.(e) The air must be still
11. Newton’s law of cooling is a special caseof Stefan’s law, because(a) This law can be derived from Stefan’slawb) Cooling should be from black surfaces.
12. When a solid sphere of radius R. densityD and specific heat ‘s’ is heated to tem-perature ‘ ’ and then cooled in an enclo-
sure to temperature 0 Then its rate offall of temperature
RDS1
dtd
13. Specific heat of given liquid comparedwith respect to water
L
LL
tsmms
= w
ww
tsmms
m = mass of the calorimeters = specific heat of calorimeter
01. The ratio of the rate of cooling (R) andrate of fall of temperature (R ) is ...
1) ms1
2) ms
3) m/s 4) s/m02. The following curve is the cooling curve
1) 2)
3) 4)
03. Equal volumes of water and turpentine oilare taken in identical calorimeters. Theseare cooled under identical circumstancesthrough same temperature range. Thequantity which is same for them will be1) Rate of fall of temperature2) Specific heat3) Rate of fall of cooling4) internal energy
04. The curve between 0elog and ‘t’will be
1) 2)
3) 4)
TimeTem
pera
ture
Tem
pera
ture
Time
Tem
pera
ture
Time TimeTem
pera
ture
t t
t t
NEWTON’S LAW OF COOLING
Page No: 48
PHYSICS05. A body of steel is heated to 100°C and is
placed in a room to cool. The followingcurves represents the correct behaviour ofthe body
1) P 2) Q3) R 4) Both P and Q
06. Two bodies A and B are kept in an evacu-ated chamber at 27°C. The temperatureof A and B are 327°C and 427°C. Theratio of rates of loss of heat from A and Bwill be1) 0.52 2) 0.25 3) 1.52 4) 2.52
07. If the rates of emission of radiation by abody at temperature T K is E, then thegraph between log E and log T will be
1) 2)
3) 4)08. Newton’s law of cooling is applicable to
1) convectional losses2) natural convectional losses3) induced convectional losses4) all the above
09. In newtons law of cooling, if the rates ofemission of radiation by a calorimeterfrom 75°C to 70°C, 70°C to 65°C, andfrom 65°C to 60°C, are 1E , 2E and 3Ethen1) 321 EEE 2) 321 EEE
3) 321 EEE 4) 321 EEE
10. A hot liquid is kept in a large enclosuremaintained at temperature 0 . Its excessof temperature over that of surroundingsis plotted as a function of time. Thefollowing curves may represent the plot.
1) a 2) b 3) c 4) d11. In the above problem, the logarithem of
temperature difference between the bodyand the enclosure is plotted against time.The probable curve is1) a 2) b 3) c 4) d
12. The rate of cooling at 600K, if surround-ing temperature is 300 K is R. The rate ofcooling at 900K is
1) R3
162) 2 R 3) 3 R 4) R
32
13. Newton’s law of cooling is used in lab fordetermination of1) specific heat of solids2) latent heat of solids3) specific heat of liquids4) latent heat of liquids
14. A block of metal is heated to a tempera-ture much higher than the room tempera-ture and allowed to cool in a room freefrom air currents. The following curvescorrectly represents the rate of cooling
1) 2)
3) 4)
TimeTem
pera
ture
PQ
R
t
dc ba
Tlog
>
Tlog
>
Tlog
>
Tlog
>
Tem
pera
ture
Time
TimeTem
pera
ture
Tem
pera
ture
Time
Tem
pera
ture
Time
NEWTON’S LAW OF COOLING
Page No: 49
PHYSICS15. The newtons laws of cooling can be derived
from1) Stefan’s law 2) Kirchoff’s law3) Faraday’s law 4) Hook’s law
16. According to Newton’s law of cooling, the rateof cooling of a body is proportional to
n where is the difference of tempera-ture of the body and the surroundings, then ‘n’is equal to1) one 2) two 3) three 4) four
17. Cooling graphs are drawn for threeliquids a,b,c. The specific heat is maxi-mum for liquid
1) a2) b3) c4) both ‘a’ and ‘b’
18. The graph between temperature differ-ence and rate of cooling1) a
2) b3) c4) all
19. A copper cube of side ‘a’ is heated andthen allowed to cool in an evacuatedenclosure. It takes time ‘t’ to cool fromtemperature 21 to . Now anothercopper cube of side ‘2a’ is placed in thesame enclosure and allowed to cool. Thetime taken by it to cool from 1 to 2 ...1) t 2) 2t 3) 3t 4) 4t
20. Of the followingA) Greater the mass of radiating bodyslower will be coolingB) Greater the temperature of surround-ings slower will be cooling1) Both are correct 2) A is correct3) Both are wrong 4) B is correct
tc ba
Temp. diff
Rat
e of
coo
ling
a b
c
KEY
01) 2 02) 1 03) 3
04) 1 05) 1 06) 1
07) 3 08) 3 09) 2
10) 4 11) 3 12) 1
13) 3 14) 2 15) 1
16) 1 17) 1 18) 2
19) 2 20) 1
Page No. 50
VELOCITY OF SOUND
DESCRIPTION:The resonance apparatus consists of a long uniform glass tube G of about one metre long and about 2 or 3
cm diameter open at both ends. Its bottom end narrows down to a nozzle. It is connected by rubber tube T to areservoir R of water . It can be raised, lowered or fixed at any position by a screw. The whole systems is fitted to avertical wooden stand. The level of water or the length of air column in the tube can be read on a metre scale S.
THEORY:
At the first resonance let the length of air column be l1 if the end correction is ‘e’, l1 + e = 4 –– (1)
At second resonance let l2 be the length of the air column. Then l2 + e = 3 4 –––– (2)
Subtracting (1) from (2) l2 - l1 =3 4 - 4
= 2
The wavelength =2( l2 - l1)
The velocity of sound at room temperature , v= n= 2n ( l2 - l1) cm/sWhere ‘n’ is the frequency of the tuning fork used.
The velocity of sound at 00 C, v0= tvt 1 em / sec
546
PROCEDURE:The tube is completely filled with water by raising the reservoir R.A tuning fork N of known frequency n is
taken and excited by rubber hammer. The vibrating fork is kept over upper open end of the tube without touching it.The level of water in the tube is slowly lowered by lowering the reservoir until a loud booming sound is heard. In thisposition, the reservoir is clamped and the length of the air column from the surface of water to the open end of thetube is measured. The experiment is repeated 2 or 3 times and the average length l1 of the air column is noted. At thisposition the frequencies of the fork and the air column are equal. Then the air column is said to be in resonance withthe frequency of the tuning fork.
The reservoir is further lowered and the experiment is repeated withthe same tuning fork. Again when a loud booming sound is heard at about alength 3l1, the reservoir is clamped and the length of air column is measured.
The experiment is repeated 2 or 3 times and the average length l2 of theair column is noted. At this position also, the air column resonates with thetuning fork for a second time. The velocity of sound in air vt at room temperaturet0 C is determined using the formula, vt = 2n ( l2-l1) cm/s
The experiment is repeated with the other tuning forks of differentfrequencies and the average velocity vt is determined. The velocity of sound voin at o0 C is determined using the formula,
v0= vt t1 cm / s
546
where t is the room temperature in oCComparison of frequencies of two tuning forks:
V= 2n1( l2 - l1) =2n2 (l21-l1
1)or n1/n2 = ( l2
1 - l11) / (l2-l1)
n1 : n2 = ( l21 - l1
1) : (l2- l1)Where n1 and n2 are frequencies of the two tuning forks. l1 and l2 are the first and second resonating lengths.
For the first fork of frequency n1 and l1’ and l2’ are the first and second resonating lengths for the second tuning forkof frequency n2 respectively.
Page No. 51
PRECAUTIONS:
1. The fork should be held only by its shank.2. The excited fork is kept horizontally just above the open end of the tube.3. The resonating lengths should be determined both while increasing and decreasing the lengths of air column.4. The lengths of air column are noted without parallax error.
Frequency First resonating length Second resonating lengthS.No n Trial Trial Mean Trial Trial Mean vt = 2n(l2-l1)
hertz I II l1cm I II l2cm cm/sec
Average value of vt= cm/s
Velocity of sound at O0 C, v0 = vt t1
546
?
RESULT :Velocity of sound at to C = cm/sVelocity of sound at O0 = cm/s
Resonating Air column Experiment
EXERCISE1. The distance between end of pipe and actual position of antinode is known as
a) correction b) end correction c) difference d) diameter (b)2. In resonating air column apparatus, the air column in resonance represents the air column in
a) only open pipe b) only closed pipec) may be open pipe or closed pipe d) not both open pipe and closed pipe (b)
3. If ‘n’ is the frequency of the tuning fork used to excite the air column in resonating air column apparatus, l1 firstresonating length and l2 is second resonating length , then the velocity of sound in air is given by the formulaa) v = n ( l2-l1) b) v = n ( l1-l2) c) v = 2n ( l2-l1) d) v = 2n ( l1-l2) (c)
4. In resonating air column apparatus while comparing the frequencies of the two tuning forks, ifl1 is the 1 st resonating length of air column with 1 st fork of frequency n1
l2 is the 2 nd resonating length of air column with 1 st fork of frequency n1
l11 is the 1 st resonating length of air column with 2 nd fork of frequency n2
l21 is the 2 nd resonating length of air column with 2 nd fork of frequency n2
then 1
2
n ______n
a) 1 2 1
1 12 2 1
n l ln l l
b) 1
1 2 21
2 1 1
n l ln l l
c)
11 1 1
12 2 2
n l ln l l
d)
1 11 2 1
2 2 1
n l ln l l
(d)
5. The average velocity of sound at room temperature is noted as V. The velocity of sound at O0 C can be foundwith the formula
Page No. 52
a) 0tv v 1 546 b) 0
tv v 1 546
c) 0tv v 546 d) 0
tv v 1 546 (b)6. Then the end correction ( e) is
a) 2 1l 3l2
b) 1 2l 3l2
c) 2 1l 2l2
d) 1 2l 2l2
(a)
7. In the resonating air column experiment, l1 represents 1 st resonating length and l2 represents 2 nd resonatinglength, The relation between 1 st and 2 nd resonating lengths isa) l2 = l1 b) l2 = 2l1 c) l2 = 3l1 d) l2 = 4l1 (c)
8. In resonating air column apparatus the approximate relation between end correction and diameter of the tubeisa) 0.3 d b) 0.2 d c) 0.6 d d) 0.4 d (a)
9. In resonance column apparatus the reason for hearing booming sound isa) The frequency of air column in the tube and the frequency of are same frequencyb) The frequency of air column in the tube is greater than the frequency of the tuning forkc) The frequency of air column in the tube is less than the frequency of the tuning fork.d) Velocity of sound in air column is greater than the velocity of sound in atmospheric air (a)
10. If you use two tuning forks of frequencies 512 Hz and 480 Hz in resonating column apparatus, the one thatproduces much velocity in the air column isa) 512 Hz b) 480 Hz c) both equally d) we can’t compare (c)
11. The resonating lengths of an air column in the first and second modes of vibration are 31 cm and 97cm. If velocity of sound is 300 ms–1 the frequency of the tuning fork isa) 1000 Hz b) 500 Hz c) 250 Hz d) 125 Hz (c)Hint : V=2n( l2- l1 )
12. The end correction of a resonance tube is 1cm . If shortest resonating length is 15cm, the next resonatinglength will bea) 47cm b)45cm c) 50 cm d) 33 cm (a)
Hint : e = 2 1l 3l2
13. A tuning fork of frequency 340 Hz is excited and held above a cylindrical tube of length 120cm. It is slowlyfilled with water. The minimum height of water column required for resonance to be first heard (Velocity ofsound = 340 ms–1) isa) 25 cm b) 75 cm c) 45 cm d) 105 cm (c)
14. If the wave length of the note emitted by a tuning fork, frequency 512 Hz in air at 170C is 66.6cm, then thevelocity of sound at O0C is nearlya) 330 m/sec b) 350 m/sec c) 320 m/sec d) 370 m/sec (a)
Hint : vt = vo t1 546
15. Two unknown frequency tuning forks are used in resonance column apparatus . When only first tuning fork isexcited the 1 st and 2 nd resonating lengths noted are 10 cm and 30 cm respectively. When only secondtuning fork is excited the 1 st and 2 nd resonating lengths noted are 30 cm respectively. The ratio of thefrequencies of the 1 st to 2 nd tuning fork isa) 1 : 3 b) 1 : 2 c) 3 : 1 d) 2 : 1 (c)
Hint :- 1 1
1 2 1
2 2 1
n l l 90 30 60 3n l l 30 10 20 1
Page No. 53
Specific heat capacity of a given (i) Solid and (ii) Liquid by method of mixture. Synopsis :- - Heat is a form of energy which is transferred from hot body to cold body until the two bodies
attain the same temperature. - CGS unit of heat is Calorie. - SI unit of heat is Joule. - Relation between them is 1 calorie = 4.2 joule. - Heat is a scalar quantity. Dimensional formula 2 2ML T . - Heat unlike temperature, pressure, volume is not an intrinsic property of a body or system. It
has meaning only as long as it describes the transfer of energy into or out of the body or system. i.e., We cannot say that a body contains certain amount of heat but that body can transfer certain amount of heat under specified conditions.
- As heat is a form of energy it can be transformed into others and vice versa Ex. Thermo couple converts heat energy into electrical energy. Resistor converts electrical
energy to heat energy. Friction converts mechanical energy to heat energy. Heat engine converts heat energy to
mechanical energy. - Whole of mechanical energy i.e., work can be converted into heat but whole of heat can never
be converted into work. - Joules law :- When mechanical energy is converted into heat the ratio of work done (W) to
heat produced (Q) always remains constant W = JQ The constant J is called mechanical equivalent of heat and has value 4.2 Joule / Calorie. J is not a physical quantity but a conversion factor. - When heat is given to a body, usually temperature increases. - The quantity of heat absorbed by a body is proportional to (i) the mass (m) of the body (ii) the
raise in the temperature of the body Q m t Q ms t Where s is a constant called specific heat of the body depends upon the material of body. - When heat is given to a body, temperature of a body may not change. Ex. In melting of solid or boiling of a liquid i.e., in change of state. - Temperature may increase without heating. Ex. In shaking a liquid in a thermo flask or compressing a gas in a cylinder i.e., by converting
work into heat. - Heat required to produce a given change in temperature of a body depends on external
conditions such as pressure or volume. - Temperature :- The degree of hotness or coldness of a body is called temperature. - Temperature is the property which determines the direction of flow of heat between two bodies
or systems in contact with each other. - The state of two bodies or systems at same temperature in which there is no net flow of heat
between them is known as state of thermal equilibrium. - For the measurement of temperature some of property of matter like volume, emf, resistance
etc. is chosen which is directly proportional to temperature. - To make a thermometer the space between the melting point and boiling points of material
used in it is divided in equal parts. - Temperature of a body is measured in different scales (1) Centigrade scale (2) Fahrenheit scale (3) Reimer scale (4) Kelvin scale. - The equation used to convert the temperature from one scale to another scale is
C F 32 R K 2735 9 4 5
Page No. 54
- The equation used to convert the temperature difference from one scale to another scale is
C F R K5 9 4 5
So C K - Among the thermometers with cylindrical and spherical bulbs the farmer is more sensitive
because cylinder is having more surface area. - Specific heat :- It is the amount of heat required to raise the temperature of unit mass of
substance through a temperature difference of one unit.
