IITJEE 2008 Physics Paper 2 Code 2 1

8/14/2019 IITJEE 2008 Physics Paper 2 Code 2 1

1/34

IITJEE 2008 PhysicsPaper 2 Code 2

SECTION IStraight Objective TypeThis section contains 9 multiple choice questions. Each question has 4 choices(A), (B) , (C) , (D) out of which ONLY

ONE is correctQ 23: A radioactive sample S1 having an activity of 5 Ci has twice the number of nuclei as another sample S2which has an activity of 10 Ci. The half lives of S1 and S2 can be

(A) 20 years and 5 years, respectively(B) 20 years and 10 years, respectively(C) 10 years each(D) 5 years eachSolution: (A)A1 =.1[N1]A2 =.2[N2]

. A[N] 5 11

. 1 = 12 ==

.2 A2[N1] 10 24

=> 20 & 4 are the answer

Q 24: Consider a system of three chargesq,qand - 2qplaced at points A, B and C respectively, as shown in

33 3

o

the figure. Take O to be the centre of the circle of radius R and angle CAB = 60Figure :

q

(A) The electric field at point O is directed along the negative x-axis

8pd R20

100Percentile Education Private Limited


2/34

(B) The potential energy of the system is Zero2

q

(C) The magnitude of the force between the charges at C and B is54pd R2

0

q

(D) The potential at point O is12pd R20

Solution: (C)

29/3 9

(i) Elective field at 0 is ==

4p tR2 6p t R2oo

L1 Field due to A & B cancel at 0

1 .(9/3)2 (29/3)(9/3) (29/3)(9/3) .

(ii) Potential energy of the system =-.- -4p to . (2R) R1(-29/ 3)(9/3)

(iii) Force between B & C =2

4p to (3R)- 92

=

54p t R2

o

(iv) Potential at 0 is =- 1 ..9/3 + 9/3 +

- 29 / 3.. = 04p to .p RR .Q 25: A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s. Thewavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular timet, the snap-shot of wave isshown in figure. The velocity of point P when its displacement is 5 cm is

Figure:


3/34

03...p3p 3p

(A)J m/s (B) -

J m/s

50 50



4/34

3p 3p

(C) i m/s (D) - i m/s 50 50

Solution: (A)

. =.p 2 - x2

p

2p V 102

.= = 2p

= 2p =p

T . 0.5 100 5Vp = 2 p 102 - 52 = 2p 3m / c

5

p 3

= m/ c

50

Hence +ve y educationQ 26: A parallel plate capacitor C with plates of unit area and separation d isfilled with a liquid of dielectric

d

constant K = 2. The level of liquid is

initially. Suppose the liquid level decreases at a constant speed V, the time

3

constant as a function of time t isFigure :

6e R

0

(A) (B)5d + 3Vt

(15d + 9Vt)eR2d 2 - 3dVt - 9V 2t2



5/34


6/34

6e 0R (15d - 9Vt)eR

(C) (D)5d - 3Vt 2d2 + 3dVt - 9V 2t2

Solution: (A)The two capricious can be considered in series

11 1

=+

CC C

12

dd

= 1 + 2

ktA ktA

1o 2o

(d /3 - vt) (2d /3 + vt)

=+

2tA tA

oo

d /3 - vt + 40 / 3 + 2vt

=

2t A

o

5d /3 + vt = 5d + 3vt2tA 6tA

oo

6t R

Time constant = CR = o

5d + 3vt

Q 27: A vibrating string of certain length L under a tension T resonates with amode corresponding to the firstovertone (third harmonic) of an air column of length 75 cm inside a tube closed atone end. The string also


7/34

generates 4 beats per second when excited along with a tuning fork of frequency n.Now when the tension of thestring is slightly increased the number of beats reduces to 2 per second. Assumingthe velocity of sound in air to be340 m/s, the frequency n of the tuning fork in Hz is

Figure:



8/34

(A) 344 (B) 336 (C) 117.3 (D) 109.3Solution: (B)T increase Vstring . T also increase

Hence, Vstring = n + 4

rd

Vstri0ng = 3Vo (ad ) (3Harmonic)

3c

=

4l

3 340

=

4 0.75

= 340

N = 340 4 = 336 MgQ 28: A light beam is traveling from Region I to Region IV (Refer Figure). Therefractive index in Regions I, II, III, andIV are n0,n0 ,n0 andn0 , respectively. The angle of incidence . for which the beam just missesentering Region

26 8

IV isFigure :

(A) sin -1 ..3 .. (B) sin -1 ..1 .. (C) sin -1 ..

