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JEE MAIN EXAMINATION - 2013

QUESTIONS WITH SOLUTIONS

PAPER CODE - S

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JEE MAIN Examination(2013) (Code - S)(Page # 2)





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PART - I [PHYSICS]Q.1 A charge Q is uniformly distributed over a long

rod AB of length l as shown in the figure. Theelectric potential at the point O lying at adistance L from the end A is:

LLA BO

(1) 2ln4 0LQ

(2) LQ

042ln

(3) LQ

08 (4) LQ

043

Sol. (2)

Formula v = KQ n2L

dxx

dv = K dQx

= K Q dxx L

v = 2L

L

KQ dxL x =

KQ n2L

Q.2 A sonometer wire of length 1.5 m is made ofsteel. The tension in it produces an elastic strainof 1%. What is the fundamental frequency ofsteel if density and elasticity of steel are 7.7 ×103 kg/m3 and 2.2 × 1011 N/m2 respectively?(1) 200.5 Hz(2) 770 Hz(3) 188.5 Hz(4) 178.2 Hz

Sol. (4)

Strain = 1

100

Stress = (strain)y = 1

100 × 2.2 × 1011

= 2.2 × 109

TA

= Stress = 2.2 × 109 T = 2.2 × 109 A

v = TM

= T

P.A =

92.2 10 .AP.A

v = 9

3

2.2 107.7 10

= 62 107

f1 = v2L

= 62 1107 2 1.5

= 178.2 Hz

Q.3 A projectile is given an initial velocity of ji

2

m/s, where i is along the ground and j

is along

the vertical. If g = 10 m/s2, the equation of itstrajectory is:(1) 4y = 2x – 5x2

(2) 4y = 2x – 25x2

(3) y = x – 5x2

(4) y = 2x – 5x2

Sol. (4)

v i 2j

vx = 1vy = 2Tan = 2

Cos = 15

y = x tan – 2

2 2

gx2u cos

y = x (2) –

2gx12 55

= 2x – 5x2

Q.4 A uniform cylinder of length L and mass M havingcross - sectional area A is suspended, with itslength vertical, from a fixed point by a masslessspring, such that it is half submerged in a liquidof density at equilibrium position. The extensionx0 of the spring when it is in equilibrium is:

(1)

MLA

kMg

21

(2)

MLA

kMg 1

(3) k

Mg(4)

MLA

kMg 1

JEE MAIN Examination(2013) (Code - S) (Page # 3)

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Sol. (1)

MLA

L2L2

Mg = Kx + B

Mg = Kx + AL g2

x = Mg ALgK 2K

= Mg AL1K 2M

Q.5 The graph between angle of deviation () andangle of incidence (i) for a triangular prism isrespresented by:

(1)

o i

(2)

o i

(3)

o i

(4)

o i

Sol. (1)

Q.6 Diameter of a plano-convex lens is 6 cm andthickness at the centre is 3 mm. If speed of lightin material of lens is 2 × 108 m/s, the focal lengthof the lens is :(1) 30 cm (2) 10 cm(3) 15 cm (4) 20 cm

Sol. (1)

RR- t

R

t d/2

d/2

R2 = (R – t)2 + 2d4

R2 = R2 + t2 – 2 tR + 2d4

(t2 is neglect)

2tR = 2d4

R = 2d

8t

R = 236 cm

8 0.3cm = 1208 = 15 cm

1f

= ( – 1) 1R

1f

= 1

15

= 130

f = 30

Q.7 The supply voltage to a room is 120 V. Theresistance of the lead wires is 6 A 60 W bulb isalready switched on. What is the decrease ofvoltage across the bulb, when a 240 W heater isswitched on in parallel to the bulb?(1) 13.3 Volt (2) 10.4 Volt(3) zero Volt (4) 2.9 Volt

Sol. (2)

120 v

+ –

v1 = 120 240246

= 117.07

120 v

v2 = 120 4854

= 106.67

v = v1 – v2 = 117.073 – 106.67 = 10.4






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Q.8 A beam of unpolarised light of intensity I0 ispassed through a polaroid A and then throughanother polaroid B which is oriented so that itsprincipal plane makes an angle of 45° relative tothat of A. The intensity of the emergent light is:(1) I0 /4 (2) I0 /8(3) I0 (4) I0 /2

Sol. (1)

45ºA B

Through A only component parallel to slit willpass so intensity after passing through A will be

0I2

.