1 QSm T
- C.G.S. unit of specific heat is 1 o 1ca l g C
- SI unit of specific heat is 1 1J K g K
- Dimensional formula is 2 2 1L T K . - Specific heat depends on nature of material of body. Dulong and petit has found that for
elemental solids (with few expectations such as carbon). - Atomic weight x specific heat = constant = 6.4 oCal / gm C . - So, heavier the element lesser will be the specific heat. (i.e.,) Hg Cu AlS S S - Specific heat of a substance also depends on temperature (particularly at low temperatures).
This temperature dependence of specific heat is usually neglected. - Specific heat is found to be maximum for hydrogen (3.5 oCal / gm C ) specific heat of water is
1 oCal / gm C . For all other substances specific heat is less than 1 oCal / gm C . and minimum
for radon and actinium ( 0.022 oCal / gm C ). - Specific heat also depends on state of substance i.e., solid, liquid or gas In the case of water Specific heat of in CGS units Ice (Solid) 0.5 Water (Liquid) 1 Steam (Gas) 0.47 - If a substance is undergoing change of state which takes place at constant temperature (called
isothermal change) specific heat
Q QSm t m 0
i.e., specific heat of a substance at its melting point or boiling point or isothermal change is infinite.
- If the temperature of a body changes without transfer of heat with the surroundings (adiabatic change) as in shaking liquid or compressing a gas
Q 0S 0m t m t
i.e., Specific heat of substance when it undergoes adiabatic change is zero. - Specific heat of a substance can also be negative. - Negative specific heat means that in order to raise the temperature, a certain quantity of heat is
to be with drawn from the body. Specific heat of saturated vapours is negative. - When specific heats are measured, the values obtained are also found to depend on the
conditions of the experiment. In general measurements made at constant pressure are different from those at constant volume.
P VC C
Page No. 55
- Specific heat of water is very large (1 oCal / gm C ). By absorbing or releasing large amount of heat its temperature changes by small amounts. This is why water is used as coolant in radiators.
- Heat capacity (or) thermal capacity :- It is the amount of heat required to raise the temperature of the body by 1 oC .
QC mst
SI unit is J/K CGS unit is oCal / C Dimensional formula 2 2 1ML T K . - Water equivalent :- Mass of water which has same thermal capacity as that of substance is
called water equivalent. - It is numerically equal to heat capacity. CGS unit is g. SI unit is Kg. - Latent heat :- It is the amount of heat lost or heat gained by a unit mass of substance to
change from one state to other state without changing temperature. - Latent heat of fusion of ice is the amount of heat required to convert 1 gram of ice at o0 C to 1
gram of water at o0 C . Its value is 6iceL 0.335 10 J / Kg or 80 Cal / g .
- Latent heat of vapourisation is the amount of heat required to convert 1 gram of water at o100 C to 1 gram of steam at o100 C .
- Its value is 62.26 10 J / kg or 540 Cal/g. - Principle of method of mixtures (or) principle of calorimetry :- When two bodies at different temperatures are mixed, heat will be transferred from hot body to
cold body until both the bodies acquire same temperature. Heat lost by hot body is equal to heat gained by cold body provided there is no exchange of heat between the body and surroundings.
- Resultant temperature of mixture is always less than temperature of hot body and is always greater than temperature of cold body.
Heat lost by hot body = heat gained by cold body 1 1 1 2 2 2m s m s 1 1 1m ,s , are mass, specific heat, temperature of hot body. 2 2 2m ,s , are mass, specific heat, temperature of cold body. And is resultant temperature of mixture. - When state of body changes, of takes place at constant temperature. Heat released or absorbed
is given by Q mL where L is latent heat. - No change of state due to same temperature. - If two bodies are at same temperature no transfer of heat between them can take place. - If water in a beaker is placed in a bath of boiling water, temperature of water in the beaker will
increase till it reaches o100 C , but never boil as the heat required for change of state will not be transferred from bath to beaker because both are at same temperature.
- Similarly if we put a water beaker in melting ice water in the beaker will cool to o0 C but will never freeze.
- No change in temperature due to in sufficient heat. If we put some ice in hot water, temperature of mixture will be greater than melting point of ice only if heat released by hot water is more than required to bring the ice to o0 C and to melt the whole ice. If heat released is insufficient to melt the whole ice, the temperature of system will be melting point of ice with some ice remaining unmelted.
Page No. 56
- To determine specific heat of solid by method of mixtures known mass of hot substance should be mixed with known mass of water in calorimeter of known temperature. After stirring resultant temperature is noted.
Heat lost by hot substance = heat gained by calorimeter and water 3 3 2 1 1 2 1 2 1 2 2 1m s t t m s t t m m s t t 1m = mass of calorimeter 2m = mass of water 3m = mass of hot solid 1s = specific heat of calorimeter 2s = specific heat of water s = specific heat of solid to be determined. 1t = initial temperature of calorimeter 2t = final temperature (resultant temperature of mixture) 3t = temperature of hot solid. - To determine the specific heat of liquid by the method of mixtures a solid of known specific
heat is used heat. Lost by solid is equal to heat gained by calorimeter and liquid. 3 3 3 1 1 1 2 1m s t t m s t t m s t t 1m = mass of calorimeter 2m = mass of liquid 3m = mass of hot solid 1s = specific heat of calorimeter s = specific heat of liquid to be determined. 3s = specific heat of calorimeter 1t = initial temperature of calorimeter and liquid t = resultant temperature 3t = temperature of hot solid. - Alternate method to determine specific heat of liquid. Take equal masses of water and liquid in identical containers. Heat them with identical lamps for equal time. Heat gained by water = heat gained by liquid w 2 1 L 3 1ms T T ms T T 1T = initial temperature of water as well as liquid. 2T , 3T = resultant (final) temperatures of water and liquid. Ls = specific heat of liquid to be determined. ws = specific heat of water
2 1L w
3 1
T Ts s
T T
Page No. 57
CONCEPTUAL QUESTIONS 1. Heat is 1. kinetic energy of molecules 2. potential and kinetic energy of molecules 3. Energy in transit 4. work done on the system 2. Heat required to raise the temperature of one gram of water through 10c is 1. 0.001 Kcal 2. 0.01 Kcal 3. 0.1 Kcal 4. 1.0 Kcal 3. Heat capacity of a substance is infinite. It means 1. heat is given out 2. heat is taken in 3. no change in temperature whether heat is taken in (or) given out 4. all of the above 4. The mechanical equivalent of heat J is 1. a constant 2. a physical quantity 3. a conversion factor 4. a dimensional quantity 5. The bullet fired from a gun gets heated on striking a target because 1. it loses energy 2. mechanical work is converted in to heat 3. of latent heat 4. specific heat 6. Which of the following states of matter have two specific heats ? 1. solid 2. gas 3. liquid 4. vapour 7. The specific heat of a gas in an isothermal process is 1. infinity 2. zero 3.negative 4. remains constant 8. In defining the specific heat, temperature is represented in 0F instead of 0C. Then the value of
specific heat will 1. decreases 2. increases 3. remain constant 4. be converted to heat capacity 9. A gas is being compressed adiabatically. The specific heat of the gas during compression is 1. zero 2. infinite 3. finite but not zero 4. undefined 10. Which of the following will extinguish the fire quickly 1. water at 1000C 2. steam at 1000C 3. water at 00C 4. ice at 00C 11. The latent heat of vapourisation of a substance is always 1. greater than its latent of fusion 2. greater than its latent heat of sublimation 3. equal to its latent latent heat of sublimation 4. less than its latent heat of fusion 12. A piece of ice at 00C is dropped into water at 00C. Then ice will 1. melt 2. be converted to water 3. not melt 4. partially melt 13. When two blocks of ice are pressed against each other then they stick together (coalesce)
because 1. cooling is produced 2. heat is produced 3. increase in pressure increase melting point 4. increase in pressure, decrease in melting point 14. In a pressure cooker cooking is done quickly because 1) the cooker does not absorb any heat 2) it has a safety valve 3) boiling point of water rises due to increased pressure 4) it is a prestige to cook in a cooker 15. A large block of ice is placed on a table when the surroundings are at 00C 1) ice melts at the sides 2) ice melts at the top 3) ice melts at the bottom 4) ice does not melt at all
Page No. 58
16. In solids, liquids and gases which of the following has largest specific heat ? 1) water 2) hydrogen 3) mercury 4) copper 17. Which of the substances has largest specific heat? 1) water 2) mercury 3) turpentine 4) kerosene 18. Amount of heat required to raise the temperature of a body through o1 K is called 1) latent heat 2) specific heat 3) thermal capacity 4) none 19. Water is used to cool the radiators of engines in motor cars because of its 1) lower density 2) easy availability 3) high specific heat 4) low specific heat 20. Equal masses of sand and water are kept in identical calorimeters. Both are heated for the same
time. Which will be at a higher temperature? 1) sand 2) water 3) both at same temperature 4) none 21. If a gm of substance at o
1t C rise of temperature requires as much heat as 'b' gm of water at o
2t C rise of temperature. The specific heat of the substance is
1) 1
2
btat
2) 2
1
btat
3) 2
1
atbt
4) 1
2
atbt
22. 1 gm of ice at o5 C and 1 gm of water are heated separately. In order to raise the temperature of both by o1 C , heat required by ice
1) less than that of water 2) more than that of water 3) same as that of water 4) none 23. The specific heat of water at the time of boiling is 1) 0 2) +ve 3) -ve 4) infinity 24. Water in a calorimeter at o0 C can be boiled in 5 min. The water is converted into steam in 1) 9 min 2) 18 min 3) 23 min 4) 27 min 25. Steam produces more severe burns on the human body than boiling water at o100 C because
steam gives out ……….. heat to the body than water 1) more 2) loss 3) equal 4) none 26. Coffee in a flask is vigorously shaken. Then the temperature of the coffee 1) increases 2) decreases 3) remains the same 4) none 27. The unit of specific heat is 1) Cal/gm/K 2) gm/cal K 3) Cal gm.K 4) cal/gm 28. Which of the following relations is correct? 1) Heat capacity = specific heat x mass 2) Specific heat = mass x heat capacity
3) Heat capacity = massspecific heat
4) specific heat = heat capacity 29. Calorie is the unit of 1) Heat 2) Specific heat 3) Mechanical equivalent of heat 4) Water equivalent 30. A piece of ice at o0 C is dropped into the water at o0 C . Then ice will 1) melt 2) be converted to water 3) not melt 4) partially melt 31. The units of the ratio of specific heat to latent heat is 1) o C 2) o K 3) o 1C 4) gram 32. The natural direction of heat flow between two reservoirs depends on their 1) Temperature 2) Volume 3) Pressure 4) Nature 33. Conversion of heat into electrical energy can be achieved by 1) Transistor 2) Voltameter 3) Thermocouple 4) Photoelectric cell 34. Which device can convert all electrical energy into heat 1) Radiator 2) Converter 3) Generator 4) Resistor
Page No. 59
35. A liquid is being converted into vapours at its boiling point ; the specific heat of the liquid at this temperature will be
1) Zero 2) Infinite 3) Positive 4) Negative 36. Specific heat of a gas undergoing adiabatic change is 1) Zero 2) Infinite 3) Positive 4) Negative 37. Steam burns are more painful than those caused by boiling water at the same temperature
because 1) Steam contains more heat than the same amount of water at the same temperature. When
steam condenses it gives out this extra latent heat. 2) Hot air is also mixed up with steam 3) Steam produces some chemical effect on the skin to cause pain 4) The question is irrelevant as steam burns are equally painful as those caused by boiling
water 38. The sprinkling of water reduces the temperature of a closed room because 1) The temperature of water is less than that of the room 2) The specific heat of water is high 3) Water has large latent heat of vaporization 4) Water is a bad conductor of heat
KEY 1) 3 2) 1 3) 3 4) 3 5) 2 6) 2 7) 1 8) 1 9) 1 10) 1
11) 1 12) 3 13) 4 14) 3 15) 3 16) 2 17) 1 18) 3 19) 3 20) 1
21) 2 22) 1 23) 4 24) 4 25) 1 26) 1 27) 1 28) 1 29) 1 30) 3
31) 3 32) 1 33) 3 34) 4 35) 2 36) 1 37) 1 38) 3
Page No. 60
RESISTIVITY OF THE MATERIAL OF A GIVEN WIRE USING METER BRIDGE
1. Dependence of resistance Resistance of a body depends on its shape, size, material and temperature, does not depend
on current and potential difference. Though resistance of a linear conductor is independent of applied voltage, for a given body it
is not unique and depends on length (l) and area of cross section (A).
bh
lR max and
blhR
min
L, b, h are length , breadth and h l b h Resistance of a conductor is directly proportional to its length and inversely proportional to
its area of cross section.
2l l lR R R r radiusA A r
is specific resistance or resistivity . Resistance varies with temperature, t oR R l t : oR and tR are resistances at temperatures 00 and t0C = temperature coefficient of resistance. Specific resistance depends on nature of the material and temperature. t 101 Specific resistance is independent of dimensions. Addition of impurity increases the resistivity Reciprocal of resistance is known as conductivity or specific conductance.
1
JE
, E= intensity of electric field and J= current density.
If wires of resistivities 21 and and lengths 1 2l and l with same area of cross section are
connected in series, equivalent resistivity 21
2211
llll
If two wires of resistivities 21 and and areas of cross sections 1 2 and AA with some
length are connected in parallel, equivalent resistivity
1221
2121
AAAA
m
dlVl
lAlR
AlR
222 when v, d, m are volume, density and mass of the
conductor. for the wires of different materials
1
2
2
1
2
2
1
2
1
2
1
mm
dd
ll
RR
2
1
2
2
1
2
1
1
2
2
1
2
1
2
1
rr
ll
AA
ll
RR
For wires made of same material, different masses and different lengths
Page No. 61
1
22
2
1
2
1
mm
ll
RR
Wires of same material, but of different dimensions 1 1 2
2 2 1
R l AR l A
Meter Bridge Works on the principle of Wheatstone bridge.