1 .. (D) sin -1 ..1 ... 4 .. 8 .. 4 .. 3 .Solution: (B)

oo oo

no sin.= n


9/34

sin.1 = nsin .2 = nsin 90 ( sin.= const.)

26 8

o

. n sin.= no

8

.. = sin -1()81Q 29: A block (B) is attached to two unstretched springs S1 and S2 with springconstants k and 4k, respectively (seefigure I). The other ends are attached to identical supports M1 and M2 notattached to the walls. The springs andsupports have negligible mass. There is no friction anywhere. The block B isdisplaced towards wall 1 by a small



10/34

distance x (figure II) and released. The block returns and moves a maximumdistance y towards wall 2.

y

Displacements x and y are measured with respect to the equilibrium position of theblock B. The ratio

is

x

Figure :

1 1(A) 4 (B) 2 (C) (D)2 4

Solution: (C)

By conservation of energy

121 2

Kx = (4K )y

22y = 1

.

x2

Note supports A/1 & A/2 are mailers and do not allow stretching . S1 does notexpand when S2 contracts.

Q 30: A glass tube of uniform internal radius (r) has a valve separating the twoidentical ends. Initially, the valve isin a tightly closed position. End 1 has a hemispherical soap bubble of radius r.End 2 has sub-hemispherical soapbubble as shown in figure. Just after opening the valve,

Figure :

(A) air from end 1 flows towards and 2. No change in the volume of the soapbubbles(B) air from end 1 flows towards end 2. Volume of the soap bubble at end 1decreases(C) no change occurs100Percentile Education Private Limited


11/34

(D) air from end 2 flows toward end 1. Volume of the soap bubble at end 1increasesSolution: (B)46

P = P +

10

r

1

46

P = P +

20

r

2

r2 > r1 P0 : atmospheric pressure

. P < P

21

Air from 1 goes to 2And since volume is constant . volume from 1 goes to 2Q 31: A bob of mass M is suspended by a massless string of length L. Thehorizontal velocity V at position A is just

sufficient to make it reach the point B. The angle . at which the speed of the bobis half of that at A, satisfiesFigure :

p ppp 3p 3p

(A) .=(B)(C)


12/34

Solution: (D)By conservation of energy

1 21 2

mv =

mv + mgl(1 - cos. ) ----(1)

o

221 21 2

Also, mg(2l) = mvo - mvtop (given) ----(2)

22

Since the vo is just sufficient

2

mv

top

= T + mg T = 0

gl

. vtop = glFrom (1) equation

1m...v2 - vo2 ... = mgl(1 - cos. )

o

2 . 4 .

3v 2

o

= gl(1 - cos. )

8

vo2 = 8gl(1 - cos. )

3

From (2) equation


13/34

mg(2l) = 1mvo2 - 1m(gl)

22vo = 5gl

. 5gt = 8gt(1 - cos. )3

7

. cos.=83p

.


14/34

SECTION -llReasoning Type

This section contains 4 reasoning type questions. Each question has 4 choices

(A) , (B) , (C) , (D), out of which ONLY ONE is correctQ 32: STATEMENT-1For practical purposes, the earth is used as a reference at zero potential inelectrical circuits.AndSTATEMENT-2The electrical potential of a sphere of radius R with charge Q uniformlydistributed on the surface is given

Q

by .

4pe R

0

(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanationfor STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correctexplanation forSTATEMENT-1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is TrueSolution: (B)Statement 1 is true since earth is used as a reference at zero potential to

measure potential difference which isindependent of the reference

Q

Statement 2 is true since the electric potential of a spherical shell is Butstatement 2 sloes not

4p tR

o

explanation statement 1

Q 33: STATEMENT-1

The sensitivity of a moving coil galvanometer is increased by placing a suitablemagnetic material as a coreinside the coil.andSTATEMENT-2

Soft iron has a high magnetic permeability and cannot be easily magnetized ordemagnetized.


15/34

(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanationfor STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correctexplanation for STATEMENT1100Percentile Education Private Limited


16/34

(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is TrueSolution: Statement 1 is true since the sensitivity of the moving willgalvanometer increases by placing magneticmaterial.

Statement 2 there is ambiguity & zero cases wises

Case 1: cannot be early permanently migrated / demagnetized soft iron has magneticpermeability and can not beeasily permanently magnetized . Hence it is the corrupt explanation of statement1. Answer is a

Case 2: can not be easily magnetized / demagnetized during galvanometersoperation. They are it does not act asa good electro magnet which is false. Answer is C

Q 34: STATEMENT-1For an observer looking out through the window of a fast moving train, the nearbyobjects appear tomove in the opposite direction to the train, while the distant objects appear to

be stationary.

andSTATEMENT-2If the observer and the object are moving at velocities V1 and V2 respectivelywith reference to a

laboratory frame, the velocity of the object with respect to the observer is V2 -V1 .

