1

Units and Measurement

1. Physical quantity, units and dimensions

Physical quantity: A quantity that can be measured by instrument, clearly defined and has proper units is called physical quantity. Physical quantities are classified as fundamental and derived quantities.

Fundamental units: The physical quantity which does not depend on any other physical quantity is called a fundamental physical quantity such as length; mass and time are called fundamental units.

Derived units: The units that can be obtained from fundamental units are called derived units.

System of units:

There are three systems of units.

Name of systemFundamental unit of

LengthMassTime

F.P.S.FootPoundSecond

C.G.S.CentimetreGramSecond

M.K.S. (S.I.)MeterKilogramSecond

In physics SI system is based on seven fundamental and two supplementary units.

(i) Fundamental units:

S.No.Basic PhysicsFundamental UnitSymbol

1.Masskilogramkg

2.Lengthmeterm

3.Timeseconds

4.Electric current ampere A

5.TemperatureKelvin or CelsiusK or (C

6.Luminous intensitycandelaCd

7.Quantity of mattermoleMol

(ii) Supplementary units:

S.No.Supplementary Physical QuantitiesSupplementary unitSymbol

1.Plane angleRadianrad

2.Solid AngleSteradiansr

Unit & Dimensions & Vectors

DEFINITIONS OF BASE UNITS:

(i) Meter:

The currently accepted definition of meter is the length of path travelled by light in vacuum in 1/299,792,458th second.

(ii) Kilogram:

Kilogram is the fundamental unit of mass. It is defined as the mass of a specific cylinder of platinum - iridium kept at the International Bureau of Weights and Measures in Paris.

.

(iii) Second:

Second is the fundamental unit of time. It is defined as 86,400th part of a mean solar day. Second is accurately measured by an atomic clock.

(iv) Coulomb:

Coulomb is the fundamental unit of charge. It is defined as the charge required to obtain 9(109 Newton of force between two equal charges separated at a distance of one meter in vacuum.

(v) Candle:

Candle is the fundamental unit of luminous intensity. It is defined as luminous intensity observed from a source of monochromatic light of frequency 540(1012 Hz, that has an intensity of 1/683 watt per steradian.

(vi) Kelvin:

Kelvin is the fundamental unit of temperature. It has value of zero where the molecular activity of gases cease.

(vii) Mole:

Mole is the fundamental unit of quantity of matter. It is defined as amount of substance of a system that contains as many elementary particle as there are in 0.012 kg of carbon-12 (C-12).

1.1 BASIC PHYSICAL QUANTITIES

PHYSICAL QUANTITYSYMBOL

DIMENSION

MEASUREMENT UNIT

UNIT

LengthsLMeterm

MassM MKilogramKg

TimetTSecondSec

Electric chargeqQCoulombC

luminous intensityICCandelaCd

TemperatureTKKelvinoK

Angle

(noneRadianNone

Mechanical Physical Quantities (derived)

PHYSICAL QUANTITY EQUATIONSYMBOLDIMENSIONMEASURMENT (in SI)UNIT

AreaAL2square meterm2

Volume

VL3cubic meter m3

velocity

vL/Tmeter per secondm/sec

angular velocity

(1/Tradians per second1/sec

acceleration

aL/T2meter per square secondm/sec2

angular acceleration(1/T2radians per square

second 1/sec2

Force

FML/T2 NewtonKg m/sec2

Energy

EML2/T2 JouleKg m2/sec2

Work

W ML2/T2JouleKg m2/sec2

Heat

Q ML2/T2JouleKg m2/sec2

Torque

(ML2/T2Newton meterKg m2/sec2

Power

PML2/T3watt or joule/secKg m2/sec3

Density

D or (M/L3kilogram percubic meterKg/m3

pressure

P ML-1/T2Newton per square meterKg m-1/sec2

impulse

pML/TNewton secondKg m/sec

Inertia

IML2Kilogram square meter

Kg m2

luminous

flux(

Clumen (4Pi candle for point source)cd sr

illumination

EC/L2lumen persquare metercd sr/m2

entropy

S ML2/T2Kjoule per degreeKg m2/sec2K

Volume

rate of flowQL3/Tcubic meter

per secondm3/sec

kinematic

viscosity(L2/Tsquare meterm2/sec

per second

dynamic

viscosity(

M/LT

Newton second

per square meterKg/m sec

specific

weight( M/L2 T2Newtonper cubic meterKg m-2/sec2

Electrical Physical Quantities (derived)