When the bridge is balanced 100
X lR l
x- unknown resistance R- known resistance l- balancing length in cm.
1 1 2 2
1 2
,100 100
X l X lR l R l
( 1 2,l l balancing lengths when X1 and X2 are connected in the
meter bridge gap one after the other separately). When they are connected to the gap at a time in series and if ls is the balancing length, then
Then 1 2
100 S
X X LsR L
2
2
1
1
100100100 ll
ll
ll
s
s
When connected in parallel.
1 2
1 2
100 100 100100
p pP
p p
l lX l lR l l l l
Meter bridge –formula to calculate Specific resistance
100 lX Rl
Unknown resistance x is placed in left gap and known resistance ‘R’ in right gap and balancing length ‘l’ is found
ALX
LX
LDXDrA
44
222
.
Knowing (Length of the wire, D diameter of the wire, of the wire can be found). 1. In a wheatstone bridge, three resistances P,Q and R are connected in the three arms and the
fourth arm is formed by two resistances 1 2S and S connected in parallel. The condition for the bridge to be balanced will be
a) 1 2
1 22R S SP
Q S S
b) 21 SS
RQP
c)
1 2
2P RQ S S
d) 1 2
1 2
( )P R S SQ S S
2. A material ‘B’ has twice the specific Resistance of ‘A’. A circular wire made of ‘B’ has twice the diameter of a wire made of ‘A’. Then for the two wires to have the same resistance, the ratio /B Al l of their respective lengths must be
a) ¼ b) 2 c) 1 d) ½ 3. AB is a wire of uniform Resistance. The galvonometer ‘G’ shows no current when the length
AC=20cm and CB=80cm. The Resistance ‘R’ is equal to
Page No. 62
a) 2 b) 8 c) 20 d) 40 4. There are two concentric spheres of radii a and b (a>b) and the space between them is filled
with a medium of resistivity , then the resistance between the inner and outer coating of the medium will be.
a) 4
b aab
b) 4
a bab
c) 2 1 1
4 b a
d) 2 2
1 14 b a
5. The masses of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1, the ratio of their electrical resistances is
a) 1:3:5 b) 5:3:1 c) 1:25:125 d) 125:25:1 6. A square aluminium rod is 1m long and 5mm on edge. What must be the radius of another
aluminium rod whose length is 1m and which has the same resistance as the previous rod. a) 5.0mm b) 4.2mm c) 2.8mm d) 1.4mm 7. The resistance of 20cm long wire is 5 ohms. The wire is stretched to a uniform wire of 40cm
length. The resistance now will be (in ohms) a) 5 b) 10 c) 20 d) 200 8. A wire of length 5m and radius 1mm has a resistance of 1ohm. What length of the wire of
the same material at the same temperature and of radius 2mm, will also have a resistance of 1 ohm?
a) 1.25m b) 2.5m c) 10m 20m 9. A metal wire of specific resistance 664 10 ohm-cm and length 198cm has a resistance of
7ohms the radius of the wire will be a) 2.4cm b) 0.24cm c) 0.024cm d) 24cm 10. A wire 50cm long 21mm in cross – section carries a current of 4A when connected to a 2v
battery. The resistivity of the wire is a) 72 10 m b) 75 10 m c) 64 10 m d) 61 10 m 11. Resistivity of iron is 71 10 ohm x met. The resistance of the given wire of a particular
thickness and length is 1 ohm. If the diameter and length of wire both are doubled, the resistivity will be in ohm x met.
a) 71 10 b) 72 10 c) 74 10 d) 78 10 12. The specific resistance and cross section area of potentiometer wire is and A
respectively. If a current I is passed through the wire, the potential gradient of the wire will be,
a) IA b) I
A c) IA
d) IA
13. What length of the wire (specific resistance 848 10 m ) is needed to make a resistance of 4.2 ? (Diameter of the wire = 0.4mm)
a) 1.1m b) 2.1m c) 3.1m d) 4.1m 14. A copper wire (specific resistance 81.7 10 m ) has a mass per unit length of 710 kg/cm.
What is the resistance of a wire 200m in length? (Density of copper = 3 -38.9 10 kg m ) a) 3k b) 0.3m c) 30m d) .3 15. Two wires of equal length and material x and y have same resistance. The ratio of the radius
of the two wires is 1:2, the ratio of the specific resistances of the two materials is a) 1:1 b) 1:2 c) 1:4 d) 4:1 16. The dimensions of a block are 1cm x 1cm x 100cm. If the specific resistance of the material
is 72 10 ohm x meter; then the resistance between the opposite rectangular faces square faces are
a) 9 92 10 , 2 10 b) 7 32 10 , 2 10 c) 5 52 10 , 2 10 d) 3 72 10 , 2 10 17. A nichrome wire 1m long and 1 2mm has cross – sectional area draws 4 amp at 2 Volts. The
resistivity of nichrome is, a) 71 10 ohm m b) 75 10 ohm m c) 74 10 ohm m d) 72 10 ohm m
Page No. 63
18. In a meter bridge, resistance in the right gap is 10 . The null point is found to occur at a distance 30cm from the left end. The uncertainity in the value of resistance in the left gap, if uncertainity in the distance of the null point is 0.5cm .
a) 1 b) 10 c) 0.01 d) 0.10 19. A coil of copper wire is immersed in a bath of water at 00 c and is connected in the left gap
of a meter bridge. After making the necessary connections; the null point is obtained at 50cm after adjusting the resistance in the resistance gap which is found to be equal to 10 ohms. When the bath is heated to boiling point, the null point occurs at 56.8cm. The average temperature coefficient of copper between 0 00 and 100c c
a) 0 10.0003 c b) 0 10.003 c c) 0 10.0039 c d) 0 10.0027 c 20. The resistance in the left and right gaps of a balanced metre bridge are 1 2 and RR . The
balanced point is 50cm. If a resistance of 24 ohms is connected in parallel to R2, the balance point is 70cm. The value of 1 2 or RR is
a) 12 b) 8 c) 17.1 d) 32 KEY 1) d 2) b 3) c 4) b 5) d 6) c 7) c 8) d 9) c 10) d 11) a 12) a 13) a 14) a 15) c 16) b 17) b 18) d 19) a 20) d
Page No. 64
Resistance of a given wire using Ohm’s law Synopsis : 1. Verification of Ohm’s law Determination of Resistance of a given wire using Ohm’s law by Voltmeter – Ammeter method. According to Ohm’s law Pd across the ends of a conductor is proportional to the current (I)
flowing through it provided the physical conditions (like temperature, pressure, strain etc) of the conductor remain unchanged.
V I V = RI R = V/I The value of R depends up on the nature of the material, its dimensions and its temperature. If
does not depends on the value of V & I. A battery of e.m.f ‘E’ is connected to a conductor XY through a rheostat, ammeter and key.
Voltmeter across the wire XY measures Pd across the wire. When key is closed, current flows through the conductor. Readings of voltmeter and Ammeter are
noted for different positions of Rheostat. It is observed that V/I = constant. The variation of V and I through a conductor is as shown in figure. It is represented by a straight
line having a constant slope V/I = R
Precautions : 1. Take the readings from voltmeter & ammeter carefully without any parallax error. 2. Connections should be done carefully. 3. Experiment should be performed as quickly as possible. Wire should not get heated up, as
it varies resistance
2. R = VI
= .
Wq I
= ML2 T– 3A– 2 Ohm
= RAl
= ML3 T– 3A– 2 Ohm – m
R = 2 .m lne t A
= lA
P = 2
mne t
3. The ratio of change of V to the change of I for a given voltage is the Dynamic resistance of the wire ( V/I)
4. For Ohmic materials the V – I graph is straight line 5. For non Ohmic resistance V – I graph is non linear
(i) (ii) (iii) (iv)
Conductor Vacuum tube Electrolyte Thermistor (i) Resistance of metal conductor is constant (ii) For vacuum tube ‘R’ decreases (iii) For liquid electrolyte R increases (iv) For Thermistor there are two different values of current for the same voltage
Page No. 65
Effect of temperature = 0
0.tR RR t
(i) For conductors temperature coefficient of resistance in positive (ii) For semi conductor is negative. (iii) is very low for manganin and constantan. So these materials are used in preparation of
standard resistances. Manganin 84% Cu, 12% Mn, 4% Ni, Constantan 60% Cu, 40% Ni The resistance of given wire can be determined by using tangent galvanometer also. In this method the principle V = K.(R+B+G). Tan B is the Resistance of Battery G is the resistance of galvanometer
By taking known values of ‘R’ corresponding cot values are noted and plot a graph between R – cot and extrapolate the graph. The unknown resistance can be obtained from the graph directly
Rheostat is used to change the strength of current in the circuit.
Commutator is used to change the direction of current in the coil of tangent galvanometer.
Q1. V – I graph for a metallic wire at three different temperatures. T1, T2, T3 as shown in figure.
Which of the three temperatures is higher 1) T1 2) T2
3) T3 4) All
Q2. If R1 and R2 are the resistances of a conductor at different temperatures t1 and t2 then its temperature coefficient of resistance is
1) 2 1
1 1 2 2
R RR t R t
2) 1 2
2 1 1 2
R RR t R t
3) 2 1
1 2 2 1
R RR t R t
4) 1 2
1 2 2 1
R RR t R t
Q3. At T degree centigrade ‘R’ is the resistance of a wire. Its resistance becomes double at a
temperature when is the temperature coefficient of resistance in /0C
1) 1
2 T
2) 2 1T
3) 2 T
T
4) 2
TT
Q4. A: The temperature below which the resistance of conductor becomes zero is called critical temperature
B: The resistance of mercury near a temperature of 4K becomes almost zero. 1) A true, B false 2) A false, B true 3) A and B true 4) A and B false
RB
- + ( )
TG
.
C
Page No. 66
Q5. In the figure the curve ‘A’ and ‘B’ may represents 1) A conductor, B semi conductors 2) A & B conductors 3) A & B semi conductors
4) A semi conductor, B conductor
Q6. Which of the following may represent temperature resistivity graph for a super conductor
1) 2) 3) 4) Q7. A tungsten coil has a resistance of 12 at 150C. If the temperature coefficient of resistance of
tungsten is 0.004 C – 1 . Then the resistance of the coil at 800 C is 1) 15 2) 25 3) 5 4) 10 Q8. Resistivity of a wire varies with 1) length 2) area of cross section 3) temperature 4) all Q9. The voltage – current graph for ohmic conductors is 1) straight line 2) parabola 3) circle 4) rectangular hyperbola Q10. The slope of voltage – current graph gives 1) Energy 2) Resistance 3) Power 4) Capacity
Q11. In addition to Tangent Galvanometer which of the following can be used to verify ohm’s law. 1) Voltmeter 2) Ammeter 3) Resistance Box 4) Commutator
Q12. Tangent Galvanometer must be connected in 1) Series 2) Parallel
3) Both series & parallel 4) Series or parallel Q13. If the number of turns in the coil of Tangent Galvanometer is increased then the deflection in it 1) decreases 2) remains same
3) increases 4) either increases or decreases Q14) If the area of the coil in Tangent Galvanometer increases then the deflection in it
1) remains same 2) increases 3) decreases 4) either increases or decreases
KEY
1) 3 2) 3 3) 2 4) 3 5) 4 6) 2 7) 1 8) 3 9) 1 10) 2 11) 2 12) 1 13) 3 14) 3
Page No. 67
COMPARING E.M.F’S OF THE TWO CELLSUSING POTENTIOMETER
In the circuit shown, the cell of emf E1 isbalanced for a length 1l on the potentiom-eter wire by connecting ‘x’ to ‘z’. In the nextevent, the cell of emf E2 is balanced for alength 2l on the potentiometer wire by con-necting ‘y’ to ‘z’. Let ‘ x ’ be the potential gra-dient on the potentiometer wire.
(i) 1 1E l x and 2 2E l x
(ii)1 1
2 2
E lE l
(iii) The e.m.f. of a cell can be determined onlywhen e.m.f. of another cell is known.
(iv) When two cells support each other, then bal-ancing length is 1l and when they opoose
each other then balancing length is 2l , then
1 2 1E E l ----(1)
1 2 2E E l ---(2)
1 2 1
1 2 2
E E lE E l
or 1 1 2
2 1 2
E l lE l l
(v) The potentiometer need not be standardisedfor comparison of e.m.f.’s of two cells but ‘ x ’must remain constant in the circuit
1. Figure shows a potentiometer with a cell of 2.0V and internal resistance 0.40maintaining apotential drop across the resistor wire AB. Astandard cell which maintains a constant e.m.f.of 1.02 V(for very moderate currents upto afew amperes) gives a balance point at 67.3cmlength of the wire. To ensure very low currentsdrawn from the standard cell, a very high resis-tance of 600k is put in series with it, which isshorted close to the balance point. The stan-dard cell is then replaced by a cell of un-known e.m.f. E and the balance point, simi-larly turns out to be at 82.3cm, length of thewire. Then the value of ‘E’ is
a) 2.147V b). 0.427Vc) 1.724V d) 1.247V
2. A battery 1E of 4 V and a variable resistance
Rh are connected in series with the wire ABof the potentiometer. The length of the wireof the potentiometer is 1 m. When a cell 2Eof e.m.f. 1.5 volt is connected between pointsA and C, no current flows through G, for60 cm length of AC.
The potential difference between theends A and B of the potentiometer wire is.a) 2.5 V b) 1.5 Vc) 3.5 V d) 0.5 V
3. In a potentiometer arrangement, a cell of e.m.f.
(.)
-+-+ 1E
BE1K hR-+
U
ZG
C
BA
Y2E
X
-+2E
-+1E
2E1E
-+ - +
A B
K1.02V
600k
G
2.0V,0.4
J
1.5V
< >100 cm
>
< >60cm
G
RhE1
B
CA
Page No. 67
COMPARING E.M.F’S OF THE TWO CELLSUSING POTENTIOMETER
In the circuit shown, the cell of emf E1 isbalanced for a length 1l on the potentiom-eter wire by connecting ‘x’ to ‘z’. In the nextevent, the cell of emf E2 is balanced for alength 2l on the potentiometer wire by con-necting ‘y’ to ‘z’. Let ‘ x ’ be the potential gra-dient on the potentiometer wire.
(i) 1 1E l x and 2 2E l x
(ii)1 1
2 2
E lE l
(iii) The e.m.f. of a cell can be determined onlywhen e.m.f. of another cell is known.