After passing through B

I = 0I2

cos2 = 0I4

Q.9 The amplitude of a damped oscillator decreasesto 0.9 times its original magnitude in 5s. In another10s it will decreases to times its originalmagnitude, where equals :(1) 0.729 (2) 0.6(3) 0.7 (4) 0.81

Sol. (1)From equation of damped oscillation

A' = bt2Ae

0.9 A = A e–2.5 b

solving b = 0.0421In another 10 seconds i.e. att = 15 seconds

A'' = 15b2Ae

A = 15b2Ae

= 0.729

Q.10 Two coherent point sources S1 and S2 areseparated by a small distance 'd' as shown. Thefringes obtained on the screen will be :

D

S1 S2

Screen

(1) semi-circles (2) concentric circles(3) points (4) straight lines

Sol. (2)

S1 S2d

BR

O

Path differnce on the circle of radius R around Oon the wall will be same hence concentric circle.

Q.11 A metallic rod of length 'l' is tied to a string oflength 2 l and made to rotate with angular speed on a horizontal table with one end of the stringfixed. If there is a vertical magnetic field 'B' inthe region, the e.m.f. induced across the endsof the rod is:

(1) 2

4 2lB(2)

25 2lB

(3) 2

2 2lB(4)

23 2lB

Sol. (2)

EMF = 2L 3L LB

2

= 25 BL2

Q.12 If a piece of metal is heated to temperature and then allowed to cool in a room which is attemperature 0, the graph between thetemperature T of the metal and time t will beclosed to:

(1)

T

OOO

0 (2)

T

0

OOO

(3)

T

OOO

(4)

T

0

OOO



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Sol. (2)Newton’s law of cooling

ddt

= – K ( – 0)

t0

d K dt

Q.13 This question has statement I and statement II.Of the four choices given after the statement,choose the one that best decribes the twostatements.Statement - I : Higher the range, greater isthe resistance of ammeter.Statement - II : To increase the range ofammeter, additional shunt needs to be usedacross it.(1) Statement-I is true, Statement-II is false(2) Statement-I is false, Statement-II is ture(3) Statement-I is true, Statement-II is true,Statement-II is the correctexplanation of statement-I.(4) Statement-I is true, Statement-II is true,Statement-II is not the correct explanation ofStatement-I.

Sol. (2)Statement-I (False)

I = Ig G1R

If R MoreI Less

Statement-II (True)Addition shunt in parallel decreases the

resistance.

Q.14 Two charges, each equal to q, are kept at x = –a and x = a on the x-axis. A particle of mass m

and charge q0 = q2

is given a small displacement

(y << a) along the y-axis, the net force actingon the particle isproportional to :

(1) y1

(2) y1

(3) y (4) –ySol. (3)

q q

q/2

a ax

F

y

F

netF 2F cosθ

2 2 2 2 1 /2

kq(q /2) y2 .(a y ) (a y )

2 2

2 2 3 /2 3kq .y kq .y~ ~

(a y ) a

netF y

Q.15 This queation has Statement I and StatementII.Of the four choices given after the Statements,choose the one that best describes the twostatements.Statement-I : A point particle of mass m movingwith speed collides with stationary point particleof mass M. If the maximum energy loss possible

is given as

2

21

mf then f =

mMm

.

Statement-II : Maximum energy loss occurswhen the particles get stuck together as a resultof the collision.(1) Statement-I is true, Statement-II is false(2) Statement-I is false, Statement-II is true.(3) Statement-I is true, Statement-II is a correctexplanation of Statement-I(4) Statement-I is true, Statement-II is ture,Statement-II is not a correct explanation ofStatement-I.