Electric

currentIQ/TAmpereC/sec

emf, voltage,

potentialEML2 /T2 QVoltKg m2/sec2C

resistance or

impedance R

ML2 /TQ2

ohm

Kgm2 /secC2

Electric

conductivity(TQ2 /M2L2mho

secC2/Kg m3

capacitance

CT2 Q2 /ML2Faradsec2C2/Kgm2

inductance

LML2 /Q2HenryKg m2 /C2

Current density

J

Q/TL2

ampere per

square meterC/sec m2

Charge density(Q/L3coulomb per cubic meterC/m3

magnetic flux,

Magnetic inductionBM/TQweber per

square meterKg/sec C

magnetic

intensityHQ/LTampere per meterC/m sec

magnetic vector

potentialAML/TQweber/meterKg m/sec C

Electric

field intensityEML/T2 Qvolt/meter or

newton/coulombKg m/sec2 C

Electric displacementDQ/L2coulomb per square meterC/m2

permeability

(ML/Q2henry per meterKg m/C2

permittivity,(T2Q2/ML3farad per metersec2C2/Kgm3

dielectric constant K M0L0T0 None None

frequency

f or (1/THertzsec-1

angular frequency(1/Tradians per second sec-1

Wave length

(LMetersM

2. Applications of Dimensional analysis

(i) To find the unit of a physical quantity

Example-1 G = [M-1L3T-2]. Its SI unit is m3kg-1s-2 or Nm2kg-2.

(ii)To convert a physical quantity from one system of units to another system of

units

n1u1 = n2u2 (1)

( Where ni and ui are numerical constant unit and dimension in a particular system)

Example-2Let us convert value of g (i.e. 9.8 m/s2) from SI system to CGS system

From eq. no. 1 [ n1u1]in SI = [n2u2]in CGS

[n2]CGS =

= n1 (

= 9.8 m/sec2 (

= 9.8 (

= 980

(iii) To check the correctness of a given physical relation

Based on principle of homogeneity, the dimensions on two sides must be same for a given relation.

Example-3 Check dimensionally

Therefore,

If dimensions are same on both sides then the relation is dimensionally correct otherwise incorrect.

(iv)To derive a relation

Example-4 Derive Plancks length in terms of G, c and h, where G is gravitation constant, c velocity of light and h is plank constant.

L= f(G, c, h), L = KGxcyhz

[L] = [M-1L3T2]x [LT-2]y [ML2T-1]z

-x + y = 0, 3x + y + 2z = 1 and 2x y z = 0

Thus, L =

If K = 1 then L 10-35 m.

The importance of Planks length is yet to be established.

Limitations of dimensional Analysis:

(i) The dimensional analysis cannot be applied to derive relations other than product of power functions, for example, s = ut + at2 or y = y0 cos (t and so on, cannot be derived directly.

(ii) The dimensional analysis cannot be applied to derive those relations that involve more than 3 unknowns, however, we can use them to check the correctness of a relation even if variables are more than 3.

(iii) Even if a physical quantity depends upon 3 quantities, out of which two have same dimension then dimensional analysis cannot be applied to derive such a formula but can be used to check the relation.

(iv) Numerical constants, trigonometric ratios and ratios which are dimensionless cannot be derived.

Physical quantities having same dimensions may not be the same. For example [ML2T-2] is a dimensional relation for torque as well as work or energy.

3.

4. Conversion FACTORS

(i) 1 A.U = 1.496(1011m

(ii) 1X-ray unit = 10-13m

(iii) 1foot = 30.48 cm

(iv) 1Chandra Shekhar limit (CSL) = 1.4 times the mass of sun

(v) 1 metric Ton = 1000kg

(vi) 1pound = 0.4537kg

(vii) 1 atomic mass unit (a.m.u) = 1.67 ( 10-27kg

(viii) 1shake = 10-8kg

(ix) 1 year = 365.25d = 3.156(107s

(x) 1 carat = 200mg

(xi) 1 bar = 0.1 M Pa = 105Pa

(xii) 1curie = 3.7(1010s-1(xiii) 1 roentgen = 2.58 ( 10-4 C/kg

(xiv) 1quintal = 100kg

(xv) 1barn = 10-28m2(xvi) 1standard atmospheric pressure = 1.013(105 Pa or N/m2(xvii) 1mm of Hg = 133N/m2(xviii) 1horse power = 746w

(xix) Gas constant, R = 8.36j/mol k = 8.36(10-7erg/mol k = 2cal/mol

(xx) 1 Weber = 108 maxwell

(xxi) 1 tesla = 1wb/m2 = 104 gauss

(xxii) 1amp turn/meter = 4((10-3 oersted

(xxiii) 1electron volt (eV) = 1.6 ( 10-19J

(xxiv) 1calorie = 4.19J

(xxv) 1watt-hour = 3.6 (103J

Example-5The density of water is equal to

Solution:Ideally speaking, the examiner should specify the temperature in this question. This is because the density of water varies with temperature. It is maximum (103 kg m-3) at 4(C.

