2004 iit mains

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PPhhyyssiiccssTime:2 hours

Note: Question number 1 to 10 carries 2 marks each and 11 to 20 carries 4 marks each.

1. A long wire of negligible thickness and mass per unit length is floating in a liquid such that the top surface of liquid dips bya distance y. If the length of base of vessel is 2a , find surfacetension of the liquid. (y < < a)

y

aa

Sol. (2T cos ) = g

T =g

2cos

T =2 2 1 / 2g(a y )

2y

+

ga

2y

.

TT

2. An ideal diatomic gas is enclosed in an insulated chamber attemperature 300K. The chamber is closed by a freely movablemassless piston, whose initial height from the base is 1m. Now thegas is heated such that its temperature becomes 400 K at constant

pressure. Find the new height of the piston from the base.If the gas is compressed to initial position such that no exchange

of heat takes place, find the final temperature of the gas.

1m

Atmosphere

Sol. Process 1 is isobaricT1= 300 K, T2= 400 K

Vconstant

T=

A 1 A h 4h m

300 400 3

= =

Process 2 is adiabatic

1

TV constant

= , ( )

7 21

75 51

53 3

A 4 4

400 T A 1 T 4003 3

= = K.

3. In Searles apparatus diameter of the wire was measured 0.05 cm by screw gauge of least count 0.001 cm.The length of wire was measured 110 cm by meter scale of least count 0.1 cm. An external load of 50 N

was applied. The extension in length of wire was measured 0.125 cm by micrometer of least count0.001 cm. Find the maximum possible error in measurement of youngs modulus.

Sol.( ) ( )

2

2

4F/ D 4FLY

L / L D L

= =

Maximum possible relative error

Solution of IIT JEE MAINS 2004 Physics paper

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( )LY L 2 D 0.1 2 0.001 0.001Y L D L 110 0.050 0.125

= + + = + + Percentage error

Y 1 4100 4

Y 11 5

= + +

= 0.8 + 4 + 0.09 = 4.89 %.

4. Two infinitely large sheets having charge densities 1 and 2respectively (1 > 2) are placed near each other separated bydistance d. A charge Q is placed in between two plates such thatthere is no effect on charge distribution on plates. Now this charge

is moved at an angle of 450with the horizontal towards plate having

charge density 2 by distance a (a < d). Find the work done byelectric field in the process.

1

d

2

A

B

450

a

Sol. E = 1 2

0

( )

2

work done by electric field, W = q E . d = aE2

q = 1 2

0

q( )a2 2

5. An -particle and a proton are accelerated from rest through same potential difference and both enter intoa uniform perpendicular magnetic field. Find the ratio of their radii of curvature.

Sol. = 2qVm

rqB

= p

p p

qr m

r m q

= 4 e1 2e = 2 : 1

6. A small ball of radius r is falling in a viscous liquid under gravity. Find the dependency of rate of heatproduced in terms of radius r after the drop attains terminal velocity.

Sol. Rate of heat produced = F.v

T T6 rv .v=

2T

dQ6 r.v

dt=

( ) 2T2

v r g /9

=

5dQ rdt

7. A syringe of diameter D = 8 mm and having a nozzle ofdiameter d = 2 mm is placed horizontally at a height of1.25 m as shown in the figure. An incompressible andnon-viscous liquid is filled in syringe and the piston is

moved at speed of 0.25 m/s. Find the range of liquid jeton the ground.

V=0.25 m/s

h=1.25 m

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Sol. AV = ConstantD2V = d2v

= =

22

2

D 8v V 0.25

2d

= =16 0.25 4m/s

= = = =2h 2 1.25 1x v 4 4 2mg 10 2

8. A light ray is incident on an irregular shaped slab of

refractive index 2 at an angle of 45with the normalon the incline face as shown in the figure. The rayfinally emerges from the curved surface in the medium

of the refractive index = 1.514 and passes throughpoint E. If the radius of curved surface is equal to 0.4 m,find the distance OE correct upto two decimal places.

45

60

E

O

R = 0.4m

1= 1 2= 2 3= 1.514

P

A

B C

D

Q

Sol. Using Snells law1sin 45= 2sin = 30.i.e. ray moves parallel to axis

3 2

OE

= 3 2

( )

R

OE = 6.056 m 6.06 m

450

600

30

E

O

R = 0.4m

1= 1 2= 2 3= 1.514

P

A

B C

D

Q

9. A screw gauge of pitch 1mm has a circular scale divided into 100 divisions. The diameter of a wire is to bemeasured by above said screw gauge. The main scale reading is 1mm and 47

thcircular division coincides

with main scale. Find the curved surface area of wire in true significant figures. (Given the length of wireis equal to 5.6 cm and there is no zero-error in the screw gauge.)