(iv) When two cells support each other, then bal-ancing length is 1l and when they opoose
each other then balancing length is 2l , then
1 2 1E E l ----(1)
1 2 2E E l ---(2)
1 2 1
1 2 2
E E lE E l
or 1 1 2
2 1 2
E l lE l l
(v) The potentiometer need not be standardisedfor comparison of e.m.f.’s of two cells but ‘ x ’must remain constant in the circuit
1. Figure shows a potentiometer with a cell of 2.0V and internal resistance 0.40maintaining apotential drop across the resistor wire AB. Astandard cell which maintains a constant e.m.f.of 1.02 V(for very moderate currents upto afew amperes) gives a balance point at 67.3cmlength of the wire. To ensure very low currentsdrawn from the standard cell, a very high resis-tance of 600k is put in series with it, which isshorted close to the balance point. The stan-dard cell is then replaced by a cell of un-known e.m.f. E and the balance point, simi-larly turns out to be at 82.3cm, length of thewire. Then the value of ‘E’ is
a) 2.147V b). 0.427Vc) 1.724V d) 1.247V
2. A battery 1E of 4 V and a variable resistance
Rh are connected in series with the wire ABof the potentiometer. The length of the wireof the potentiometer is 1 m. When a cell 2Eof e.m.f. 1.5 volt is connected between pointsA and C, no current flows through G, for60 cm length of AC.
The potential difference between theends A and B of the potentiometer wire is.a) 2.5 V b) 1.5 Vc) 3.5 V d) 0.5 V
3. In a potentiometer arrangement, a cell of e.m.f.
(.)
-+-+ 1E
BE1K hR-+
U
ZG
C
BA
Y2E
X
-+2E
-+1E
2E1E
-+ - +
A B
K1.02V
600k
G
2.0V,0.4
J
1.5V
< >100 cm
>
< >60cm
G
RhE1
B
CA
Page No. 68
1.25 V gives a balance point at 35 cm length ofthe wire. If the cell is replaced by another cellthen the balance point shifts to 63 cm, then e.m.fof the second cell isa) 4.25 V b) 2.25 Vc) 3.25 V d) 1.25 V
4. In a potentiometer, a standard cell of emf5V and of negligible resistance maintains asteady current through the potentiometerwire of length 5m. Two primary cells ofemf’s E1 and E2 are joined in series with (i)same polarity and (ii) opposite polarity. Thecombination is connected through a galva-nometer and a jockey to the potentiometer.The balancing lengths in the two cases arefound to be 350 cm and 50 cm respectively.The values of E1 and E2 are respectivelya) 3 V and 2 V b) 2.5 V and 1.5 Vc) 2 V and 1.5 V d) 1.5 V and 1 V
5. Two cells of e.m.f. E1 and E2 (E1>E2) areconnected as shown in figure. When a po-tentiometer is connected between A and B,the balancing length of the potentiometerwire is 300 cm. When the same potentiom-eter is connected between A and C, the bal-ancing length is 100 cm. The ratio of E1 andE2 is
a) 3 : 2 b) 2 : 3c) 2 : 1 d) 1 : 2
6. There exists a constant potential differencebetween ends of a potentiometer wire. Twocells are connected, in turn, in such a waythat they help each other and are balancedon the potentiometer wire at 60 cm. Whenthe two cells oppose each other, then theyare balanced for a length of 30 cm. The ratioof e.m.f’s of the two cells isa) 4 : 1 b) 1 : 4c) 1 : 3 d) 3 : 1
7. Potentiometer wire, PQ of 1m length is con-nected to a standard cell E1. Another cell, E2of emf 1.02 V is connected as shown in thecircuit diagram and null position is obtainedat a distance of 51 cm from P. the value ofE1 is
a) 1 V b) 2 V c) 3 V d) 4 V8. In a potentiometer experiment, two cells of emf
E1 and E2 are used in series and in conjunctionthe balancing length is found to be 81cm of thewire. If the polarity E2 is reversed, then the bal-ancing length becomes 27 cm. The ratio E1/E2isa) 1 : 1 b) 2 : 1 c) 3 : 1 d) 4 : 1
9. A potentiometer wire of length 100 cm hasa resistance of 10 . It is connected in se-ries with resistance (shown in figure) and acell of emf 2V and negligible resistance. Asource of emf 10 mV is balanced against alength of 40 cm of potentiometer wire.Thevalue of R1 is
a) 526.67 b) 790 c) 1580 d) Zero10. The potentiometer wire AB is 100 cm long.
The value of R for which the galvanometershows no deflection, when AC=40cm(shown in figure) is
a) 13 b) 17 c) 15 d) 2111. The potentiometer wire AB is 600 cm long.
The distance from A, the jockey J shouldtouch the wire to get zero deflection in thegalvanometer, is
a) 320 cm b) 120 cmc) 20 cm d) 450 cm
KEY1) d 2) a 3) b 4) c 5) a 6) d7) b 8) b 9) a 10) c 11)a
CE1B E2
A + - - +
r1.02V
E1
SE2
P Q
G
> > > > > > > >> > > >
100cm40cm
10mV
R1
R1R1
2V
> > > > > > > >G
(R
A C B
10
> > > >
> > > > G
2E
rJ B
rE
ARAB=15 r
Page No. 68
1.25 V gives a balance point at 35 cm length ofthe wire. If the cell is replaced by another cellthen the balance point shifts to 63 cm, then e.m.fof the second cell isa) 4.25 V b) 2.25 Vc) 3.25 V d) 1.25 V
4. In a potentiometer, a standard cell of emf5V and of negligible resistance maintains asteady current through the potentiometerwire of length 5m. Two primary cells ofemf’s E1 and E2 are joined in series with (i)same polarity and (ii) opposite polarity. Thecombination is connected through a galva-nometer and a jockey to the potentiometer.The balancing lengths in the two cases arefound to be 350 cm and 50 cm respectively.The values of E1 and E2 are respectivelya) 3 V and 2 V b) 2.5 V and 1.5 Vc) 2 V and 1.5 V d) 1.5 V and 1 V
5. Two cells of e.m.f. E1 and E2 (E1>E2) areconnected as shown in figure. When a po-tentiometer is connected between A and B,the balancing length of the potentiometerwire is 300 cm. When the same potentiom-eter is connected between A and C, the bal-ancing length is 100 cm. The ratio of E1 andE2 is
a) 3 : 2 b) 2 : 3c) 2 : 1 d) 1 : 2
6. There exists a constant potential differencebetween ends of a potentiometer wire. Twocells are connected, in turn, in such a waythat they help each other and are balancedon the potentiometer wire at 60 cm. Whenthe two cells oppose each other, then theyare balanced for a length of 30 cm. The ratioof e.m.f’s of the two cells isa) 4 : 1 b) 1 : 4c) 1 : 3 d) 3 : 1
7. Potentiometer wire, PQ of 1m length is con-nected to a standard cell E1. Another cell, E2of emf 1.02 V is connected as shown in thecircuit diagram and null position is obtainedat a distance of 51 cm from P. the value ofE1 is
a) 1 V b) 2 V c) 3 V d) 4 V8. In a potentiometer experiment, two cells of emf
E1 and E2 are used in series and in conjunctionthe balancing length is found to be 81cm of thewire. If the polarity E2 is reversed, then the bal-ancing length becomes 27 cm. The ratio E1/E2isa) 1 : 1 b) 2 : 1 c) 3 : 1 d) 4 : 1
9. A potentiometer wire of length 100 cm hasa resistance of 10 . It is connected in se-ries with resistance (shown in figure) and acell of emf 2V and negligible resistance. Asource of emf 10 mV is balanced against alength of 40 cm of potentiometer wire.Thevalue of R1 is
a) 526.67 b) 790 c) 1580 d) Zero10. The potentiometer wire AB is 100 cm long.
The value of R for which the galvanometershows no deflection, when AC=40cm(shown in figure) is
a) 13 b) 17 c) 15 d) 2111. The potentiometer wire AB is 600 cm long.
The distance from A, the jockey J shouldtouch the wire to get zero deflection in thegalvanometer, is
a) 320 cm b) 120 cmc) 20 cm d) 450 cm
KEY1) d 2) a 3) b 4) c 5) a 6) d7) b 8) b 9) a 10) c 11)a
CE1B E2
A + - - +
r1.02V
E1
SE2
P Q
G
> > > > > > > >> > > >
100cm40cm
10mV
R1
R1R1
2V
> > > > > > > >G
(R
A C B
10
> > > >
> > > > G
2E
rJ B
rE
ARAB=15 r
Page No. 69
INTERNAL RESISTANCE OF A CELL: It is the resistance offered by the electrolyte of the
cell. It depends on area of the electrodes used distance between the electrodes nature of electrolyte area of cross section of the electrolyte throughwhich the current flows and age of the cell.
Internal resistance of an ideal cell is zero. The power transferred to the load is maximum when
external resistance becomes equal to the internalresistance by maximum power transfer theorem.
When a cell of EMF ‘E’ and internal resistance‘r’ is connected to an external resistance ‘R’ asshown, where i = current in the circuitV = potential difference across the externalresistanceV1 = Voltage across internal resistance (or) lostvoltsThen, EMF of the cell, E = V + V1
From Ohm’s law, i = rRVV
rRE '
V = i R = rRER
Fractional energy useful = rRR
EV
% of fractional useful energy =
100
EV
= 100
rRR
Fractional energy lost, rR
rEV'
% of lost energy, 100rR
r100EV'
internal resistance, r = RV
VE
When the cell is charging, the EMF is less thanthe terminal voltage (E < V) and the directionof current inside the cell is from + ve terminalto the –ve terminal.
V = E + i r When the cell is discharging, the EMF is greater
than the terminal voltage (E >V) and thedirection of current inside the cell is from – veterminal to the +ve terminal.
V = E – i r
Hence E V
If external resistance (R) is equal to the internalresistance (r) then the source delivers maximumpower and the terminal voltage across the cell
2E
rRERV
Hence the % of energy lost and energy usefulare each equal to 50%
POTENTIOMETER: It is a device which is used to
a) Compare EMFS of two cellsb) determine the current in a circuitc) determine the internal resistance of a celld) determine the unknown resistance.e) measure thermo EMFS
A cell of emf E and internal resistance ‘r’ in theprimary circuit maintains uniform potential gradientalong the length of its wire.Current through the potentiometer wire,
i = RrE
.
Potential gradient or potential drop per unit length
= liR
where ‘l’ is the total length of potentiometer
wire, ‘R’ is the total resistance of the wire and ‘i’is the current through potentiometer wire due toprimary circuit.
If a resistance RS is connected in series with the
potentiometer wire then i = SRRr
E and
Page No. 70
potential drop per unit length = lR
RRrE
S
Comparison of EMFS using potentiometer :- Ifarell 21 & balancing lengths when two cells of
EMFS E1 & E 2 are connected in the secondarycircuit.
one after the other then, 21
21
ll
EE
Internal resistance of a cell
r = Rl
llRV
VE
221
Where 1l = balancing length for the cell connectedin the secondary circuit.
2l = balancing length when a resistance R isconnected in parallel to the cell.E = emf of the cellV = Terminal voltage.
Page No. 71
DEFLECTION MAGNETOMETERTo compare the magnetic moments of two short bar magnets and to verifythe inverse square law.
Description: The deflection magnetometer consists of a short magnetic needlepivoted at the centre of a circular scale so that it can rotate freely in ahorizontal plane. The scale is divided into four quadrants and each isgraduated from 00-900. A thin long aluminium pointer is rigidly fixed atright angles to the needle at its mid point. The whole system is enclosedin a brass or wooden case provided with a glass top, this is called compassbox. It is placed at the centre of a rectangular wooden plank. Two halfmetre scales are fixed, one on either side of the compass so that theirzero starts from the centre of the needle.
D.M.M. works on the principle of tangent law in magnetism, B=BH TanB = Magnetic induction field due to bar magnet.BH = Earth’s horizontal magnetic field.
1. The ratio of the magnetic moments of the two magnets in equi distance method is
TanTan
MM
2
1 where is mean deflection due to first magnet.
is mean deflection due to second magnet.
2. The ratio of the magnetic moments of the two magnets in Null-method is 32
31
2
1
dd
MM
where d1 is the distance of the first magnet from the centre of the compass box. where d2 is distance of the second magnet from the centre of the compass box.
3. To verify the inverse square law we can show that 2B
A
TanTan
Where A is mean deflection in Tan A position.
B is mean deflection in Tan B position.Procedure: Equal distance method - Tan A position:All magnets and magnetic substances were kept away from the deflection magnetom-eter. The D.M.M. was rotated until the arms were perpendicular to the magnetic merid-ian (In E-W direction). In this position, the compass was rotated so that the aluminiumpointer reads 00 - 00 on the circular scale.
A bar magnet of magnetic moment M1 was placed on one of the arms so that itsaxis is along east - west direction. The centre of the magnet was kept at a distance from
the centre of the needle. The deflections 1 and 2 of the two ends of the pointer werenoted. The magnet was reversed end to end in the same position and the deflections
Page No. 72
3 and 4 were noted. Keeping the magent at the same distance on other arm the
deflections 5 and 6 were noted. The magnet was reversed end to end and the deflec-
tions 7 and 8 were noted. Take the average of eight deflections as .The experiment was repeated for different distances. Then, the experiment was
repeated for the second magnet of magnetic moment M2and average diflection is noted.
The ratio of the magnetic moments was given by
TanTan
MM
2
1.
The experiment is repeated for different distances, the readings are noted inthe following table.
The D.M.M. was kept in Tan A position and a bar magnet of moment M1 was placed onone of the arms with its centre at a distance d1 from the centre of the needle. Thesecond magnet of moment M2 is kept on the other arm and similar poles were keptfacing each other. Their magnetic axes passed through centre of the needle. Its posi-tion was adjusted so that pointer read 00 - 00.
The distance (x1) of the centre of the magnet from the centre of the needle wasnoted. The two magnets are reversed end to end. The first magnet was kept at the samedistance. The position of the second magnet was again adjusted until the deflectionbecome zero. The distance (x2) was noted. The two magnets were interchanged and theexperiment was repeated with the first magnet at the same distance from the centre ofthe needle. For null deflection, before and after reversing the magnets end to end, thedistance of x3 and x4 for the second magnet were noted. The average of the four dis-tances (x1, x2, x3, x4) d2 was found. The ratio of the magnetic moments of the two mag-
net is given by 32
31
2
1
dd
MM
.
The experiment was repeated for different distances of d1 and every time d2 wasdetermined.
Equal Distance Method - Tan B position:Procedure:The deflection magnetometer was rotated until the arms were parallel to the magneticmeridian. In the position, the compass was rotated so that the aluminium pointer read00 - 00 on the circular scale.
Page No. 73
CONVEX MIRRORTHEORY :
Principal focus : It is the point on the principal axis at which a narrow beam of light rays, paralleland close to principal axis appears to diverge after reflection.