5.

6.

(a)

(b)

(c)

(i) (ii) (iii)

Example-6

One atmospheric pressure is equal to

Solution:1 atmospheric pressure = 76 cm of Hg

= 76 ( 13.6 ( 981 dyne cm-2

= 1.01 ( 106 dyne cm-2 = 1.01 ( 105 N m-2

Example-7 If C is the capacity and R is the resistance, then the dimensional formula of is

Solution:(B)CR is time constant of CR circuit.

7. Experiments based on vernier CALIPERS & screw gaugeA meter scale can measure accurately up to onetenth part of one cm. Its least measurement 0.1 cm, is called least count of scale. There is limitation of meter scale that the meter scale cannot measure the value less than 0.1 cm. For greater accuracy measurement we have devices such as,

(i) Vernier Callipers

(ii) Screw Gauge

(i) Vernier Callipers: A vernier callipers provides with an auxiliary (or vernier) scale in addition to the main scale. The vernier scale can slide along the main scale. The vernier scale is so graduated (or marked) that the length of total number of divisions on it is smaller by length of one division on main scale.

The least count of vernier scale is calculated by using the following formula

Least count of vernier scale (or vernier constant)=

or

Least count (vernier constant) = 1 M.S.D.(Main scale division) 1 V.S.D.(vernier scale division).

Example-8 If N division of vernier coincides with (N 1) division of main scale. Given one main scale division is equal to a unit, find the least count of the vernier.

Solution:Vernier constant = 1MSD 1VSD = MSD = ,

Generally, the value of 1 main scale division on vernier callipers is 0.1 cm and there are 10 divisions on the vernier scale, i.e., x = 0.1 cm and n = 10.

( Least count of vernier callipers = = 0.01 cm.

Zero error of vernier callipers: If the zero marking of main scale and vernier callipers does not coincide, necessary correction has to be made for this error which is known as zero error of the instrument. If the zero error of the vernier scale is to the right of the zero of the main scale the zero error is said to be positive & the correction will be negative otherwise vice versa.

(ii) Screw gauge:Least count =

Zero error of screw gauge: In a perfect instrument the zeros of the main scale and circular scale coincides with each other, In this condition screw gauge has zero-error, otherwise the instrument is said to have zero-error which is equal to the cap reading with the gap closed. This error is positive when zero line or reference line of the cap lies above the line of graduation and corresponding corrections will be just opposite otherwise vice-versa. Example-9 What will be the measurement of following screw gauge position?

Solution: Reading = Main scale reading + Number of circular scale division(or screw gauge reading)(least count

3mm + 45( 0.01mm = 3.45 mm

Example-10 What will be the measurement of following screw gauge position?

Solution: Reading = Main scale reading + Number of circular scale division(or screw gauge reading)(least count

5.5mm + 16( 0.01mm = 5.66 mm

OBJECTIVE 1.

Which of the following sets cannot enter into the list of fundamental quantities in any system of units?

(A)length, time and mass(B)mass, time and velocity

(C)length, time and velocity(D)gravitational constant

Solution:(C)Since velocity is derivable from length and time therefore it cannot be grouped with length and time as fundamental quantity.

2.Sleman is S.I unit for

(A) Specific-Conductance(B) Inductance

(C) Capacitance(D) Pressure

Solution 2:(A)

3.A science student takes 100 observations in an experiment. Second time he takes 500 observations in the same experiment. By doing so the possible error becomes

(A) 5 times(B) 1/5 times

(C) Unchanged(D) None of these

Solution 3:(B) (1/5 times)

4.The unit of surface energy per unit area may be expressed

(A)Nm2(B) Nm1

(C) Nm(D) Nm2Solution 4:(B) Surface energy per unit

Surface energy per unit area =

5.Density of a liquid is 13.6 gcm3. Its value in SI units is

(A) 136.0kgm3

(B) 13600kgm3

(C) 13.60kgm3

(D) 1.360kgm3

Solution 5:(B) Density = 13.6 g cm3

=

= 13600 kg m3 [ 1 g = 103 kg, 1 cm = 102 m]

6.