Sol. (2)Statement–I (False)

For max energy loss, e = 0

V = mv

m M

loss = 12

mv2 – 12

(m + M)2mv

m M

= 12

mv2 – 12

2 2m vm M

=

2 2 2mv m M m v12 m M

= 21 mMv

2 m MStatement–II (True)






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Q.16 The I – V characteristic of an LED is:

(1)

I

OOO V

(2) I

OV

RRYGB

(3)

I

O

(R)(Y)(G)(B)

Red

Yello

wGre

enBl

ue

V

(4)

I

O V

BGYR

Sol. (3)LED work in forrward biased

Q.17 Assume that a drop of liquid evaporates bydecrease in its surfaceenergy, so that its temperature remainsunchanged. What should be the minimum radiusof the drop for that to be Possible? The surfacetension is T, density of liquid is and L is itslatent heat of vaporization.(1) T/L (2) 2T/L

(3) L/T (4) LT /

Sol. (2)

L)dm(r4)drr(4s 22

drr4dm 2

2Tr

L

r

Q.18 Two capacitors C1 and C2 are charged to 120 Vand 200 V respectively. It is found that byconnecting them together the potential on eachone can be made zero. Then:(1) 3C1 + 5C2= 0(2) 9C1 = 4C2(3) 5C1 = 3C2(4) 3C1 = 5C2

Sol. (4)

200V

120V

C1

C1

2211 vcvc

)200(c)120(c 21

21 c5c3

Q.19 What is the minimum energy required to launcha statellite of mass m from the surface of a planetof mass M and radius R in a circular orbit at analtitude of 2R?

(1) R

GmM2

(2) RGmM3

(3) RGmM6

5

(4) RGmM3

2

Sol. (3)Energy = u + KE

R3GmM

21

R3GMm

RGMm

= 65

R

GMm



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Q.20

2p0

p0

v0v2v0

p

The above p-v diagram represents thethermodynamic cycle of anengine, operating with an ideal monoatomicgas. The amount of heat, extracted from thesource in a single cycle is:

(1) 00211 vp

(2) 4p0v0

(3) p0v0 (4) 00213 vp

Sol. (4)

B

AD

C

V0 2V0

V

1122vAB vPVP23TnR

23TncQ

= 2VP3 00

TnR25TnCQ pBC

1122 VPVP25 = 5P0V0

Total = ( 1- 5 + 5)P0V0 = 6.5 P0V0

21. A circular loop of radius 0.3 cm lies parallel to amuch bigger circular loop of radius 20 cm. Thecentre of the small loop is on the axis of thebigger loop. The distance between their centresis 15 cm. If a current of 2.0 A flows through thesmaller loop, then the flux linked with bigger loopis :(1) 3.3 × 10–11 weber(2) 6.6 × 10–9 weber(3) 9.1 × 10–11 weber(4) 6 × 10–11 weber

Sol. (3)

= 2

03 /22 2

IR

2 R x

× r2

= 2 2

03 /22 2

IR r

2 R x

=

2 27

3 / 22 2

4 10 2 0.2 0.03

2 0.2 0.15

= 9.09 × 10–11

22. A diode detector is used to detect an amplitudemodulated wave of 60% modulation by using acondenser of capacity 250 pico farad in parallelwith a load resistance 100 kilo ohm. Find themaximum modulated frequency which could bedetected by it.(1) 5.31 MHz (2) 5.31 kHz(3) 10.62 MHz (4) 10.62 kHz

Sol. (4)

V<1

RC < 12 5

12.5 10 10 F <40 KHz

From given options the highest frequency , 40KHz is 10.62 KHz.

23. An ideal gas enclosed in a vertical cylindricalcontainer supports a freely moving piston of massM. The piston and the cylinder have equal crosssectional area A. When the piston is in equilibrium,the volume of the gas is V0 and its pressure isP0. The piston is slightly displaced from theequilibrium position and released. Assuming thatthe system is completely isolated from itssurrounding, the piston executes a simpleharmonic motion with frequency :

(1) 0

02

MVPA

21 (2)

0

0

PAMV

21

(3) MVPA

21

0

0 (4)

0 02

V MP12 A

Sol. (1)

A

A

P V0 0

M

Adiabatic

ttanconsPvr

0dp.vdvprv r1r






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)dp(v)dx.A(vPr r1r

r

1r

vAdxrpvdp

dx.

vrPAdP

dx.kdx.V

rPAdF2

2rpAmv2

km2T

20

0

P A1f2 mV

24. A hoop of radius r and mass m rotating with anangular velocity 0 is placed on a rough horizontalsurface. The initial velocity of the centre of thehoop is zero. What will be the velocity of thecentre of the hoop when it cases to slip ?