The distance between the pole of the mirror and the principal focus is called focal length ‘f’. Let Uand V be the distances of the object and its image from the pole of the convex mirror of focal length ‘f’.Convex mirror always forms a virtual image for all the distances of the object. Therefore the distances of theobject U are + ve and the distances of the image V are always-ve. The focal length is determined by followingformula.
1 1 1f U V
UVf(U V)
The focal length calculated from the above is always ‘–ve’.
PROCEDURE :
a) Direct method: The given convex mirror P is mounted on a V-shaped stand and placed on the table ata convenient height. A short bright pin ‘O’ is placed on the the table in front of the convex mirror.Now a virtual image, small in size is observed behind the mirror. Another long bright pin is placedbehind the mirror so that the two pins and the mirror lie along the same straight line. The position ofthe long pin behind the mirror is adjusted so that it coincides with the image of the short pin withoutany parallax i.e., there is no separation between the long pin and the image of the short pin when thehead with one eye closed is moved towards left or right perpendicular to the principal axis. Now theposition of the long pin gives the position of the image. The distances of the short pin O and the longpin I from the mirror are measured as U and V respectively.
PO = U cm
PI = V cm
then f = (UV)
U V
cm
Page No. 74
The experiment is repeated as above for different positions of the pin ‘O’ and the observations are noted inthe tabular from.Observations : Direct method
S.No Distance of the short pin PO Distance of the long pin Focal length f =
U V
U V
cm
from the mirror U cm PI from the mirror V cm
1.
2.
3.
4.
5.
6.
Average focal length f = .......... cm
b) Plane mirror method : The given convex mirror is mounted on a V- shaped stand and placed on thetable . A long bright pin O is placed in front of the mirror at a convenient long distance. A plane mirrorM is placed between the convex mirror P and the long pin ‘O’ so that it ( plane mirror) covers thelower half of the convex mirror. The height of the pin is so adjusted as that the image of the upper halfof the pin is formed in the convex mirror and that of the lower half is formed in the plane mirror. Theposition of the plane mirror is adjusted so that there is no parallax between the images formed by theplane mirror and convex mirror. The positions of the pin O, the plane mirror M and the convex mirrorP are noted. Then
PO = U cmPO-PM = V cm
The focal length f is calculated from the following formula f = (UV)
(U V) cm
Page No. 75
The experiment is repeated as above for different positions of the pin ‘O’ and the observations are noted inthe tabular form.Plane mirror method :
SNo Position of the Position of the Position of the Distance of Distance of Focal length
long pin plane mirror convex mirror the object the image
U V
fU V
O cm M cm P cm V = PO cm U = PO – PM cm
1.
2.
3.
4.
5.
6.
Average focal length ‘f’ = .......... cm
PRECAUTIONS :1. The pins and the mirror should be along the same straight line2. The pins should be bright and vertical3. The plane of the mirror should be vertical in its stand4. The pins and the eye should be in the straight line.
EXERCISE
1) Choose the correct statement among the followinga) focal length of convex mirror is +veb) Convex mirror always forms a virtual image for all the distances of the objectc) Convex mirror forms a virtual image not for all the distances of the objectd) It is a convergent mirror (b)
2) In the direct method of finding focal length of the convex mirror, A short bright pin placed on the tablein front of it. Another long bright pin is placed behind the mirror so that the two pins and the mirror liealong the same straight line. The position of the long pin behind the mirror is adjusted so that itcoincides with the image of the short pin without any parallax. Then the image is represented by.a) short pin placed in front of the mirrorb) long pin placed behind the mirrorc) both the pinsd) none of the two pins (b)
3) In finding the focal length of the convex mirror by plane mirror method, choose the following correctstatementa) plane mirror full length of the convex mirrorb) plane mirror has to place behind the convex mirrorc) plane mirror covers only the lower of the convex mirrord) The image forms in front of the convex mirror (c)
Page No. 76
4) In finding the focal length of the convex mirror by plane mirror method , the image due to convex mirror andplane mirror coincides without parallax at some distance from the pole of the convex mirror. The distance ofthe image isa) The distance of the image from the pole of the convex mirrorb) The distance of the image from the plane mirrorc) The difference in distances of the two mirrorsd) The difference in distances of the object distance from the convex mirror and distance of the planemirror from the convex mirror. (d)
5) U-is the object distance, V - is the image distance , f- is the focal length of the convex mirror.Choosethe correct relation among them
a) f = uv
u vb) f =
uvu v
c) f = uv
u v d) f = uv
u v
(d)
6) An object is placed in front of the convex mirror at a distance of 50 cm. A plane mirror is introducedcovering the lower half of the convex mirror. If the distance between the two mirrors is 10cm, it isfound that there is no parallax between the images formed by the two mirrors. Then the focal length of theconvex mirror isa) b) c) d) (b)
Hint: u=50cm, v = 40 – 10 =30 cm. f = uv cm
u v
= -75cm (ans)
7) The image of an object in a convex mirror is 4cm from the mirror. If the mirror has focal length 12cm,the position of the object is.a) 8cm b) 4 cm c) 6 cm d) 12 cm (c)
Hint: f = uv and
u v
u(4)is ve 12
u 4
u = 6 cm
Page No. 77
FOCAL LENGTH OF CONCAVE MIRROR SYNOPSIS A mirror which is a part of sphere is called spherical mirror. There are two types of spherical mirrors. Concave mirror is a conversing mirror In case of concave mirror, outer surface is polished. The mid point of the mirror is called pole(p) of the mirror. Center of curvature (c) is the center of the sphere of which the mirror is a part. The line joining the pole and center of curvature is called the principal axis. Principal focus (F) It is a point on the principal axis of a convex mirror at which a parallel beam after reflection appears to diverge from that point. The distance between pole and principal focus is called focal length. Concave mirror – Image formation To locate the position of the image following rules are followed a) A ray of light parallel to the principal axis, after reflection passes through the focus. b) A ray of light originating from the focus after reflection becomes parallel to the principal axis c) A ray of light passing through the center of curvature will be reflected back,ie retraces its path. Tabular form Position of Object Position of image Nature of Image
1) At infinity 2) Beyond C
3) At C
4) Between F & C
5) At F
6) Between F & P
At F Between ‘F’ and ‘C’ At C Beyond C At infinity Image is formed behind the concave mirror
Real, inverted and point size Real inverted and diminished in size Real, inverted and same size as that of the object Real, inverted and magnified Virtual, erect and magnified
C
A
F P
B
Incident light
Page No. 78
It forms both real and virtual images Its magnification is equal to ‘1’ or less than ‘1’ or greater than ‘1’. As the object approaches the concave mirror, the real image moves away from the
mirror, but the virtual image moves towards the mirror. The minimum distance between the object and real image is zero. They are used as reflectors, in solar cookers, in opthalmoscope by dentists, by
eye specialists, as shaving mirrors and in reflecting telescopes. Formulae For any spherical mirror R = 2f (R = Radius of curvature)
For all spherical mirrors R2
f1
v1
u1
Where u and v are the object and image distances from the pole of the mirror Sign convention a) The distances of real object and real images are taken as positive b) The distances of virtual object and virtual image are taken as negative. c) Focal length of concave mirror is taken as positive
Linear magnification m = objecttheofSizeimagetheofSize =
cetandisObjectcetandisageIm =
uv
For real image m = fu
ff
fvuv
v = f (1+m) ;
m11fu
For a virtual image m = uf
ff
fvuv
v = f (m-1) ; 11u fm
Areal magnification M = 2m = 2
2
uv =
objecttheofAreaimagetheofArea
Newton’s formula xyf Where x and y are the distances of real object and real image from the focus Graphs : 1. U-V graph F=
4OA OB cm
Page No. 79
2. 1 1
u v graph F = 2 cm
OA OB
MULTIPLE CHOICE QUESTIONS 1) A concave mirror of focal length ‘f’ produces an image 1/nth of the size of the
object. The distance of the object from the mirror is
1) (n-1)f 2) fn
1n
3) f
n1n
4) (n+1) f
2. A concave mirror has radius of curvature of 1m. light from a distant star is incident on the mirror. The distance of the image of the star from the mirror is
1) 0.5m 2) 1m 3) 2m 4) 0.25m 3. A Candle flame 3cm is placed at a distance of 3m from a wall. How far from wall
must a concave mirror be placed in order that it may form an image of flame 9cm high on the wall ?
1) 225cm 2) 300cm 3) 450cm 4) 650cm 4. A plane mirror is approaching you at 5cm/sec. At what speed will your image
approach you ? 1) 5cm/sec 2)10cm/sec 3) 15cm/sec 4) 20cm/sec 5. A man 160cm high stands in front of a plane mirror. His eyes are at a height of
150cm from the floor. Then the minimum length of the plane mirror for him to see his full length image is
1) 85cm 2) 170cm 3) 80cm 4) 340cm 6. An object is placed at a distance of 40cm in front of a concave mirror of focal
length 20cm. The image produced is 1) Virtual and invested 2) Real and erect 3) Real, inverted and diminished 4) Real, inverted and same size as the object
45°
V
B
O A U
B
O A
1V
1u
Page No. 80
7. Which of the following graph is applicable between v1 and
u1 for a spherical
mirror ? 1) 2) 3) 4) 8. If magnification of a concave mirror is 4 times as great when the object was 25cm
from the mirror as it was with the object at 40cm from the mirror and image formed was real in both cases, then focal length of the mirror should be
1) –20cm 2) –50cm 3) +20cm 4) +50cm 9. An object is situated at a distance of 15cm from a convex lens of focal length
30cm. The position of the image formed by it will be 1) –5cm 2) 10cm 3) –10cm 4) +5cm 10. A concave mirror has a focal length of 0.1m. The radius of curvature of the
mirror is 1) +0.1m 2) -0.1m 3) +0.2m 4) -0.2m 11. A concave mirror of focal length ‘f’ forms an image of the same size as the
object. The distance of the object from the mirror is 1) f 2) f / 2 3) 2f 4) 4f 12. A point object is placed at a distance of 500mm from a concave mirror of focal
length 200mm. If the object is moved towards the mirror by 100mm, the image moves by a distance of
1) 67mm towards mirror 2) 67mm away from the mirror 3) 10mm towards mirror 4) 10mm away from the mirror 13. A candle is placed in front of a concave mirror. The image of candle is formed on
a wall at a distance of 2m from the candle. The size of the image is 6 times that of the candle. The distance of the candle from the mirror is
1) 1.6m 2) 2.4 m 3) 2m 4) 0.4m 14. A concave mirror forms, on a screen a real image of twice the linear dimensions
of the object. Object and screen are then moved until the image is three times the size of the object. If the shift of the screen is 25cm, then the focal length of the mirror is
1) 5cm 2) 16.6cm 3) 25cm 4) 37.5cm 15. An object of length 6cm is placed on the principal axis of a concave mirror of
focal length ‘f’ at a distance of 4f. The length of the image will be 1) 2cm 2) 6cm 3) 10cm 4) 12cm 16. The focal length of a concave mirror is 7.5cm. If the object is at 15cm from the
mirror then the position of the image formed is 1) At the same point of the object 2) 15cm behind the mirror 3) 7.5cm behind the mirror 4) 7.5cm in front of mirror
KEY 1-10 1 1 3 2 3 4 2 1 2 3 11-16 3 2 4 3 1 1
v1
u/1
v1
u/1
v1
u/1
v1
u/1
Page No. 81
Convex lens
Lens formula for convex lens is 1 1 1f u v ;
Where ‘ u ’ is the object distance and ‘ v ’ is the image distance. Sign convention 1. All the distance are measured from the optic center of lens. 2. Distances of real objects, real images are ve . 3. Distances of virtual objects, virtual images are ve . 4. Focal length of convex lens is ve and focal length of concave lens is ve .
Linear magnification size of the imagemsize of the object
tantan
image dis ce vobject dis ce u
,
Areal magnification 2
2
image area vobject area u
For a real image and for a convex lens.
1 1 1 v v f fm mf u v u f f u
If ‘ x ’ and y are the Distances of real object And real image measured from
the focus of a convex lens then f xy The minimum distance between a real object and its real image formed by a single lens is ‘ 4 f ’.
As every point of a lens forms complete image, if a portion of it is obstructed or covered, full image will be formed but intensity of the image will be reduced.
Focal length of the Convex lens by u v method: In this method, the object
y I
I O F F
G x f
f
Page No. 82
Distance is varied in steps, and
O
G O
object image
u v The corresponding, image distance are measured. Then each time the Focal length of the lens is measured,
From the formula 1 1 1f u v .
or uvfu v
.
The average value of ‘ f ’ gives the focal length of the given lens. u v graph:-
If a graph is plotted between u & v ,
the nature of the graph is a rectangular ) 45º
v
u
C A
O B hyperbola. From the graph, the focal length of the convex lens can be measured.
4OB OCf
1 1u v graph
-------------------
From 1 1u v graph, the focal length of the
O
B
A
1v
1u
gives lens is measured from the formula 2fOA OB
Page No. 83
Lens displacement method or conjugate foci method In this method, when the distance between The object and the screen (L) is greater than
1 I 2 I
d
u v
u v
J
O
L 1 L
2 L
object screen
1. position
2. position 4 f , then there are two positions of the lens for which the image of the object on the screen is distinct and clear one is a magnified image and the other is a diminished one. In these two positions of the lens, the distance of the object and image from the lens are interchanged. Hence it is also called as conjugate foci method.
If 1I and 2I are the sizes of images in the first and 2nd positions of the lens. ‘OJ ’ is the size of the object and 1m and 2m are their magnifications of the lens for first and second
position of the lens. Then 1 21 2,I Iv vm m
u OJ u OJ
21 2 1 21m m OJ I I
1 2OJ I I
From fig. ,2 2
L d L du v
2 2
1 1
2 2
m I v L dm I u L d
substituting u and v values in 1 1 1f u v
then 2 2
4L df
L
Page No. 84
PROBLEMS 1. The image of a square hole on a screen illuminated by light is obtained on another screen
with the help of a converging lens. The distance of the illuminated square from the lens is 40cm. The area of the image is 9 times that of the square hole. The position of the image and focal length of the lens are
1) 120cm, 30cm 2) 120cm, 40cm 3) 130cm, 30cm 4) 130cm, 40cm 2. An object is kept at a distance of 4cm from the first focus of a convex lens. A real image
is formed at a distance of 9cm from its 2nd focus. Then focal length of that lens is 1) 6.5cm 2) 13cm 3) 6cm 4) 36cm 3. A lens forms a real image of an object on a screen placed at a distance of 100cm from the
screen. If the lens is moved by 20cm towards the screen, another image of the object is formed on the screen. The focal length of the lens is
1) 12cm 2) 24cm 3) 36cm 4) 48cm 4. The distance at which an object should be placed in front of a convex lens of focal length
10cm to obtain a real image double the size of the object is 1) 15cm 2) 5cm 3) 10cm 4) 20cm 5. Where should an object be placed in front of a convex lens of focal length 30cm to form
a virtual image magnified 3 times. 1) 20cm 2) 30cm 3) 40cm 4) 50cm 6. A lens forms image on a screen placed at a distance of 50cm from an object. When the
lens is shifted towards the screen by 10cm, another image is formed on the screen. Find the focal lengths of the lens
1) 5cm 2) 7cm 3) 9cm 4) 12cm 7. A plane object 2cm long and 10cm broad is placed parallel to a thin convex lens of focal
lengths 10cm. Find the area of the image when the object is at a distance of 20cm from the lens.