If the size of a unit be represented by k and is numerical value as n, then

(A)n ( k(B)n (

(C)n ( k2(D)n (

Solution:(B)Value = nk. Since value is fixed therefore nk = constant.7.The SI unit of the universal gas constant R is

(A) Erg K1 MOL1(B) Watt K-1 MOL-1

(C) Newton-1 MOL-1(D) Jule1 MOL1

Solution 7:(B)

8.The maximum error in the measurement of mass and density of the cube are 3% and 9% respectively. The maximum error in the measurement of length will be

(A) 9%(B) 3%

(C) 4%(D) 2%

Solution 8:(C) Density = =

V =

=

Max. fractional error

Percentage error

% = 3% + 9%

= 4 %

9.

The SI unit of electrochemical equivalent is

(A)kg C(B)C kg-1

(C)kg C-1(D)kg2C-1Solution:(C) According to Faraday's first law of electrolysis, m = ZQ or . So, SI unit of Z is kg C-110.Which of the following has a dimensional constant

(A) Refractive index(B) Passions ratio

(C) Relative velocity (D) Gravitation of constant

Solution 10:(D) All of physical quantity has no dimension except gravitational force so correct.

11. The dimensions of surface tension ( length are

(A)ML0T-2(B)MLT-2

(C)ML-1T2(D)ML2T2Solution 11:(B) ML0T2 ( L= MLT-2

12Pick the odd man out

(A)Weight (B)Thrust

(C)Electromotive force (D)Force

Solution 12:(C)

13Dimension formula for luminous flux is

(A)ML2T-2

(B)ML2T-3

(C)ML2T1

(D)None of these

Solution 13:(D)

14.

If w, x, y and z are mass, length, time and current respectively, then has dimensional formula same as

(A)electric potential(B)capacitance

(C)electric field(D)permittivity

Sol. : (A)

15. MLT-1 ( T-1 are the dimension of

(A)Power

(B)Momentum

(C)Force

(D)Couple

Solution 15:(C)16. The unit of impulse is the same as that of

(A) Moment of force (B) Linear momentum

(C) Rate of change of linear momentum(D) Force

Solution 16:(B) Impulse = Force ( time

= MLT2 ( T

= MLT1

i.e. Dimension of linear momentum

17. The dimensions of capacitance are

(A)M-1L-2TI2(B)M-1L-2T2I-2

(C)ML-2T-2I-1(D)M-1L-2T4I2

Solution 17:(D) C =

C =

=

= M1L2T4I2

18. The dimension of angular momentum ( length are

(A)MLT-1 (B) ML3T-1

(C)ML-1T(D) ML0T-2Solution 18:(B) ML3T1

19. The SI unit of the universal gas constant R is

(A)erg K-1mol-1(B)watt K-1mol-1

(C)newton K-1mol-1(D)joule K-1mol-1

Solution 19:(D)

20. The dimension of planks are the same as those of

(A)energy(B)power

(C)angular frequency (D)angular momentum

Solution 20:(D) E = h(

Planks constant h =

Dimension of (h) =

h = ML2T1

Dimension of angular momentum = ML2T1

21.The volume V of water passing any point of a uniform tube during t seconds is related to the cross-sectional area A of the tube and velocity u of water by the relation

V ( A(u(t(Which one of the following will be true?

(A)( = ( = ((B)( ( ( = (

(C)( = ( ( ((D)( ( ( ( ((Solution 21:(B) V = k. (( u( t(

L3 = k (L2)(. (LT1)(. (T)(

L3 = k . L(2( + () T( + (

2( + ( = 3

( + ( = 0

( = (, 2( + ( = 3

so are can conclude that

( ( ( = (22. Which one of the following relations is dimensionally consistent where h is height to which a liquid of density ( rises in a capillary tube of radius, r, T is the surface ension of the liquid, ( the angle of contact and g the acceleration due to gravity

(A)

(B)

(C)

(D)

Solution 22:(A)

23The dimension of calories are

(A)ML2T-2 (B)MLT-2

(C)ML2T1(D)ML2T-1Solution 23:(A) Calories is unit of energy so dimension of calories is = ML2T2

24. The dimension of potential difference ( length are

(A)ML3T-3I-1(B)MLT-2I-1

(C)ML2T-2I(D)MLT-2I

Solution 24:(A) V =

= ML3T3I1

25.