Sol. Least count =1mm

0.01mm.100

=

Diameter = M. S. + No. of division coinciding with main scale Least count.= 1mm + 47 0.01 mm= 1.47 mm = 0.147 cm.

Curved surface area = d=22

0.147 5.67

= 2.6 cm2

10. The age of a rock containing lead and uranium is equal to 1.5 109yrs. The uranium is decaying into leadwith half life equal to 4.5 109yrs. Find the ratio of lead to uranium present in the rock, assuming initiallyno lead was present in the rock. (Given 21/3= 1.259)

Sol.1/ 2t / T 1/ 3

U

O

N 1 1 1

N 2 2 1.259

= = =

U

Pb U

N 1

N N 1.259=

+

Pb

U

N0.259

N= .


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11. An inductor of inductance (L) equal to 35 mH

and resistance (R) equal to 11 are connectedin series to an AC source. The rms voltage ofa.c. source is 220 volts and frequency is 50 Hz.

V

O t

(a) Find the peak value of current in the circuit.

(b) Plot the current (I) vs (t) curve on the given voltage vs (t) curve. (Given =22

7)

Sol. Z = 2 2( L) R +

I0=0

3 2 2

V 220 220Amp

Z (100 35 10 ) (11)= =

+ I

VL

VR

2 2L RV V+

tan =L

R

=

3100 35 10

11

= 1

= 450

I = I0sin (t - )4

= 20 sin (100 t - )4

V

O/4

I

t

12. Two identical blocks A and B are placed on a rough inclined plane ofinclination 45

0. The coefficient of friction between block A and incline is

0.2 and that of between B and incline is 0.3. The initial separation between

the two blocks is 2 m. The two blocks are released from rest, then find

(a) the time after which front faces of both blocks come in same line and

(b) the distance moved by each block for attaining above position.45

0

AB

B=0.3

A=0.2

2 m

Sol. aA= g sin 45 0.2g cos 45=2

4 2 m / s

aB= g sin 45 0.3 g cos 45=27 2 m/s

2

aAB= 0.5 2 m/s2

sAB=2

AB

1a t

2

t2=2 2

40.5 2

=

t = 2 sec.

sB= 2B1 a t 7 2 m2 =

sA=2

A

1a t 8 2

2= m

13. In a photoelectric setup, the radiations from the Balmer series of hydrogen atom are incident on a metalsurface of work function 2eV. The wavelength of incident radiations lies between 450 nm to 700 nm. Findthe maximum kinetic energy of photoelectron emitted.

(Given hc/e = 1242 eV-nm).

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Sol. E = 13.62

1 1 hc 1242

4 en

= =

=2

2

1242 4n

13.6(n 4)

minwhich lies between 450 nm and 700 nm is for transition from n = 4 to n = 2 and is equal to 487.05 nmFor maximum K.E. of photoelectron

min

hc

= K.E.max

K.Emax=

=

13.6 122 0.55eV.

4 16

14. A spherical ball of radius R, is floating in a liquid with half of its volume submerged in the liquid. Now theball is displaced vertically by small distance inside the liquid. Find the frequency of oscillation of ball.

Sol. Restoring force = R2xg (for small x)

m2

2

2

d xR x g

dt

=

2

2

gd x 3x

2 Rdt= , (as

34 Rg mg

3 2

= )

Motion is SHM

x

2=g3

2 R

f =3g1

2 2R.

15. The two batteries A and B, connected in given circuit, have equale.m.f. E and internal resistance r1and r2respectively (r1> r2). The

switch S is closed at t = 0. After long time it was found that terminalpotential difference across the battery A is zero. Find the value of R.

S

RR

R

R

R

R

L

C

E, r1 E, r2

A B

Sol. Since average voltage across capacitor and inductor for D.C. sources will be zero at steady state.

I =eq 1 2

2E

(R r r )+ +=

1 2

2E

3R(r r )

4+ +

. . . (i)

P.D. across the battery A = E Ir1= 0

I = E/r1 . . . (ii)

From (i) and (ii),

R = 1 24(r r )

3

16. A point object is moving with velocity 0.01 m/s on principal axis towards a convex lens of focal length 0.3m. When object is at a distance of 0.4 m from the lens, find(a) rate of change of position of the image, and(b) rate of change of lateral magnification of image.