1) 10 sq.cm 2) 20 sq.cm 3) 40 sq.cm 4) 80 sq.cm 8. A convex lens of focal length ‘f ’ is placed some where between an object and the screen.
The distance between the object and the screen is ‘x’. It the magnification is ‘m’ the focal length of the lens is given by
1) 21
mxm
2) 21m x
m
3) 21
mxm
4) 21m x
m
9. The graph is drawn plotting 1u
along x-axis and 1v
along y-axis the scale is 0.01m. If
the intercept on the x-axis is 0.5m, the focal length of the lens is (in meter) 1) 2.00 2) 0.50 3) 0.20 4) 1.00 10. A convex lens form an image of an object on a screen 0.20 m from the lens on moving
the lens 0.60m towards the object an image is again formed on the screen. The focal length of the lens is
1) 0.08 m 2) 0.16 m 3) 0.24 m 4) 0.32 m 11. If in a convex lens image and object distance are each equal to ‘d’ then the focal length of
the lens in the case of real image is
1) 2d 2) d 3) 2d 4)
4d
12. A slide projector give a magnification of 10. If slide of dimensions 3x2 cm is projected on the screen, the area of the image on the screen
Page No. 85
1) 6 2cm 2) 60 2cm 3) 600 2cm 4) 6000 2cm 13. A luminous object is separated from a screen by a distance D. What is the greatest focal
length of a lens could have to focus the object on the screen.
1) D 2) 4D 3) 4D 4)
2D
14. How will the image formed by a convex lens be effected, if the central portion of the lens is wrapped in black paper, as shown in figure.
1) Full image will be formed but without central portion
2) Two images will be formed, one due to each exposed half. 3) No image will be formed. 4) Full image will be formed but is less bright 15. A converging lens is used to form an image on a screen. When the upper half of the lens
is covered by an opaque screen 1) Half the image will disappear 2) Complete image will be formed. 3) Intensity of image will increase 4) Intensity of image will decrease
KEY 1) 1 2) 3 3) 2 4) 1 5) 1 6) 4 7) 2 8) 1 9) 1 10) 2 11) 3 12) 3 13) 3 14) 4 15) 4
Page No. 86
i – d. curve 1. What is the shape of the graph between angle of incidence (i) and angle of deviation(d)? Ans. The i – d curve is a parabola 2. What is angle of prism(A)? Ans. The angle between two refracting faces of a prism where they meet. It is opposite to the base of the
prism. Angle of prism for an equilateral prism is 600. 3. What is the formula for refractive index of the material of the prism?
Ans. sin
2
sin2
A D
A
A = Angle of the prism D = minimum Deviation 4. What is the path of refracted ray inside the prism in minimum deviation position? Ans. The refracted ray is parallel to base of the prism in minimum deviation position 5. What are the condition to obtain the minimum deviation position Ans. 1) The refracted ray inside the prism is parallel to the base of the prism 2) The angle of incidence(i1) is equal to angle of emergence (i2) 3) The two angles of refraction are equal 1 2r r 6. What is the formula for angle of deviation? Ans. Angle of deviation 1 2d i i A 7. What is the relation between 1 2,r r and angle of the prism(A)? Ans. 1 2A r r 8. How are 1 2&i i and 1 2&r r are related in minimum deviation position? Ans. 1 2i i i and 1 2r r r
1 2 22AA r r r r r r (1)
and we know that 1 2d i i A
22
A DD i i A D i A i (2)
9. Define refractive index Ans. Refractive index is defined as the ratio of velocity of light in vacuum to velocity of light in the
medium
0( )( )m
velocity of light in vacuum VVelocity of light in metium V
It has no units of dimensions. Its value is always greater than or equal to 1 10. What is Snell’s law? An s. In refraction “The ratio of sine of angle of incidence sin i to the sine of angle of refraction sin r
is constant” for a given pair of media for a given wavelength.
sinsin
ir
11. What is the relation between refractive index , angle of prism A and angle of minimum deviation?
Page No. 87
Ans. sin
2sin
2
A D
A
12. How does the refracted ray bend when a ray of light goes from rarer medium to denser medium? Ans. It bends towards the normal drawn to the boundary at the point of incidence i r 13. How does the refracted ray bend when a ray of light goes from denser to rarer medium. Ans. It bends away from the normal drawn to the boundary at the point of incidence i r 14. Which colour of light has largest value of refractive index?
Ans. Violet ( 1 1V
& ,V R V R )
15. Which colour of light travels fastest in the medium? Ans. Red travels fastest in the medium V As , ReR V d V ioletV V 16. If the wavelength of incident light increases does the refractive index of medium remains constant.
Ans. Refractive index is not constant . We know that 1
. If increases, the refractive index of given
pair of media decreases. 17. For small angled prism, what is the angle of deviation?
Ans. We know that sin
2sin
2
A D
A
For small values of ,sin
/ 1 12 2
A D A A D D DA A A
1D A 18. Which colour of light is deviated least and which colour is deviated largest in prism? Ans. deviation 1D As V R ,
deviation of violet is largest and deviation of red is least 19. When white light passes through prism, dispersion takes place. Why? Ans. The difference in angle of deviation of different colours and also the shape of prism make the light to
travel different distances inside the prism will lead to dispersion of light. 20. A completely transparent material is invisible in vacuum. What is its refractive index.? Ans. Unity. 21. What does the colour of a star indicate? Ans. Its temperature 22. Can you use a hollow prism filled with water to get a spectrum? Ans. Yes. [Spectrum can’t be formed with a hollow prism] 23. What is absolute refractive index? Ans. It is the refractive index of a medium w.r.t vacuum. 24. When a light ray enters into a different medium which physical quantity of the light changes? Ans. The frequency (colour) doesn’t change, the velocity changes hence the wavelength changes. 25. What is a prism? Ans. It is a Transparent medium like glass bounded by two plane surfaces. Two surfaces are transparent
(called refracting surfaces) and the third surface is not transparent (called the base) 26. What is the difference between telescope and microscope?:
Page No. 88
Ans. A telescope is used to see the magnified images of distant objects. Where as the microscope is used to obtain the magnified images of very small objects placed near to it .
27. Define the edge of prism? Ans. The line along which the surfaces meet is called refracting edge of the prism. 28. What do you mean by pure spectrum ? Ans. A pure spectrum is one in which there is no overlapping of colours. In this spectrum each colour
occupies a separate position and distinctly visible. 29. What is meant by impure spectrum? Ans. The spectrum in which there is no overlaping and mixing of colours is called impure spectrum. 30. What do you mean by monochromatic source of light ? Give examples. Ans. A source of light which gives only one wavelength of radiation is called monochromatic source Ex : Sodium light is nearly monochromatic
Page No. 89
( Refractive index of a glass slab )
1. A mark is made at the bottom of a beaker and a microscope is focussed on it. The microscope is the risedthrough 0.015m. To what height water must be poured into the beaker to bring the mark again into focus
( Given 43
aw )
1) 0.06 2) 0.05 3) 0.04 4) 0.032. What is the apparent position of an object below a rectangular block of glass 6 cm thick, it a layer of water
4 cm thick is on top of the glass 3 4;2 3g w
1) 2 cm 2) 3 cm 3) 4 cm 4) 1 cm3. One face of a glass cube of side 0.06 m is silvered. An object is placed at a distance of 0.07 m from the face
opposite to the silvered face. Looking from the object side, the image of the object appears to be 0.11 mbehind the silvered face. Calculate the refractive index of the material of glass.1) 1.6 2) 2.5 3) 1.5 3) 1.4
4. If a plane glass plate is placed on letters of different colours, then which coloured letters appears to be moreraised1) blue 2) green 3) voilet 4) red
5. Calculate the index of refraction of a liquid from the following into glass :Reading for the bottom of an empty beaker : 11.324 cmReading for the bottom of the beaker, when partially filled with the liquid : 11.802 cmReading for the upper level of the liquid in the beaker : 12.895 cm1) 2.437 2) 3.436 3) 2.436 4) 1.437
6. While determining the refractive index of a liquid experimentally, the microscope was focussed at the bottomof a beaker, when its reading was 3.965 cm. On pouring liquid upto a height of 2.537 cm inside the beaker,the reading of the refocussed microscope was 3.348 cm. Find the refractive index of the liquid.1) 1.321 2) 3.322 3) 4.323 4) 2.321
7. The bottom of a container is a 4.0 cm thick glass ( 1.5) slab. The container contains two immiscibleliquids A and B of depths 6.0 and 8.0 cm respectively. What is the apparent position of a scratch on the outersurface on the bottom of the glass slab, when viewed through the container ? Refractive indices A and B are1.4 and 1.3 respectively1) 4.79 2) 4.80 3) 4.89 4) 5.89
8. A beaker containing liquid is placed on the table underneath a microscope which can be moved along avertical scale. The microscope is focussed, through the liquid onto a mark on the table when the reading onthe scale is a. It is next focussed on the upper surface of liquid and the reading is b. More liquid is added andthe observations are repeated. The corresponding readings are c and d. The refractive index of liquid is
1) d b
d c b a
2)
d c b ad b
3) b d
d c b a
4)
d c b ab d
9. A person looking through a telescope ‘T’ just sees the point ‘A’ on the rim at the bottom of a cylindrical vesselwhen the vessel is empty. When the vessel is completely filled with a liquid ( 1.5), he observes a mark atthe centre B, of the bottom without moving the telescope or the vessel. What is the height of the vessel if thediameter of its cross section is 10 cm.1) 8.45 cm 2) 8.55 cm 3) 7.45 cm 4) 7.55 cm
10. A small object is placed 20 cm in front of a block of glass 10 cm thick and its farther side silvered. The imageis formed 23.2 cm behind the silvered face. Find the refractive index of glass.1) 3.52 2) 4.53 3) 1.51 4) 2.52
11. A rectangular glass block of thickness 10 cm and refractive index 1.5 placed over a small coin. A beaker isfilled with water of refractive index 4/3 to a height of 10 cm and is placed over the glass block. Find theapparent position of the object when it is viewed at near normal incidence
Page No. 90
1) 12.2 2) 13.2 3) 14.2 4) 15.212. A bird in air looks at a fish vertically below it and inside water ‘h1’ is the height of the bird above the surface
of water and ‘h2’, the depth of the fish below the surface of water. If refractive index of water with respect toair be ‘ ’, then the distance of the fish as observed by the bird is,
1) 1 2h h 2) 21
hh
3) 1 2h h 4) 1 2h h
13. A vessel of depth ‘t’ is half filled with oil of refractive index 1 and the other half is filled with water of
refractive index 2 . the apparent depth of the vessel when viewed from above is
1) 1 2
1 22t
2) 1 2
1 22t
3) 1 2
1 2
2t
4) 1 2
1 2
2t
14. A plane mirror is placed at the bottom of a tank containing a liquid of refractive index ‘ ’. ‘P’ is small objectat a heigth ‘h’ above the mirror. An observer ‘O’, vertically above ‘P’, outside the liquid sees ‘P’ and itsimage in the mirror. The apparent distance between these two will be
1) 2 h 2) 2h 3)
2( 1)
h 4)
11h
15. An air bubble in glass slab ( 1.5) from one side is 6 cm and from other side is 4 cm. The thickness ofglass slab is1) 10 cm 2) 6.67 cm 3) 15 cm 4) 20 cm
KEY
01-10 1 2 3 4 4 1 3 1 1 311-15 3 2 1 2 3
Page No. 91
P-N JUNCTION DIODE a) P-N junction diode: An electronic device consisting of a p – n junction is called a p – n junction
diode. b) Depletion layer: A thin layer in the middle of the p – n junction which is void of the charge carriers
is called a depletion layer. This layer is formed due to diffusion of electrons and holes at the junction. Thickness of the depletion layer is about 10-6m Depletion layer acts as a di – electric. Thickness of depletion layer can be increased or reduced by the applied voltage.
c) Potential barrier BV : The potential created across the depletion layer is called the potential barrier
BV
BV 0.3 V for Ge p – n junction,
BV 0.7 V for Si p – n junction,
BV can be increased or decreased by an applied voltage.
BV depends (a) on temperature of the junction, b) doping concentration and (c) the nature of the semi conductor
Forward bias:-
a) When the battery is connected to p – n junction diode, with its p-side to +Ve terminal, and n-side to the –Ve terminal, it is called a forward bias.
b) This bias opposes the potential barrier and BV almost reduces to zero c) Depletion layer is reduced but never becomes zero. It offers low resistance (from 1 to 25 ) d) It acts as a conductor (or an on – switch)
e) Current is of the order of mA Reverse bias:-
a) When the battery is connected to p – n junction diode, with its p side to –Ve terminal, and n-side to +Ve terminal, this is called reverse bias
Page No. 92
b) BV Increases. In this bias supports the potential barrier c) Depletion layer increases and resistance to the order of 10 (A small amount of reverse current may
leak to drift of minority charge carriers) this is due to thermal agitation of the atoms. d) This acts as an off – switch e) Current is of order of 610A A
Knee voltage KV : The forward voltage at which the current through the junction starts increasing
rapidly. Rectification: The conversion of AC voltage into DC voltage is called rectification. A device used
for such a purpose is called rectifier. Half wave rectifier: A single diode that produces d.c. during half cycle of ac is called half wave
rectifier. 1. During the +Ve half cycle of ac input, the diode acts as a forward bias and allows the current through it.
During the –Ve half cycle of ac input, the diode acts as a reverse bias and does not allow the current through it.
2. Efficiency = DC power outputAC power input
0.406dc L
ac L f
P RP R R
( where LR - Load resistance fR - forward
resistance of diode)
3. Max Current mm
f L
VIR R
4. 2m
rms ACII I
5. D.C output current Im 0.318ImdcI
6. D.C output voltage Vdc = 0.318Vm Vm
Full wave rectifier:
Page No. 93
1) In this case two diodes, and a centre tap (T) Trans former are used. 2) The centre tap (T) of the secondary coil can be considered as a reference and is taken as zero for the AC
voltage in the secondary. The AC voltages at A and B will be out of phase ( 0180 ) with each other as shown in figure.