(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanationfor STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct

explanation for STATEMENT1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is TrueSolution: (B) Statement 1 is true since when we obscure from a train we allcomparing the angular velocity whichis more for nearer objectsStatement 1 is true by definition of relative velocityBut it is wrong explanation of statement 2Q 35: STATEMENT-1It is easier to pull a heavy object than to push it on a level ground.andSTATEMENT-2

The magnitude of frictional force depends on the nature of the two surface incontact.



17/34

(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanationfor STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correctexplanation for STATEMENT1(C) STATEMENT-1 is True, STATEMENT-2 is False(D) STATEMENT-1 is False, STATEMENT-2 is TrueSolution: (B)

Statement 1 is true (by observation & explained also)

11

SinceN>N=>f>f=>itiseasiertopullthantopushStatement 2 is also true but it is not the correct explanation

SECTION -lllLinked Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph , 3 multiple choicequestions have to be answered. Each question has 4 choices (A) , (B) , (C) ,(D),outof which ONLY ONE is correctQ 36-38:The nuclear charge (Ze) is non-uniformly distributed within a nucleus ofradius R. The charge density . (r)

[charge per unit volume] is dependent only on the radial distance r from thecentre of the nucleus as shown infigure. The electric field is only along the radial direction.Figure :



18/34

Q 36: The electric field at r =R is

(A) independent of a (B) directly proportional to a2

(C) directly proportional to a(D) inversely proportional to aQ 37: For a = 0, the value of d (maximum value of . as shown in the figure) is

3Ze 3Ze 4Ze Ze(A)34 Rp(B)3Rp(C)33 Rp(D)33 Rp

Q 38: The electric field within the nucleus is generally observed to be linearlydependent on r. This implies

R 2R

(A) a=0 (B) a =(C) a=R (D) a =23

Solution: (A) The electric field at r = R depends on total nuclear change inside r= R which is correct irrespectiveof P

Solution: (B)

Net charge = Ze =.. dv

=.. Adr (radial symmetry)

=.. (4pr 2 )dv

= 4p. . r 2dr

. r

Also,

+

= 1

dRrd

..= d -R

dr

R . r 3d .


19/34

. Ze = 4p.o .. dr 2 - ..

. R .

.dR3 dR4 .

= 4p. -

.

. 3 R4 .



20/34

4p 3 p 3

dR = dR

12 33Ze

. d =pR3

Solution: (C) We know that for a uniformly charged sphere

r

.

E = 0 = r = R r t3 o

Hence we can conclude a = R for . to be cones

Q 39-41:A uniform thin cylindrical disk of mass M and radius R is attached to twoidentical massless springs ofspring constant k which are fixed to the wall as shown in the figure. The springsare attached to the axle of the disksymmetrically on either side at a distance d from its centre. The axle is masslessand both the springs and the axleare in a horizontal plane. The unstretched length of each spring is L. The disk isinitially at its equilibrium position

with its centre of mass (CM) at a distance L from the wall. The disk rolls withoutslipping with velocity V0 = V0 i .The coefficient of friction is .

Figure :

.Q 39: The net external force acting on the disk when its centre of mass is atdisplacement x with respect to itsequilibrium position is

2kx 4kx

(A) - kx (B) - 2kx (C) - (D) -33

Q 40: The centre of mass of the disk undergoes simple harmonic motion with angularfrequency . equal to

k 2k 2k 4k

(A)(B)(C)(D)MM 3M3M


21/34



22/34

Q 41: The maximum value of V0 for which the disk will roll without slipping is

M M 3M5M

(A) g (B) g (C) g (D) g k 2k k2k

Solution: (D)

By Newton s law:2Kx - f = ma (1)

fr =. I cm (2)

12

I cm = mr (3)

2

a = r . (4)1a

. fr =

mr

2r1

. f =

ma

2

ma

. 2Kr = ma +24Kr

.= a3m

4Kr

=> Net fraternal force = (-r direction as assumed)

3

Solution: (D)



23/34


24/34

d 2x 4kx

m += 0 (Equation of SHM)dt2 3

4k

.. =

3m

Solution: (C)By conversation of energy

111

mvo2 + Iw2 = (2K)x 2

2R 2

3m

. x = v

o

4Kma

Also f = ( for rolling )

2

ma

f max = ( for rolling )

2

ma

. mg =2m . 4Kx .