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Sol.1 1 1

f v u=

2 2

1 dv 1 du0

dt dtv u + =

2

2

dv v du

dt dtu= .. . . (i)

1 1 1

30 v 40=

v =120 cm.

f = 0.3m

O

0.4 m

u0 = 0.01 m/s

m =22

2

dv v v1

du fu

= =

dm 2 v dv1

dt f f dt

=

12 1201 0.09 1.8 s0.3 30

= =

17. An experiment is performed to verify Ohms law using a resister of resistance R = 100, a battery of

variable potential difference, two galvanometers and two resistances of 6 -310 and 10 are given. Drawthe circuit diagram and indicate clearly position of ammeter and voltmeter.

Sol.

G

G

R = 10-3

R = 106

100

Voltmeter

Ammeter

18. A uniform rod of length L, conductivity K is connected from one endto a furnace at temperature T1. The other end of rod is at temperatureT2and is exposed to atmosphere. The temperature of atmosphere is

Ts. The lateral part of rod is insulated. If T2 Ts

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19. A cubical block is floating inside a bath. The temperature of system

is increased by small temperature T. It was found that the depthof submerged portion of cube does not change. Find the relation

between coefficient of linear expansion () of the cube and volumeexpansion of liquid ().

x

Coeff. Of linear

expn. =

Coeff. of volume

expn. =

Sol. At initial temperature for the equilibrium of the block

ALbg = Axg

Lb= x . . . (i)

At final temperature

A= A(1 + 2T) = (1 T)

For the equilibrium of the block

A(1 + 2T)(x) (1 T) = ALb= Ax 1 + 2T T = 1 = 2

20. In a Youngs double slit experiment light consisting of two wavelengths 1= 500 nm and 2= 700 nm isincident normally on the slits. Find the distance from the central maxima where the maximas due to two

wavelengths coincide for the first time after central maxima. (GivenD

1000d

= ) where D is the distance

between the slits and the screen and d is the separation between the slits.

Sol. y1=1

nD

d

y2=2

mD

d

y1= y2 n =7

m5

For the first location, m = 5, n = 7

y = 7 1000 5 10-7= 35 10-4 = 3.5 mm.


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CChheemmiissttrryyTime: 2 hours

Note: Question number 1 to 10 carries 2 marks each and 11 to 20 carries 4 marks each.

1. For the given reactionA + BProductsFollowing data were givenInitial conc. (m/L). Initial conc. (m/L) Initial rate [mL

-1s

-1]

[A] [B]0.1 0.1 0.050.2 0.1 0.10.1 0.2 0.05a) Write the rate equation.b) Calculate the rate constant.

Sol. a) Let the order w.r.t A & B are x any y respectivelyr = K[A]x[B]y

0.05 = K[0.1]x[0.1]y

0.1 = K[0.2]x[0.1]y

or 2 = [2]x

x =10.05 = K[0.1]

x[0.1]

y

0.05 = K[0.1]x[0.2]y

1 = [2]y

y=0b) rate equation = r = K[A] [B]

0

0.1 = K[0.2]

K = 0.5 Sec-1

2. 100 ml of a liquid contained in an isolated container at a pressure of 1 bar. The pressure is steeply increasedto 100 bar. The volume of the liquid is decreased by 1 ml at this constant pressure. Find the H & U.

Sol. H =0, qp= U-WW = PdV

= 1001 atmmL= 10-2KJ = U

3. Draw the shape of XeF4 and OSF4according to VSEPR theory. Show the lone pair of electrons on thecentral atom

Sol.

F

F F

F

Xe(square planar)

(sp3d2)F

O

F

S

F

F

(Trigonal bipyramidal)

(sp3d)

Solution of IIT JEE MAINS 2004 Chemistry Paper

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4. The structure of D-Glucose is as followsCHO

H OH

HOH

H OH

OHH

OH

a) Draw the structure of L Glucose.b) Give the reaction of L Glucose with Tollens reagent.

Sol.

CHO

H OH

HOH

H OH

HOH

OH

(L glu cos e)

H OH

HOH

H OH

HOH

OH

O-

O

3 2Ag (NH )+

5. a) Draw New mann`s projection for the less stable staggered form of butane.b) Relatively less stability of the staggered form is due to

i) Torsional strain.ii) Vander Waals strain.iii) Combination of the above two.

Sol. a) CH3

HH .

