Indian Institute of Technology New Delhi - (IIT) 2011 B.Tech with solutions IIT - Question Paper

The difference between the O.S of two different types qfiuiphur atoms = +5

21.    A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt. 75), alanine and phenylalanine. GlycinecontribuSesft/- 0% to the total weight of the hydrolysed products. The number of glyGinlwiiispresent in the decapeptide is

Ans : 6

Sol : 100_958

₄₇ ? x 958 = 450.12 - ' 100

The weight of glycine present in 958 decapeptide is 450.12.

the molecular weight of one glycine is 75

450.12

no. of glycines = = ⁶ 00 = ⁶

22.    The total number of alkenes possible by dehydrobromination of 3-bromo 3-cyclopentylhexane using alcoholic KOH is

Ans : 5

Sri Chaitanya IIT Academy.A.P. Sol :

Br

C - C - C - C = C - C
	+
C - C - C = C - C - C

cis + trans

%

23. The maximum number of electrons that can have principal quantum number, n=3, and spin quantum number, m_s=-1/2, is

Ans : 9

Sol : In third orbit 3s, 3p, 3d sub levels are present and a total of nine orbitals are present. Each orbital has one electron with m_s = -1/2

SECTION - I (Total Marks : 21)

( SINGLE CORRECT CHOICE TYPE )

This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct

24. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A,

The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

Ans : A

Sol

c) 27

a) 9 Ans : D

_ 1 1 ₂

Sol : 324 _-x-w²

22

26. A meter bridge is set-up as shown, to determine an unknown resistance X using a standard 10 ohm resistor. The galvanometer shows null point when tapping - key is at 52 cm mark. The end-corrections are 1 cm and 2cm respectively for the ends A and B. The determined value of X is

b) 10.6 ohm c) 10.8 ohm

a) 10.2 ohm Ans :B

10

Sol :

(J

2

Y 10

52 +1 48 + 2

A' = 10'v3iJ2

X = 10.6Q

a) 0% Ans : D

27. A 2F capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is
b)
c) 75%

Sol : q₀ = 2V

^qL= ^q2 2 8

4V² = V² +V²

25

4V² _ 1 -¹ -

2 25

4V² _ ^25-1 -4 _ ²⁰x100 = 80 25 25

28. A police car with a siren of freqjftenckHz is moving with uniform velocity 36km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is

a) 8.50 kHz    b) 8.25 kHz    c) 7.75 kHz    d) 7.50 kHz

Ans : A

Aol : frequency reflucted by wall f

f' _ 8 x10³

frequency heared by car driver ^f" _

₃ 32 33

f" _ 8 x10³ x 31 32

33

_ 8 x10³ x

31

8.50 kHz

29. 5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the initial temperature to be T, the work done in the process is

Ans: A

R

Sol: W = -(T - 72)

T_lV_l^r = k

\7

t vr=t₂ i 8

T₂ = 471

9 RT

W=

30. Consider an electric field E = E₀x, where kP*a constant. The flux through the shaded area (as shown in the figuieiSuPfiothis field is

	y
a

z

^area = 42a² EA = E2a² cos 45⁰ = Ea²

SECTION - II (Total Marks : 16)

( MULTIPLE CORRECT CHOICE TYPE )

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) for its answer, out of which ONE or MORE may be correct

31. A metal rod of length L and mass m is pivoted at one end. A thin disk of mass M and radius R(<L) is attached at its center to the free end of the rod. Consider two ways the disc is attached; (case A) the disc is not free to rotate about its center and (case B) the disc is free to rotate about its center. The rod - disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true ?

a)    Restoring torque in case A = Restoring torque in case B

b)    Restoring torque in case A < Restoring torque in case B

c)    Angular frequency for case A > Angular frequency for case B

d)    Angular frequency for case A < Angular frequency forcase B

ncy for cas

Ans : AD

Sol : Torque about point of suspenstion is same in both case
case A : T =
a = kd

case B ; In case B ; Disc is not going to get any torque about its centre.

all of its point have acceleration that is equal to acceleration of end point of rod.

ML³ + m

/////

3

k

k

MR² , ml²

-+ ML +-

2

w₂ > w

32. A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state

a)    heat flow through A and E slabs are same

b)    heat flow through slab E is maximum

c)    temperature difference across slab E is smallest

st

d) heat flow through C = heat flow through B + heat flow through D

Ans :ABCD

O

33. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true ?

a)    They will never come out of the magnetic field region

b)    They will come out travelling along parallel paths

c)    They will come out at the same time

d)    They will come out a different times

Ans : BD

mv

^So' :^r _ qB

They go in different direction but come out along parallel paths.

_T _ 2nm

qB

as m is different T also different

34. A spherical metal shell A of radius R_A and a solid metal sphere B of radius R_B (< R_A) are

kept far apart and each is given charge +Q. Now they are connected by a thin metal wire. Then

_A _ r_b*JSS$

a) _E^ide _ 0    b) Q_a > Q_b    c) g_    d) E_A^nsurface < E⁰B^nsurface

* W

Ans :ABCD

Sol : a) E_a = 0

,ft _ Qbl

^b) Ra R_b

RA > RB

^Qa > ^Qb

c) R _ V

d) ^E _

^2s0

so Ea

SECTION - III (Total Marks : 15)

(COMPREHENSION TYPE)

This section contains 2 paragraphs. Based upon one of the paragraphs 2 multiple choice questions and based on the paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question Nos. 35 to 36

A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m. When the electrons are subjected to an to electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency 'o_p' which is called the palsma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency o, where a part of the energy is absorbed and a part of it is reflected. As o approaches o_p, all the free electrons a set to resonance together and all the energy is reflected. This is the explanation of hjgh reflectivity metals.