..

2 . 3 .3

. x =g

23m 3


25/34

. v=

g

o

4K 2

3m

. vo = gK



26/34

SECTION lVMatrix Match Type

This section contains 3 questions. Each questions contains statements given in twocolumns, which have to bematched . Statements in Column I are labelled as A, B, Cand D whereas statements

in COLUMN II are labelled as p,q, r and s . The answers to these questions have to be appropriately bubbled asillustrated in the followingexample .

If the correct matches are A-q,B-p,C-r,D-q,then the correctly bubbled matrix willlook like the following

Q 42: Column I gives a list of possible set of parameters measured in someexperiments. The variations of theparameters in the form of graphs are shown in Column II. Match the set ofparameters given in Column I

with the graphs given in Column II. Indicate your answer by darkening theappropriate bubbles of the 4 X4 matrix given in the ORS.

(A) Potential energy of a simple pendulum (y axis) as a function of displacement(x axis)(B) Displacement (y axis) as a function of time (x axis) for a one dimensionalmotion at zero or constantacceleration when the body is moving along the positive x-direction(C) Range of a projectile (y axis) as a function of its velocity (x axis) whenprojected at a fixed angle(D) The square of the time period (y axis) of a simple pendulum as a function ofits length (x axis)

(P)(Q)(R)100Percentile Education Private Limited


27/34

(S)Solution:

A P, SB Q, R ,SC SD Q

(i) Potential energy of simple Pendulum . (displacement from mean position)2. Both p and s can be correct depending on the displacement of mean position(ii) Displacement . t2 (for constant 0). t (fora=0)At t = 0, displacement may be nor-zero. q, r, s are correct.

(iii) Range v/s velocityu sin 2.

R =

g. R . u (starts from origin)=> s is correct.

22

(iv) t = (.p )l(starts from origin)g

. q is correct.

Q 43: Column I contains a list of processes involving expansion of an ideal gas.Match this with Column IIdescribing the thermodynamic change during this process. Indicate your answer bydarkening the appropriatebubbles of the 4 X 4 matrix given in the ORS.



28/34

(A) An insulated container ha as two chambers separated by a valve. Chamber Icontain ns an ideal gas andthe chamber II has vacuum m. The valve is opened.

(B) An ideal monoatomic gas expands to twice its original volume such that itspressu rewhere V is the volume of the gas

(C) An ideal monoatomic gas expands to twice its original volume such that itspressu rewhere V is itits volume

(D) An ideal monoatomic gas expands such that its pressure P and volume V followsthhe behavior shownin the graph

(p) The tempera ature of the gas decreases(q) The tempera ature of the gas increases or remains constant(r) The gas lose es heat(s) The gas gain ns heatSolution:

A PB P, RC P, SD P, S

v -1

(i) The process is adiabatic TV = cons st.=> V increa ase & T decrease. p is corr rect(ii) PV= const.dv

C = C ++ P

v

. T

1 1-n

PV =const t.=K &PV =RT=>KV=RT

100Percentile Edu ucation Private Limited


29/34

K dV

C = C +

vn

V . T

K RVn

= C +

vn

V K(1 - n)RR

=+ 1 - n

v - 1R . 1 .

Hence C =- R = R.-1.

v -1 . v -1 .. 2 - v . R

= R. .=

. v -1 . 2

2

VT = const since PV= const

. v increases => T decreasesAs T decreases & C is + ve

. Q = C%T < O=> Gas loses heat. P, r are correct

(C) PV4/3 = const=> TV1/3 = const. V .. of V increases => T decreases => P is correct

RR

C =+

.- 11 - nRR

=+

35-11 - 43


30/34

3R=- 3R

2

31T100Percentile Education Private Limited


31/34


32/34

(D)(p) Real image(q) Virtual image(r) Magnified image(s) Image at infinitySolution:

A P, Q, R, SB QC P, Q, R, SD P, Q, R, S

11 1

(i) Concave mirror+

= f = O

v u |f|

u f

. v =u - f

When u = f=> image at infinity

f

M =. can be greater than 1u - f

It can formReal image u > fVirtual image u < f

. p, q, r, s correct

(ii) Convex Mirror Always forms virtual image11 1

+

=

f > O

uv f



33/34

uf

v =-

u + f. forrealobjectu>O &f>O. v .8

. for real object v < O . Virtual image

v

f

M =

=

< 1

uu + f

. M R1 . f>O. it behaves as convex lens same as c



34/34

Documents

IITJEE 2008 Physics Paper 2 Code 2 1