H

.CH3

.H

b) Less stability is due to Vander Waals strain

6. Arrange the following oxides in the increasing order of Bronsted basicity.

2 7 3 2 2 3Cl O , BaO, SO , CO , B O

Sol. 2 7 3 2 2 3Cl O SO CO B O BaO< < < 7 (a4. b4. c4)1/7

(1 + a) (1 + b) (1 + c) > 7 (a4. b4. c4)1/7

(1 + a)7(1 + b)7(1 + c)7> 77(a4. b4. c4).

16.

1

ax

2

x c 1bsin , x 0

2 2

1f (x) , x 0

2

e 1 1, 0 x

x 2

+ <

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Sol. Let f (x) = 3x2+ (3 - 2) x - sin x

f (0) = 0, f2

= - ve

f(x) = 6x + 3 - 2- cos xf(x) = 6 + sin x > 0

f(x) is increasing function in 0,2

there is no local maxima of f(x) in 0,2

graph of f(x) always lies below the x-axis

in 0,2

.

f(x) 0 in x 0,2

.

/2

O

y=f(x)

x

y

3x2+ 3x 2x + sinx sinx + 2x ( )3x x 1+

.

18. A =

2a 0 1 a 1 1 f a

1 c b , B 0 d c , U g , V 0

1 d b f g h h 0

= = =

. If there is vector matrix X, such that AX = U has

infinitely many solutions, then prove that BX = V cannot have a unique solution. If afd 0 then prove thatBX = V has no solution.

Sol. AX = U has infinite solutions |A| = 0a 0 1

1 c b

1 d b

= 0 ab = 1 or c = d

and |A1| =

a 0 f

1 c g

1 d h

= 0 g = h; |A2| =

a f 1

1 g b

1 h b

= 0 g = h

|A3| =

f 0 1

g c b

h d b

= 0 g = h, c = d c = d and g = h

BX = V

|B| =

a 1 1

0 d c

f g h

= 0 (since C2and C3are equal) BX = V has no unique solution.

and |B1| =

2a 1 1

0 d c

0 g h

= 0 (since c = d, g = h)

|B2| =

2a a 1

0 0 c

f 0 h

= a2cf = a2df (since c = d)

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|B3| =

2

2

a 1 a

0 d 0 a df

f g 0

=

since if adf 0 then |B2| = |B3| 0. Hence no solution exist.

19. A bag contains 12 red balls and 6 white balls. Six balls are drawn one by one without replacement of whichatleast 4 balls are white. Find the probability that in the next two draws exactly one white ball is drawn.

(leave the answer in terms ofnCr).

Sol. Let P(A) be the probability that atleast 4 white balls have been drawn.

P(A1) be the probability that exactly 4 white balls have been drawn.P(A2) be the probability that exactly 5 white balls have been drawn.

P(A3) be the probability that exactly 6 white balls have been drawn.P(B) be the probability that exactly 1 white ball is drawn from two draws.

P(B/A) =

( ) ( )

( )

3

i i

i 1

3

i

i 1

P A P B / A

P A

=

=

=

12 612 6 10 2 11 11 52 4 1 1 1 1

18 12 18 126 2 6 2

12 6 12 612 6

1 5 0 62 418 18 18

6 6 6

C CC C C C C C. .

C C C C

C C C CC CC C C

+

+ +

=

( )

12 6 10 2 12 6 11 12 4 1 1 1 5 1 1

12 12 6 12 6 12 62 2 4 1 5 0 6

C C C C C C C C

C C C C C C C

+

+ +

20. Two planes P1and P2pass through origin. Two lines L1and L2also passing through origin are such that L1lies on P1but not on P2, L2lies on P2but not on P1. A, B, C are three points other than origin, then prove

that the permutation [A, B, C] of [A, B, C] exists such that(i). A lies on L1, B lies on P1not on L1, C does not lie on P1.

(ii). Alies on L2, Blies on P2not on L2, Cdoes not lie on P2.

Sol. A corresponds to one of A, B, CandB corresponds to one of the remaining of A, B, CandC corresponds to third of A, B, C.Hence six such permutations are possible

eg One of the permutations may A A; B B, C CFrom the given conditions:A lies on L1.B lies on the line of intersection of P1and P2and C lies on the line L2on the plane P2.

Now, Alies on L2C.Blies on the line of intersection of P1and P2BClie on L1on plane P1A.Hence there exist a particular set [A, B, C] which is the permutation of [A, B, C] such that both (i) and(ii) is satisfied. Here [A, B, C] [CBA].


2004 iit mains

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