35. Taking the electric charge as '_e' and the permittivity as 's₀', use dimesional analysis to determine the correct expression for o_p.

^p

Ans : C

n

Sol : o _ t = T^-

ms_n

D.f. = _T-i

36. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N - 4 x 10²⁷ m ³. Take s₀ -10^-11 and _m -10^-30, where these quantities are in proper SI units.

a) 800 nm    b) 600nm    c) 300nm    d) 200nm

Ans : B

o

S_ol _: o _ 2nn n _ 2n

, , c c x 2 , c x 2n

c = nA A = =- A = -

Paragraph for Question Nos. 37 to 39

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative

T

e

o

S

o

S

&

37. The phase spacem diagram for a ball thrown vertically up from ground is

Momentum

b)

d)

Ans : D

Sol : Vertically thrown particle comes back to original position till the particle reaches the highest position of its path, its momentum is +ve. After that, it terns to zero and them negative

38. The phase space diagram fro simple harmonic motion is a circle centrred at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical energies respectively. Then

Momentum

	Position
c)Ej =4 E2

4

thespring-mass system, with the mass submerged in water,

b)

d)

Ans : A

Sri Chaitanya IIT Academy.A.P.    >22<    :: IIT JEE 2011 Paper - I ::

Sol : As the block oscillates, due to viscous effects, its total energy decreases continuously and so its amplitude decreases continuously. Assuming that the block is initially pulled down and released, its momentum will increase upwards till it reaches the mean position

SECTION -IV

( INTEGER ANSWER TYPE)

This section contains 7 questions . The answer to each of the questions is a single -digit integer, ranging

from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.

40. Steel wire of length L at 40C is suspended from the ceilling and then a mass m is hung from its free end. The wire is cooled down from 40C to 30C to regain its original length L. The coefficient of liner thermal expansion of the steel is 10-5/C, Youngs modulus of steell is 10¹¹ N/m² and radius of the wire is 1mm. Assume that L>> diameter of the wire. Then the value of m in kg is nearly

Ans : 3

mgL

Sol : Y= , e=aL At Ae

_Y = . ^mS^L

AaLAL

M x10

10¹¹ =

x1x10 x10" m = 3.14 * 3

41. The activity of a freshly prepared radioactive sample is 10¹⁰ disintegrations per second, whose mean life is 10⁹s. The mass of an atom of this radioisotope is 10^-23 kg. The mass (in mg) of the radioactive sample is

Ans : 1

Sol : 10¹⁰ = x N

:.N = 10¹⁹ total mass

= 10²⁵ x10¹⁹ kg = 10 ⁶ kg = 1 mg

42. A block moving on an inclined plane making an angle 45 with the horizontal and the coefficient of friction is . The force required to just push it up the inclined plane is 3 times the force required to just prevent from sliging down. If we define N = 10, then N is

Ans : 5

^mg ₊ U^mg 1

V2 V2J V2 J ³[i ^- = [i+

3 3u = 1 + u 4/u = 2

U = 1/2 10u = 5

A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick a shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3m/s². The coefficient of friction between the ground and the ring is large enough that olling always occurs and the coefficient of friction between the Jjifk and the ring is (P/ 10). The value of P is

Stick

4

F = N²+ ²N² (i)

r_p = NR /uNR = 2mR^{2 a} j (ii)

solving (i) and (ii),

= 2.6 and 0.36

but = 2.6 is not a possible solution = 0.36

p = 3.6 * 4

iu₀1₀nR² (300) .

V = -V-Zsin300t

l

. = V

¹ = r m = inR²

46. Four point chargs, each of +q, are rigidly fixed at the four corners of a square planar soap film of side a . The surface tension of the soap film is y. The system of charges and

1/N

planar film are in equilibrium, the ^a

Ans : 3

t/

Sol

&

PART III : MATHEMATICS

SECTION - I (Total Marks : 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct

47. Let P = {# : sin#-cos# = V2cos#j and Q = {#: sin# + cos# = V2sin#| be two sets. Then

b) Q P

d) P = Q

c) P Q Ans: D

^Sol: P = sin#- cos# =V2cos# sin# = V2cos# + cos# sin # = (V2 +1) cos#

1) cos#

.

V2sin# = cos# + sin#

P = Q

48. Let the straight line * = b divide the area enclosed by y = (1 - x²), y = 0, and * = 0 into two parts R₁ (0 < x < b) and R₂ (b < x < 1) such that ^R1 ^- R2 = 4. Then b equals

Ans : B

b 1 1 Sol : j⁽¹ -x)2dx-j⁽¹ -x)2dx = 4

Sri Chaitanya IIT Academy.A.P.    >27<    :: IIT JEE 2011 Paper - I ::

49. Let a and ( be the roots of _x² _-6_x-2 _ 0, with a > ( . If a_n _aⁿ -(ⁿ for _n > 1, then the

a_w - 2a_s value of 2_a is

a) 1    b) 2    c) 3    d) 4

Ans : C

Sol : a₁₀ _a¹⁰ -( _ (a + ()(a⁹ -(⁹)-a((a⁸ -(⁸)

a₁₀ _ 6a₉ + 2a₈

50. A straight line L through the point (3,-2) is inclined at an angle 60⁰ to the line + y _ 1. also intersects the x-axis, then the equation of L is

a) y + V3x + 2 - 343 _ 0    b) y-V3x + 2 + 33 _ 0_

c) y - x + 3 + 2>/3 _ 0    d) 43y + x - 3 j Ans : B

Sol : Let m be the slope of the required lin

ne

1 - mV3

m _ 0 or m_

51. Let (x₀, y₀) be the solution of the following equations

(2x)^ln2 _(3y)^h3

3lnx _ 2y