(1) 2

r 0 (2) r0

(3) 4

r 0 (4) 3r 0

Sol. (1)C.O.A.M. (abont bottom)

rvmr2mr cm2

02

2rV 0

cm

25. Two short bar magnetic of length 1 cm eachhave magnetic moments 1.20 Am2 and 1.00 Am2

respectively. They are placed on a horizontaltable parallel to each other with their N polespointing towards the South. They have a commonmagnetic equator and are separated by adistance of 20.0 cm. The value of the resultanthorizontal magnetic induction at the mid - pointO of the line joining their centre is close to :(Horizontal component of earth's magneticinduction is 3.6×10–5 Wb/m2)(1) 3.50 × 10–4 Wb/m2

(2) 5.80 × 10–4 Wb/m2

(3) 3.6 × 10–5 Wb/m2

(4) 2.56 × 10–4 Wb/m2

Sol. (4)

S

S

N

N

20cm S

B)r(M

)r(M

4MBBBB 23

0221

53

7 106.3)1.0(12.110

54 106.3102.2 = 2.56 × 10–4

26. The anode voltage of a photocell is kept fixed.The wavelength of the light falling on thecathode is gradually changed. The plate currentI of the photocell varies as follows:

(1)

I

O

(2)

I

O

(3)

I

O

(4)

I

OSol. (2)

Energy 1

27. Let [0] denote the dimensional formula of thepermittivity of vacuum. If M = mass, L = length,T = time and A = electric current, then :(1) [0] = [M–1 L2 T–1 A–2](2) [0] = [M–1 L2 T–1 A2](3) [0] = [M–1 L–3 T2 A](4) [0] = [M–1 L–3 T4 A2]

Sol.EnergyVolume =

120E

2 = 120

2Fq

2 2

3

ML TL

= 0 22MLT

AT

0 M–1 L–3 T4 A2



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28. In a hydrogen like atom electron make transitionfrom an energy level with quantum number n toanother with quantum number (n – 1). If n >> 1,the frequency of radiation emitted it proportionalto

(1) 2/3n1

(2) 3n1

(3) n1

(4) 2n1

Sol. (2)

Energy 2n1

22 n1

)1n(1h

2222

22

)1n(n1n2

)n()1n()1n(nh

22 )n(nn2

3n1h

29. In an LCR circuit as shown below both switchesare open initially. Now switch S1 is closed, S2kept open. (q is charge on the capacitor and =RC is Capacitive time constant). Which of thefollowing statement is correct ?

S1

S2

V

C

L

R

(1) At t = 2, q = CV (1 – e–2)

(2) At t = 2

, q = CV (1 – e–1)

(3) Work done by the battery is half of the energydissipated in the resistor(4) At t = , q = CV / 2

Sol. (1)

S1

S2

V

C

L

R

q = CV t

RC1 e

At t = 2RCq = CV [1 – e–2]

30. The magnetic field in a travelling electromagneticwave has a peak value of 20 nT. The peak valueof electric field strength is :(1) 9 V/m (2) 12 V/m(2) 3 V/m (4) 6 V/m

Sol. (4)

EB

= c

E = B.c = 20 × 10–9 × 3 × 108

= 6 v/m






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PART - II [MATHEMATICS]31. The real number k for which the equation,

2x3 + 3x +k = 0 has two distinct real roots in [0, 1](1) lies between –1 and 0.(2) does not exist(3) lies between 1 and 2(4) lies between 2 and 3

Sol. 2f'(x) = 6x2 + 3 so f(x) is increasing xequation can not have 2 distinct roots