What is the power of a 100 W bulb in cgs units?

(A)106 erg/s(B)107 erg/s

(C)109 erg/s(D) 1011 erg/s

Solution:(C)100 W = 100 J s-1= 100 ( 107 ergs-1.

26.A quantity X is given by (0l, where (0 is the permittivity of free space, l is the length, is a potential difference and is a interval. The dimensional formula for X is the same as that of

(A)Resistance(B)Charge

(C)Voltage(D)Current

Solution:(D) x = = =

27.Let ((0) denote the dimensional formula for the permittivity of the vacuum, and ((0) that of the permeability of the vacuum. If M = mass, L = length, T = time and A = electric current

(A)[(0] =[M1L3T2A](B)[(0] = [ML3T4A2]

(C)[(0] = [MLT2A2](D)[(0] = [M 1L3T2A]

Solution:(B)

28.The dimensions of are

(A)(A2L3T4M4)(B)(A2T4L3M)

(C)(A0M0L0T0)(D)(AT2L3M1)

Solution:(C)

29.

Density of liquid is 15.7 g cm-3. Its value in the International System of Units is

(A)15.7 kg m-3(B)157 kg m-3

(C)1570 kg m-3(D)15700 kg m-3Solution:(D)15.7 g cm-3 = 15.7 ( 10-3kg(10-2m)-3= 15700 kg m-330.On the basis of dimensional equation, the maximum number of unknown that can be found is

(A)One(B)Two

(C)Three(D)Four

Solution:(C) 31.If v stands for velocity of sound, E is elasticity and d the density, then find x in the equation v =

(A)1(B)

(C)2(D)

Solution:(D) V =

LT1 =

LT1 = L3x + x T2x

1 = 2x

x =

32. The dimension of (0E2 ((0 is permittivity of free space and E is electric field) are

(A)MLT-1(B)ML2T-2

(C)ML-1T-2(D)ML2T-1Solution:(B) Dimension of energy = ML2T233.

A weber is equivalent to

(A)A m-2(B)A m-1

(C)A m2(D) T m-2Solution:(D)1 T = 1 Wb m-2

34.With the usual notation, the equation said to give the angle of banking ( is

(A) Numerical correct only

(B) Dimensionally correct only

(C) Both numerical & dimensionally correct

(D) Neither numerical nor dimensionally correct

Solution:(C)

35.When light travels through glass, the refractive index ( is found to vary with the wavelength ( as ( = A + B/(2 , what is dimension of B ?

(A)L (B)L2

(C)L-1 (D)L-2

Solution:(B) Dimension of wavelength = L2

Dimension of refractive index = M0L0T0

(Dimension of B is L236.The dimension of (0E2 ((0 is permittivity of free space and E is electric field) are

(A) MLT-1(B) ML2T-2

(C) ML-1T-2(D) ML2T-1Solution: (B)

37. A travelling wave in a stretched string is described by the equation y = A sin (kx-(t)

The dimension of k is

(A) M0L-1T0(B) M0L0T0

(C) M0L2T0(D) MLT-1Solution:(A) k =

(Dimension of k = L138.Dimension formula of Stefans constant

(A)ML2T-2(-4(B)ML2T-3(-4

(C)ML0T-3(-4(D)M0LT-1

Solution:(C)39.Of the following quantities, which one has dimensions different from the remaining three

(A)Energy per unit volume

(B)Force per unit area

(C)Product of voltage and charge per unit volume

(D)Angular momentum

Solution:(C)

40. The dimension equation for magnetic flux is

(A)ML2T-2I-1(b) ML2T-2I-2

(C)ML-2T-2I-1(d)ML-2T-2I-2

Solution:(A) ( = B. A

= . A =

= ML2T2I1

41The dimension of the Rydberg constant are

(A) M0L-1T(B) MLT1

(C) M0L1T0(D) ML0T2

Solution: (C) M0L1T042. The pairs of physical quantities which have same dimension are

(A)Reynolds number and coefficient of friction

(B)Latent heat and gravitational potential

(C)Curie and frequency of light wave

(D)Planks constant and torque

Solution:(B)

43.In the relation x = 3yz2, x and z represent the dimensions of capacitance and magnetic induction respectively. What will be the dimension of y

(a) M-3L-2T4Q4(B) M2L-2T4Q4

(c) M-2L-2T4Q4(D) M-3L-2T4QSolution:(A) x = 3yz2

dimension of y =

=

=

= M-3 L-2 T+4 Q4

44.