3) During the +ve half cycle of AC (at A) diode 1D is forward biases and diode 2D is under reverse bias.
Current flows through 1D only. 4) During the +ve half cycle of AC (atB) diode 2D is under forward bias and conducts currents and diode
1D is under reverse bias and does not conduct current. 5) Hence there is unidirectional current in the total resistance as pulsations (DC). So it is called a full wave
rectifier. 10.812
r L
RDC power outputinput AC power R R
6)
7) Ripple factor of rectifier circuit r = ..
output A C voltageInput A C voltage
Modulation: The process of superposing audio sound frequency on radio frequency (electro magnetic
waves) is called modulation. Demodulation: (Detection): The process of separating audio waves from the radio waves (electro magnetic
waves) is called de-modulation. Note:- a) The knee voltage does not depend on the current through the junction. b) The semi conductor devices are temperature sensitive devices c) A semi conductor holes is a characteristic feature of a semi conductor. d) A semi conductor is a non-ohmic resistor e) Silicon is preferred to germanium as it has small leakage current f) Ge. Si belong to 4th group (tetra valent)
Zener Diode 1. In the reverse bias, when the applied voltage is increases beyond a critical value. There is a marked
increase in the reverse current. The junction diode is said to be in a condition of breakdown. If the p.d is further increases, the diode is likely to burn out.
2. Break Down Voltage (or Zener voltage) zV : The critical value of the reverse p.d applied to a junction diode at which the reverse current suddenly suddenly increases is called breakdown voltage (or zener voltage) zV
Zener break down:- 1) Zener breakdown takes space in a very thin junction i.e when both sides of junctions are very
heavily doped 2.) When a small reverse bias voltage is applied, a very strong electric field is setup across the thin
depletion layer. This field is enough to break the covalent bonds. Now a large no of electrons and holes are produced which constitute the reverse saturation current (Zener current)
Page No. 94
3) Zener current is independent of the applied voltage and depends only on the external resistance. This process is reversible
Avalenche breakdown:- 1) This occurs in diodes at a low level of doping in p-n junction diode. 2) When a diode is subjected to high reverse p.d the highly accelerated minority charge carries collide
the valence electrons ( in the covalent bonds) in the depletion layer. This releases massive no of electrons.
3) This is called Avalanche breakdown. This process is not reversible. Zener Diode: Zener diode is reverse – biased heavily doped silicon (or germanium) P-N junction, diode
which is operated in the breakdown region. a) It can be used as a voltage regulator (or stabilizer)
b) In the forward bias, it acts as an ordinary diode ` c) At a particular reverse voltage, the current increases abruptly. This is called breakdown voltage. d) The total current (I) through series resistance R=Zener current 2I +load current 2 Z LI I I I
from Kirchoff’s 2nd law in ZV iR V
Z in outV V iR V 1;in s Z Z L Z ZV V V V I R I R The voltage can be regulated as follows: A) Due to change in input voltage: i) If the input voltage increases (at fixed R), there is an increase in I and IR increases the increase in
current is absorbed by the zener diode so that LI remains constant. Therefore.
L LI R constant out ZV V as both are in parallel combination.
ii) If I decreases, ZI drops to smaller value and LI is adjusted to give a constant output voltage L LI R B) Due to changes in load Resistor LR (at fixed I): i) If LR decreases, LI increases but ZI decreases so as to keep I and IR constant. In this Way…
Remains constant at ZV . ii) If LR increases, LI decreases but ZI increases to deep I and IR constant. Therefore outV remains
constant at ZV
PROBLEMS
1. The dominant mechanisms for motion of charge carriers in forward and reverse diased silicon p-n
junction are. 1) drift in forward bias, diffusion in reverse bias 2) diffusion in forward, drift in reverse bias 3) diffusion in both forward and reverse bias 4) drift in both forward and reverse bias 2. Terminal potentials of two diodes P and Q are as shown in figure.
Page No. 95
Then which of the following is correct.
1) Both are reverse biased 2) P is forward biased and Q is reverse biased 3) P is reverse biased and Q is reversed biased 4) both are forward biased 3. The width of forbidden gap in silicon crystal is 1.1 e.V. When the crystal is converted into p-type
semiconductor, the distance of fermilevel from conduction band is 1) greater than 0.55 eV 2) equal 0.55 eV 3) less than 0.55 eV 4) equal to 1.1 eV 4. Pick out the incorrect statement the reverse current in p-n junction diode 1) will be minimum and constant 2) remains constant even after the breakdown voltage 3) becomes infinity at breakdown 4) reverse current is controlled by external resistance 5. The cause of the potential barrier in a p-n diode is 1) depletion of positive charges near the junction 2) concentration of +ve charges near junction 3) concentration of negative charge near the junction
4) concentration of positive and negative charges near the junction. 6. At which reverse voltage current increases steeply 1) threshold voltage 2) knee voltage 3) break down voltage 4) stopping voltage 7. The forbidden energy gaps in the Ge and Si are 0.7 eV and 1.1 eV respectively. It implies that
1) both Ge and Si are perfect conductiors at very low temperature but very good insulator at room temperature 2) both Si and Ge are perfect insulators at all temperature 3) both Si and Ge are good insulators at low temperatures but start conduction at room temperature with Si some what better conductor than Ge 4) same as (3) but with Ge showing better conductivity at room temperature.
8. In which case in the P-N junction forward biased
1) 2)
3) 4) 9. The no. of minotiry carriers crossing the junction of diode depends primarily on the 1) concentration of dopping impurities 2) magnitude of potential barrier 3) magnitude of the forward bias voltage 4) rate of thermal generation of an electron-holepair 10. Four silicon diodes are connected as shown in the fig. Assuming the diodes to be ideal. Find the
current through the resistor ‘R’
1) 0.02 amp 2) 0.08 amp 3) 0.04 amp 4) 0.0005 amp
11. In a full waverectifier, without filter, the frequency of the input voltage is 60Hz. The frequency of the output voltage is
1) 60 Hz 2) 120 Hz 3) 30Hz 4) 0 Hz 12. Find the current from below circuit if forward resistance of the diode is 30ohm.
1) 0.05 A 2) 0.01 A 3) 0.05 A 4) 0.01 A 13. Fig. Shows Zener diode with a break down voltage of 40V connected to 100 V.D.C. source with a
series resistance SR and a load resistor LR . Find the i output voltage across LR
Page No. 96
1) 40V 2) 20V 3) 50V 4) 60V 14. Zener break down occurs at a junction which is 1) Heavely doped and have wide depletion layer 2) Lightly doped and have wide depletion layer 3) Moderately doped and have narrow depletion layer
4) Heavely doped and have narrow depletion layer 15. At break down region of zener diode which of the following does not change much? 1) voltage 2) current 3) dynamic impedance 4) capacitance 16. The circuit shown in the fig. Contain two diodes each with a forward resistance of 50 ohm and with
infinite reverse resistance. If the battery voltage is 6V, find the current through the 100 ohm resistance.
1) 0.02 A 2) 0.01 A 3) 0.05 A 4) 0.1 A 17. The forward biased characteristics of a p-n junction are :
18. The reverse biased characteristics of a p-n junction are
19. The value of barrier potential depends on a) Doping density b) temperature c) both d) None 20. Depletion layer consists of : a) electrons b) protons
c) mobile charge carries d) immobile ions 21. Which is ‘in correct’ statement regarding reverse saturation current in p-n junction diode
a) Reverse saturation current is also known as leakage current b) Current doubles for every 1000 C rise in temp c) Current carries are produced by thermal agitation d) Current is due to minority carries
22. If in a p-n junction diode, a square in put signal of 10 V is applied as shown
Page No. 97
then out put signal across RL will be,
23. The depletion layer in the p-n junction is caused by a) drift holes b) drift electron
c) diffusion of carrier ions d) migration of impurity ions 24. When two semi conductors p and n-type are brought into contact theory form a p-n junction which
acts like a an a) rectifier b) amplifier c) conductor d) oscillator 25. Two identical p-n junctions may be connected in series with a battery in three ways as shown, The
potential drop across the two p-n junctions are equal
a) circuit (i) & (ii) b) circuit (ii) & (iii) c) circuit (iii) & (i) d) circuit (i) only Explanation:-
In the circuit(iii) both the p-n junction are forward biased and hence have same junction resistance and same potential drop similarly in the first circuit both the p-n junction are reverse biased and have equal junction resistances.
26. The avalanche breakdown occurs at 1) higher reverse voltage 2) Lower reverse voltage
3) Lower forward voltage 4) Higher forward voltage 27. ‘P’ – type semi conductor is 1) +ve charged 2) –ve charged 3) Electrically neutral 4) may be +ve (or) –ve 28. Always zener diode is used in 1) Forward Bias 2) Reverse Bias 3) may be forward (or) reverse 4) None 29. A device which converts AC to DC 1) Regulator 2) Transformer 3) Rectifier 4) Oscillator 30. At ‘O’ K (Absolute zero), P-N Junction diode acts as 1) Insulator 2) conductor 3) semi conductor 4) super conducted
Page No. 98
01 - 10 1 2 3 2 4 3 4 2 4 1 11 - 20 2 1 1 4 1 1 3 3 3 4 21 - 30 2 3 3 1 3 1 3 2 3 1
KEY
Page No. 99
ZENERDIODE SYNOPSIS
1. Zener diode:- A heavily doped p-n junction diode that is operated in the reverse bias break down region , with out being damaged. 2. BreakdownVoltage or zener voltage:- The value of reverse voltage at which the reverse current increases abruptly is known as zener voltage. This breakdown caused by two mechanisms 1. Zener break down 2. Avalanche break down 3. Reason for zener break down:- As the zener diode is heavily doped the depletion layer will be very thin . As such even for small
reverse bias voltage the electric field across the junction will be very large
cmv
dvE /105 and
will be capable of breaking the covalent bonds and releasing electrons and holes. The Voltage at which break down occurs is less than 6v . Avalanche breakdown:- diode is lightly doped , width of the depletion region is wider, low intensity electric field at junction and reverse voltage is more than 6v. 4. Available zener voltages(Range) :- V4.2 to V200 5. The break down voltage depends on doping concentration 6. The main feature that distinguishes a zener diode from ordinary p-n junction diode is that it is heavily dopped and has a sharp break down voltage 7. Silicon is preferred to Germanium in the construction of zener diode because 1). Si has higher thermal conductivity and consequently higher power dissipation capacity than Ge 2). It has low reverse saturation current. 3). Si devices are less sensitive to temperature variations. 8. In forward bias, zener diode behaves just as ordinary diode 9. The series resistance R limits the current through the zener diode with in the maximum allowable current through the zener diode and hence protects the diode from the damage. 10. Uses of zener diode 1. As voltage regulators, clippers and square wave generators. 2. As fixed reference voltage for biasing. 3. For protecting meters against damage from accidental applications of high voltage. 4. Power dissipated in zener diode is zizVzP and this can never be more than the specified wattage rating of the zener diode. 11. In V-i characteristic curve, there is a point
Where further increase of negative voltage results in a sharp increase in current. The reverse bias potential that results in this sudden change is called Zener potential zV
12. The maximum reverse bias potential that can be applied before commencement of zener region is called peak Inverse Voltage.
DI
iV
I LI
ZI
+
_
LR
R
zVV 0
DV ZV
o
Page No. 100
MULTIPLE CHOICE QUESTIONS 1. The main cause of Zener break down is 1. Low doping 2 .High doping 3. Ge being base semiconductor 4. Production of electron –hole pairs due to thermal excitation 2. The main cause of avalanche breakdown is 1. High doping 2. Recombination of electrons and holes 3. Collision ionization 4. Shifting of electrons from conduction band to valance band 3. The current flow in a zener diode is mainly due to 1. Thermally generated charge carriers 2. Minority charge carriers 3. Collision generated charge carries 4. Holes 4. Zener diode is used as a 1. Amplifier 2. Rectifier 3. Oscillator 4. Voltage Regulator 5. Zener breakdown will occur at low voltages if, 1. Impurity level is low 2. Impurity level is High 3. Impurity is less in‘n’side 4.Impurity is less in‘p’side 6. Zener diode is used as Voltage Regulator when connected in 1. Forward bias 2. Reverse bias 3. Parallel to load 4. Series with the load 7. In the Zener diode circuit, total current is I , load current is LI ,Zener current is ZI then 1. LZ II 2. LZ III 3. LZ III 4. III LZ 8. In a circuit zener diode is connected parallel to a load resistance of k5 in reverse bias . A k15 resistance is connected in series current through the zener diode is [input voltage 100v and vVz 10 ] 1. mA2 2. mA4 3. mA6 4. mA9 9. In above problem P.D across series resistance is 1. v100 2. v90 3. v10 4. v80 10. In above problem vVz 110 the current passing through the zener diode is 1. 0 2. mA2 3. mA4 4. mA6
KEY
1-10 4 3 3 4 2 2 3 2 2 1
P ageN o101 P age
p a p w e r e f e
Input characterestics
Fig.11.19 p-n-p Transistor (C.E) inputcharacteristicsThe output voltage ceV is fixed (say at zero
volts). The input voltage beV is changed insteps (say 0.1V) upto 1 volt and the corre-sponding base current bI is noted down. Thisprocess is repeated for different values (say10V, 20V etc.) of ceV .Output characterestics
1) Saturation :- In this region the collectorcurrent becomes almost independent of thebase current. This happens when bothjunctionis are forward biased.2) Cut off region:- In this region the collec-tor curent is almot zero. Thin happens whenboth the junctionis are reverse biased.3) Active region:- In this region collector
current cI is many times greater than base
current bI . A small change in input current
bI produces a large change in the output
current cI . This happens when emitter
junction is forward biased and collectorjunction is reverse biased. The transistorworks as an amplifier when operated in theactive region.
Hybrid or ‘h’ parameters of transistor:
1) Input impedance, (h ie) :
b
beie
ce I
VhI
constant.
Unit for hie is Ohm (same as resistance)2) Reverse voltage ratio, (hre)
b
bere
ce I
VhV
constant.
It is a dimensionless constant.3) Forward current ratio, (hfe)
ceVb
cfe I
Ih
constant
It is a dimensionless constant.
4) Output admittance, oeh
bICE
Coe V
Ih
Unit for oeh is siemen. 1) In put characterstics are drawn between VBE
verses IB at constant VCE2) Out put characterstics are drawn betweenVCE verses IC at constant IB
Advantages of a semi conductor device overvaccum tube device:a) Filaments are not required in a transistor,where as it is a must in a vaccum tube triode.b) Very low operating voltages (0-8V) are re-quired for a transistor, where as very high oper-ating voltages are required for vaccum tube tri-ode (0-300V)c) Creation of vaccum is not necessary in a tran-sistor where as it a must in a vaccum tube triode.d) A transistor is more compact, economical andhas a long life when compared to a vaccum tubetriode.