32. The number of values of k, for which the systemof equations :(k + 1) x + 8y = 4kkx + (k + 3)y = 3k – 1has no solution, is :(1) 2 (2) 3 (3) infinite (4) 1

Sol. 4(k + 1)x + 8y = 4k

kx + (k + 3)y = 3k – 1

– 81k

= – 3kk

k2 + 4k + 3 = 8kk2 – 4k + 3 = 0k = 3, k = 1for k = 1 same lines k = 3 is only solution

33. If P =

1 31 3 32 4 4

is the adjoint of a 3 × 3 matrix

A and |A| = 4, then is equal to :(1) 5 (2) 0 (3) 4 (4) 11

Sol. 4

4244231331

131 = 42

12 + 6 + 12 – 18 – 12 – 4 = 16 = 11

34. Let Tn be the number of all possible trianglesformed by joining vertices of an n-sided regularpolygon. If Tn+1 – Tn = 10, then the value of nis:(1) 10 (2) 8 (3) 7 (4) 5

Sol. 4Tn = nC3n + 1C3 – nC3 = 10nC2 = 10 nCr + nCr – 1 = n + 1Cr n = 5 nCr – 1 = n + 1Cr – nCr

35. At present, a firm is manufacturing 2000 items.It is estimated that rate of change of productionP w.r.t additional number of workers x is given

by dP 100 12 xdx

. If the firm employs 25 more

workers, then the new level of production ofitems is :(1) 3500 (2) 4500 (3) 2500 (4) 3000

Sol. 1

dxdp

= 100 – 12 x

P = 100x – 2/3x12 2/3

+ c

If x= 0 then P = 2000 P = 100 x – 8x3/2 + 2000If x = 25P = 2500 – 1000 + 2000 = 3500

36. ABCD is a trapezium such that AB and CD areparallel and BC CD. If ADB = , BC = p and CD= q, then AB is equal to :

(1) 2 2

2 2p q

p cos q sin

(2) 2

22

)sinqcosp(sin)qp(

(3) 2 2(p q )sin

pcos qsin (4)

2 2p q cospcos qsin

Sol. 3

B

p

C

x A

q

90º–

22 qp

D

sinAB

= BD

sin ( )

AB = ))sin(sinBD

=

sincoscossin

sinBD

= 2222

22

qp

pcosqp

qsin

sinqp

=

cospsinqsin)qp( 22



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37. All the students of a class performed poorly inMathematics. The teacher decided to give gracemarks of 10 to each of the students. Which ofthe following statistical measures will not changeeven after the grace marks were given ?(1) mode (2) Variance(3) mean (4) median

Sol. 2variance does not change

38. A ray of light along x + y3 = 3 gets reflectedupon reaching x-axis, the equation of thereflected ray is :

(1) y = 3x – 3 (2) y3 = x – 1

(3) y = x + 3 (4) y3 = x – 3Sol. 4

x-axis

(0,1)

(0,–1)

3y3x

3( ,0)

so equation is

11

y3x

39. The area (in square units) bounded by the curvesy = x , 2y – x + 3 = 0, x -axis, and lying in thefirst quadrant is:

(1) 18 (2) 427

(3) 9 (4) 36

Sol. 3

y

(0,0)

2y–x+3=0

(9,3)y= x

(3,0)x

(9,0)

y2 = 2y + 3y2 – 2y – 3 = 0y = 3 y = – 1

Area = 3.6.21dxx

9

0

= 93

x29

0

2/3

= 3

2.27 – 9

= 9 sq. units.

40. If z is a complex number of unit modulus and

argument , then arg

z1z1

equals.