A sextant is used to measure

(A)area of hill(B)height of an object

(C)breadth of a tower(D)volume of the building.

Solution:(B)The height of a tree, building, tower, hill etc. can be determined with the help of a sextant.45.

What is the dimensional formula of coefficient of linear expansion?

(A)[ML2T-2K-1](B)[MLT-2K-1]

(C)[M0L0TK-1](D)[M0L0T0K-1]

Solution:(D)lt = l0(1 + (t) or

46.

A pressure of 106 dyne cm-2 is equivalent to

(A)105 N m-2(B)104 N m-2

(C)106 N m-2(D)107 N m-2Solution:(A)Remember the conversion factor of 10.47. The Vander Waals equation for a gas is (P+a/v2)(V-b) = nRT. The ratio b/a will have the following dimensional formula

(A) M-1L-2T2 (B) M-1L-1T-1

(C) ML2T2(D) MLT-2Solution:(A) Dimension of (b) = L3

Dimension of (a) = ML5 T-2

( Dimension of =

48. If the time period of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula and if a = 1, c = -1, then b is

(A)1(B)2

(C)3(D)4

Solution:(C) T = (M L-3)a/2 Lb/2 (ML0T-2)c/2

M0L0 T = Ma/2 + c/2 L-3a/2+b/2 T-c

- 3 a + b =0

b = 3 a

b = 3 (1

b = 3

49.If P represents radiation pressure, C represents speed of light and Q represents radiation energy striking a unit area per second, then the non-zero integers x, y and z, such that PxQyCz is dimensionless are

(A) x=1, y=1, z=1 (B)x=1, y=-1, z=1

(C)x=-1,y=1,z=1(D)x=1, y=1, z=-1

Solution:(B) M0L0T0 = Px Qy Cz

= (M L-1 T-2)x (ML2T-2)y (LT-1)z

x + y = 0 ( x = -y

-x + 2y +z = 0

-2x 2y z = 0

x = -y

50. In the relation ,P is pressure Z is distance k is Boltzman constant and ( is the temperature. The dimension formula of ( will be

(A)M0L2T0 (B)M1L2T-1

(C)ML0T-1(D)M0L2T1

Solution:(A) is dimension less quantity

dimension of ( =

=

( = MLT2

Dimension of is equal to dimension of pressure P

P =

ML1T2 =

( =

( = M0L2T051. Velocity v, acceleration a and force f are taken as fundamental quantities, then angular momentum will have the dimension

(A) fv2a-2 (B) f2v2a-2

(C) fv3a-2 (D) None of theseSolution:(D) Angular momentum (L) ( vx ay fz

MLT2 =

MLT2 = Mz Lx +y + z Tx 2y 2z

1 = z

1 = x + y + z

2 = x 2y 2z

(z = 1

x + y = 0

( x = y

2 = x + 2x 2

0 = x, y = 0

(Angular momentum(L) = f

52. Fund the unit of acceleration ( time?

(A) ms1(B) ms3

(C) ms+1(D) ms+2

Solution:(A) Acceleration =

53.What is the unit of current ( Resistance.

(A) amps(B) volt

(C) coulomb(D) faradSolution:(B)

54.What will be equivalent energy of 5eV in joule?

(A) 8.0 1022J(B) 8.0 1019J

(C) 8.0 1025J(D) 8.0 1026JSolution:(B)

55.One joule is the equivalent of ?

(A)

(B)

(C)

(D)

Solution:(B)

56.Least count of screw gauge depend on ?

(A)Main scale division

(B)circular scale

(C) no. of circular scale division

(D) Main scale division & no. of circular scale divisionSolution:(D)

57.Least count of vernier calipers depend on?

(A)Main scale division

(B) vernier scale

(C) no. of vernier scale division

(D) Main scale division & no. of vernier scale divisionSolution:(D)

58.Least count of spherometer depend on ?

(A)Main scale division(B)circular scale

(C) no. of circular scale division

(D) Main scale division & no. of circular scale divisionSolution:(D)

59.What is the dimension of angular frequency ( time?

(A)Dimension less(B) sec2

(C) sec3(D) sec+1 Solution:(A)

60.What is the dimension of wave length ( Frequency?

(A)M(B) LT1

(C) T(D) MT1 Solution:(B)

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