I i i i i
i i i i i
Ic
Page No. 102
Using multimeter – Identification of base of a transistor Synopsis
1. Transistor is a device used to send the signal from low resistance path to high
resistance path
2. It is a 3 terminal device
3. Physically the middle terminal of the transistor is known as base
4. The base region is thin in size in comparison to other two regions
5. Doping of the base region is very low
6. In open circuit when the resistance is measured with multimeter between two
terminals if the resistance is low the terminals are emmiter and base
7. If the resistance is high the terminal are base and collector
8. Inclosed circuit if the current to the terminal is very low that is the base
9. In the transistor equation E B CI I I
10. Base current is always measured micro amperes E C BI I I
11. For N-P-N transistor base is positive with respect other two terminals
12. For P-N-P transistor base is negative with respect other two terminals
Questions
1. The size of the base region
a) Thin b) wide c) long d) all the above
2. Doping in the base region
a) low b) high c) moderate d) Infinite
3. In the transistor currents base current BI a) smaller than EI and CI b) greater than EI and CI c) B EI I d) B CI I
4. In the circuit emitter base junction is always
a) Forward biased b) Reverse biased
c) Un-biased d) All above
5. In the circuit the collector-base junction in always a) Forward biased b) Reverse biased
c) Un biased d) All above * * * * * *
Page No. 103
Page No103
Using multimeter – Identification of base of a transistor
Synopsis 1. Transistor is a device used to send the signal from low resistance path to high
resistance path 2. It is a 3 terminal device 3. Physically the middle terminal of the transistor is known as base 4. The base region is thin in size in comparison to other two regions 5. Doping of the base region is very low 6. In open circuit when the resistance is measured with multimeter between two
terminals if the resistance is low the terminals are emmiter and base 7. If the resistance is high the terminal are base and collector 8. Inclosed circuit if the current to the terminal is very low that is the base 9. In the transistor equation E B CI I I 10. Base current is always measured micro amperes E C BI I I 11. For N-P-N transistor base is positive with respect other two terminals 12. For P-N-P transistor base is negative with respect other two terminals
Questions
1. The size of the base region a) Thin b) wide c) long d) all the above
2. Doping in the base region a) low b) high c) moderate d) Infinite
3. In the transistor currents base current BI a) smaller than EI and CI b) greater than EI and CI c) B EI I d) B CI I
4. In the circuit emmitor base junction is always a) Forward biased b) Reverse biased c) Un-biased d) All above
5. In the circuit the collector-base junction in always a) Forward biased b) Reverse biased c) Un biased d) All above
* * * * * *
Page No 104
TRANSISTOR 1) Transistors are three terminal solid state devices just like triode. Transistors can be npn
or pnp depending on nature of the components used. 2) A pnp transistor consists of silicon (or Germanium) bar crystal in which a layer of N-
type silicon or Ge is sandwiched between two layers of P-type silicon. 3) An npn transistor consists of a layer of P-type between two layers of N-type material. 4) The three layers of a transistor forms three components namely Emitter (E), Base (B),
& collector (C). 5) The size and impurity concentration is different for these components and is as given
below. Emitter (E) : It supplies charge carriers (electrons in npn transistor and holes in pnp transistor) It has high density of impurity concentration, i.e, it is highly doped. Its size is moderate when compared to other sections. It is always forward biased. Collector (C) : It is moderately doped, its size is large when compared to other sections, to dissipate more heat. It is always connected in reverse bias. Base (B) : It is middle region. It is thin and is very lightly doped. How to check whether the given transistor is pnp or npn using multimetre A transistor can be assumed to consist of two back to back connected PN junction diodes. So by using the concept that a PN junction conducts current only when it is
Emitter (E)
Base (B)
Collector (C) n n p
Emitter (E)
Base (B)
Collector (C) p p n
Base (B)
Emitter (E)
Collector (C)
n
p
n
Base (B)
Emitter (E)
Collector (C)
p
n
p
Page No 105
forward biased, and it blocks the current in reverse bias, we can check whether the given transistor is npn or pnp. The forward bias resistance for a transistor is low. Usually it is around 900 ohms or 1k ohm. And reverse bias resistance of a transistor in very high, its value is around Mega ohm’s (M ). First adjust the multimetre in resistance mode & choose two terminals among 3 terminals of a transistor to be connected to the two prober of multimetre, so that multimetre shows low resistance (around 1k ohm). Then those terminals selected are said to be forward biased and terminal connected to red probe is p-type and that connected to black probe is n-type. Now reconnect the black probe to third terminal of transistor and check the resistance, we will study two cases. Case (1) : If the resistance is low (around 1k ohm) then again the connection is said to be forward biased, then the terminal connected to black probe is n-type and hence we can say that the given transistor is npn. Case (2) : If the resistance is high (around M ohm’s) then the terminal connected to black probe in also said to be p-type, & we can say that the given transistor is pnp. OBJECTIVE TYPE QUESTIONS 1) While checking the nature of components of a transistor, if multimetre shows low
resistance then nature of the components connected to the +Ve and Ve terminals of multimetre respectively are
1) p-type, n-type 2) n-type, p-type 3) n-type, n-type 4) p-type, p-type 2) While checking the nature of components of a transistor, if multimetre shows high
resistance, then nature of the components connected to the Ve & Ve terminals of multimetre respectively can be
1) p-type, n-type 2) n-type, p-type 3) n-type, n-type 4) p-type, p-type 3) A transistor is essentially 1) A current operated device 2) power driven device 3) A voltage operated device 4) Resistance operated device. 4) In a transistor, the base is 1) A conductor of low resistance 2) A conductor of high resistance 3) An insulator 4) an extrinsic semiconductor. 5) The main difference between voltage & power amplifier is that 1) power amplifier handles current 2) power amplifier handles large voltage 3) power amplifier handles large power 4) None of the above. 6) Negative feed back 1) Increases stability 2) Decreases stability 3) Producer oscillation 4) Stops current in the tube.
Page No 106
KEY 1) 1 2) 2,3,4 3) 1 4) 4 5) 2 6) 1 QUESTIONS AND ANSWERS 1) Why are transistors and semiconductor diodes generally metal – capped ? Ans All transistors and semiconductor diodes are photosensitive. To prevent photo electron
emission they are generally metal – capped. 2) Is the n-type semiconductor positively charged (or) negatively charged ? Ans It is electrically neutral. 3) All intrinsic semiconductors are insulators at absolute zero. Is this true (or) false ? Ans This is true, because all electrons are tightly bound at absolute zero. 4) Two p-n junction diodes are joined in series with n-joined to n, will the combination
conduct if the first in forward biased ? Ans No, because the second will be reverse biased. 5) Can holes be treated as antiparticles of electron ? Ans No, because an electron is a material particle but not a hole. In fact the antiparticle of
the electron is the positron. 6) Explain how a reverse biased current is formed in a semiconductor diode ? Ans The reverse biased current is formed by the movements of thermal electrons & holes.
Page No 107
TO CHECK THE CORRECTNESS OF A DIODE AND
TRANSISTOR USING MULTIMETER
Multimeters
Multimeters are very useful test instruments. By operating a multi-position switch on the meter they can be quickly and easily set to be a voltmeter, an ammeter or an ohmmeter. They have several settings (called 'ranges') for each type of meter and the choice of AC or DC. Some multimeters have additional features such as transistor testing and ranges for measuring capacitance and frequency.
Digital multimeters
All digital meters contain a battery to power the display so they use virtually no power from the circuit under test. This means that on their DC voltage ranges they have a very high resistance (usually called input impedance) of 1M or more, usually 10M , and they are very unlikely to affect the circuit under test. Typical ranges for digital multimeters like the one illustrated: (the values given are the maximum reading on each range)
DC Voltage: 200mV, 2000mV, 20V, 200V, 600V.
AC Voltage: 200V, 600V. DC Current: 200µA, 2000µA, 20mA, 200mA,
10A*. *The 10A range is usually unfused and connected via a special socket.
AC Current: None. (You are unlikely to need to measure this). Resistance: 200 , 2000 , 20k , 200k , 2000k , Diode Test.
Digital meters have a special diode test setting because their resistance ranges cannot be used to test diodes and other semiconductors.
Liquid-Crystal Display
(LCD)
Digital Multimeter
Photograph © Rapid Electronics
Page No 108
Analogue multimeters
Analogue meters take a little power from the circuit under test to operate their pointer. They must have a high sensitivity of at least 20k /V or they may upset the circuit under test and give an incorrect reading. See the section below on sensitivity for more details. Batteries inside the meter provide power for the resistance ranges, they will last several years but you should avoid leaving the meter set to a resistance range in case the leads touch accidentally and run the battery flat. Typical ranges for analogue multimeters like the one illustrated: (the voltage and current values given are the maximum reading on each range)
DC Voltage: 0.5V, 2.5V, 10V, 50V, 250V, 1000V. AC Voltage: 10V, 50V, 250V, 1000V. DC Current: 50µA, 2.5mA, 25mA, 250mA.
A high current range is often missing from this type of meter. AC Current: None. (You are unlikely to need to measure this). Resistance: 20 , 200 , 2k , 20k , 200k .
These resistance values are in the middle of the scale for each range.
It is a good idea to leave an analogue multimeter set to a DC voltage range such as 10V when not in use. It is less likely to be damaged by careless use on this range, and there is a good chance that it will be the range you need to use next anyway!
Q1. How do you test a diode with a multimeter?
The techniques used for each type of meter are very different so they are treated separately: Q2. How do you test a diode with a DIGITAL multimeter ?
Digital multimeters have a special setting for testing a diode, usually labelled with the diode symbol.
Connect the red (+) lead to the anode and the black (-) to the cathode. The diode should conduct and the meter will display a value (usually the voltage across the diode in mV, 1000mV = 1V).
Reverse the connections. The diode should NOT conduct this way so the meter will display "off the scale" (usually blank except for a 1 on the left).
Analogue Multimeter
Photograph © Rapid Electronics
Diodes
a = anode k = cathode
Page No 109
Q3. How do you test a diode with an ANALOGUE multimeter ?
Set the analogue multimeter to a low value resistance range such as × 10. It is essential to note that the polarity of analogue multimeter leads is reversed on
the resistance ranges, so the black lead is positive (+) and the red lead is negative (-)! This is unfortunate, but it is due to the way the meter works.
Connect the black (+) lead to anode and the red (-) to the cathode. The diode should conduct and the meter will display a low resistance (the exact value is not relevant).
Reverse the connections. The diode should NOT conduct this way so the meter will show infinite resistance (on the left of the scale).
Q4d. How do you test a transistor with a multimeter?
Set a digital multimeter to diode test and an analogue multimeter to a low resistance range such as × 10, as described above for testing a diode. Test each pair of leads both ways (six tests in total):
The base-emitter (BE) junction should behave like a diode and conduct one way only.
The base-collector (BC) junction should behave like a diode and conduct one way only.
The collector-emitter (CE) should not conduct either way.
The diagram shows how the junctions behave in an NPN transistor. The diodes are reversed in a PNP transistor but the same test procedure can be used.
Testing an NPN transistor
Page No 110
M ULTIM ETER (ANALOGUE MULTIMETER)
Practical verification with analogue multimeter :_ 1. Multimeter is used to (i) Distinguish between p – n – p and n – p – n transistors
(ii) Check whether given diode is in working condition or not.
(iii) Check the working condition of transistor
(iv) Identify the base of a transistor
(v) Identify the terminals of an I.C.
2. To Check whether diode is in working order (or) not, the following procedure is adopted (i) The multimeter must be set as resistance meter
(ii) note the deflection by the touching its probes to two ends of the given diode.
(iii) Now, by reversing the probes, deflection must be observed.
(iv) If the deflection in multimeter is large in both the cases (or) small in both the cases then diode is
spoiled.
(v) If the deflection is large in one case and small in the other case, the diode is in working order.
3. To check whether a transistor is working or spoiled, the following procedure is carried out.
(i) The resistance of transistor between emitter and base should be measured and deflection must be
noted.
(ii) By reversing the probes deflection must be observed.
(iii) The procedure above should be repeated by touching the base and collector.
(iv) The resistance in the both cases is low (i.e. Large deflection), transistor is spoiled.
(v) If the resistance is low in one direction (i.e. large deflection) and high in the other direction (i.e small
deflection), the transistor is in working order.
MULTIPLE CHOICE QUESTIONS 1. If the Resistance of a diode is , low then multimeter deflection is a) Less b) More c) deflection does not depend d) less (or) more 2. How do we find whether a diode is working or not by using a multimeter. a) Forward bias resistance is more b) Reverse bias resistance is less c) both the cases resistance is more d) Forward bias resistance is less and reverse bias resistance more 3. If the diode is spoiled then multimeter shows a) Large deflection in one case b) small deflection is reverse case c) Deflection is large in one case and small in reverse case d) Large (or) small in both cases
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4. If a semi conductor device has more than three legs (or) pins, it is a a) transistor b) junction diode c) zenor diode d) integrated circuit 5. Multimeter black wire is connected to base and red wire is connected to either of the other legs, if the resistance is Low, it is a a) p – n – p transistor b) n – p – n transistor c) p – n – p (or) n – p – n transistor d) both 6. How to observe unidirectional flow of current in case of junction diode using a multimeter a) In the forward bias more deflection in multimeter b) In the reverse bias, less defelction in multimeter c) current flows when it is forward biased and does not flow when it is reverse biased d) none 7. A multimeter connected to L.E.D then deflection before and after reversing probes is a) both cases same deflection b) large deflection and then falls to zero c) almost no deflection d) very small in one case and very large in reverse case or vice versa and emits light 8. Connect extreme terminals of a transistor with two probes of multimeter. If it show less deflection (or)
more resistance then central terminal is a) Base b) emitter c) collector d) any one 9. Multimter is used to a) check whether a given diode (or) transistor is in working order b) identify the base of a transistor and terminals of an IC. c) distinguish between p – n – p and n –p –n transistor d) All the above 10. Multimeter ends are connected to emitter base of working order transistor then multimeter shows a) Large deflection in one direction and small deflection in other direction b) Both cases more deflection c) both cases small deflection d) all the above 11. Black wire terminal is connected to base and red wire terminal is connected to either of the other legs of
transistor, multimeter deflection is low then it is a) p – n – p transistor b) n – p – n transistor c) p – n – p and n – p – n d) none
KEY 1 – 10 b d d d a c d a d a 11-20 b