(1) (2) – (3) – (4) 2

–

Sol. 1|z| = 1

arg 1 z1 z

= arg (1 + z)2

= 2 arg(1 + z) =

41. If f(x)dx (x) , then 5 3x f(x )dx is equal to :

(1) 3 3 2 31 x (x ) x (x )dx C3

(2) 3 3 3 31 x (x ) x (x )dx C3

(3) 3 3 2 31 x (x ) x (x )dx C3

(4) 3 3 3 31 x (x ) 3 x (x )dx C3

Sol. 1

dx)x(fx 35

x3 = t 3x2 dx = dt

t f(t) 3dt

= 31

t f(t)dt (t)dt

= 31

[x3 (x3) – (x3) 3x2dx] + c






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42. Let A and B be two sets containing 2 elementsand 4 elements respectively. The number ofsubsets of A x B having 3 or more elements is :(1) 219 (2) 211 (3) 256 (4) 220

Sol. 1n(A × B) = 8number of subsets = 8C3 + 8C4 + .... + 8C8

= 28 – 8C0 – 8C1 – 8C2= 256 – 1 – 8 – 28 = 219

43. If the lines x 2 y 3 z 4

1 1 k

and

x 1 y 4 z 5k 2 1 are coplaner, then k can

have :(1) exactly two values (2) exactly three values(3) any value (4) exactly one value

Sol. 1S.D = 0

k4j3i2a1 , k5j4ia2

k)k(jib1

, kj2i)k(b2

12 aa

. 21 bb

= 0

12kk11111

= 0

1(1 + 2k) + 1(1 + k2) –1(2 - k) = 0k2 + 3k = 0 k = 0, –3

44. The x-coordinate of the incentre of the trianglethat has the coordinates of mid points of itssides as (0, 1) (1, 1) and (1, 0) is :(1) 21 (2) 1 2

(3) 2 2 (4) 2 2Sol. 4

(2, )A(x , y )1 1

(0, )A(x , y )3 3C(0,1)

(0, )B(x , y )2 2

B

(1,0) (1,1)

222

2x2 + 1 = 1 + 0x1 + 0 = 2x3 + 1 = 1x-coordinate of incentre

= 2244

= 22

2

× 2222

45. Consider :Statement - I : (p ^ ~ q) ^ (~ p ^ q) is afallacy.Statement - II : (p q) (~ q ~ p) is atautology.(1) Statement - I is true; Statement - II is false.(2) Statement - I is false; Statement - II istrue.(3) Statement -I is true; Stetement - II is true;Statement - II is a correct explanation forStatement - I(4) Statement - I is true; Stetement-II is true;Statement - II is not a correct explanation forStatement -I

Sol. 4

p q p q ~ p ~ q ~ q ~ p S II p ^ ~q ~ p ^ q S IT T T F F T T F F FT F F F T F T T F FF T T T F T T F T FF F T F T T T F F F

Both S-I and S-II are true but S-II does notexplan S-I

46. If the equations x2 + 2x + 3 = 0 and ax2 + bx +c = 0 , a, b, c, R, have a common root, thena : b : c is :(1) 1 : 3 : 2 (2) 3 : 1 : 2(3) 1 : 2 : 3 (4) 3 : 2 : 1

Sol. 3First equation has imaginary roots

Both roots are common a : b : c = 1 : 2 : 3

47. The sum of first 20 terms of the sequence 0.7,0.77, 0.777, ..... is :

(1) 817

(179 + 10–20) (2) 97

(99 + 10–20)

(3) 817

(179 – 10–20) (4) 97

(99 – 10–20)

Sol. 1

S = 7 0.1 0.11 0.111 .....

= 97

0.9 0.99 0.999 ........

= 97

(1 0.1) (1 0.01) (1 0.001) .....

= 97

2010.1 110

201 110

= 97

20( 10 1)209

= 817

{179 + 10–20}



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48. The term independent of x in expansion of10

2/13/13/2 xx1x

1xx1x

is :

(1) 210 (2) 310 (3) 4 (4) 120Sol. 1

10

2/12/1

2/12/1

3/13/13/2

3/1

)1x(x)1x()1x(

)1x()1xx()1x()1x(

=

10

2/1

2/13/1

x)1x(1x

= 102/13/1 )x1(1x = 102/13/1 xx Tr + 1 = 10Cr

(x1/3)r (–x–1/2)10–r

= 10Cr (–1)10–r xr/3 . 210r

x

= 10Cr (–1)10–r . 210r

3r

x

for independent term 3r

+ 210r

= 0

2r + 3r – 30 = 0 r = 6

Hence T7 = 10C6 (–1)4

= 10C6 = 1.2.3.47.8.9.10

= 210

49. If the vectors AB = i3

+ k4 and AC

=

k4j2i5 are the sides of a triangle ABC, thenthe length of the median through A is :

(1) 33 (2) 45 (3) 18 (4) 72Sol. 1

m(4,–1,4)

C(5,–2,4)

(0,0,0) A B (3,0,4)

AB = k4i3

CA = k4j2i5

| AM | = 16116 = 33

50. If x, y, z are in A.P and tan–1x, tan–1y and tan–

1z are also in A.P., then :(1) 6x = 3y = 2z (2) 6x = 4y = 3z(3) x = y = z (4) 2x = 3y = 6z

Sol. 32tan–1 y = tan–1 x + tan–1 z

2y1y2

= xz1zx

y2 = xz ( 2y = x + z) x = y = z

51. The intercepts on x-axis made by tangents to

the curve, y = x

0

Rx,dt|t| , which are parallel

to the line y = 2x, are equal to :(1) ±3 (2) ±4 (3) ±1 (4) ±2

Sol. 3

dxdy

= x = 2. 2

0

dtt = 2t2

= 2

y – 2 = 2(x – 2)y = 2x – 2x – intercept = ± 1

52. Distance between two parallel planes 2x + y +2z = 8 and 4x + 2y + 5 = 0 is :

(1) 27

(2) 29

(3) 23

(4) 25

Sol. 1Distance between parallel planes

D = 22221

cba

dd

D = 41425)8(

D = 621

D = 27






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53. The circle passing through (1, –2) and touchingthe axis of x at (3, 0) also passes through thepoint :(1) (5, –2) (2) (–2, 5)(3) (–5, 2) (4) (2, –5)

Sol. 1

r

(3,0)

(3,r)

(1,–2)

(x – 3)2 + (y – r)2 = r2

passes through (1, –2)(1 – 3)2 + (–2 – r)2 = r2

4 + 4 + r2 + 4r = r2

r = – 2centre : (3, –2)r = 2Hence circle : [(x – 3)2 + (y + 2)2 = 4]

54. The equation of the circle passing through the

foci of the ellipse 2 2x y 1,

16 9 and having centre

at (0, 3) is :(1) x2 + y2 – 6y – 5 =0(2) x2 + y2 – 6y + 5 = 0(3) x2 + y2 – 6y – 7 = 0(4) x2 + y2 – 6y + 7 = 0

Sol. 3

9 = 16(1 – e2) 1 – e2 = 169

e2 = 167

e = 47

so foci are (± 7 , 0)clearly (B) satisfy the point

55. If y = sec(tan–1x), then dxdy

at x = 1 is equal to:

(1) 1 (2) 2 (3) 21

(4) 21

Sol. 3y = sec(tan-1 x)

y = 2x1

dxdy

= 2x1

x

1

x2x1

1xatdxdy

=

21

56. The expression Acot1

Atan

+ Atan1

Acot

can be

written as :(1) tanA + cotA (2) secA + cosecA(3) sin A cos A + 1 (4) secA cosecA + 1

Sol. 4

Acot1Atan

+ Atan1

Acot

)AcosA(sinAcosAsin2

+ )AsinA(cosAsin

Acos2

= AcosAsin

AcosAsinAcosAsin 22

= 1 + secA cosec A

57. Given: A circle, 2x2 + 2y2 = 5 and a parabola, y2

= 54 x.Statement - I: An equation of a common tangent

to these curves is y = x + 5 .

Statement - II: If the line, y = mx + m5

(m 0)

is their common tangent, then m satisfies m4 –3m2 + 2 = 0,(1) If Statement-I is true but Statement - II is false.(2) If Statement-I is false but Statement-II is true.(3) If both Statement - I and Statement - II aretrue, and Statement - II is the correctexplanation of Statement- I.(4) If both Statement-I and Statement - II aretrue but Statement - II is not the correctexplanation of Statement-I.

Sol. 1

y = mx + m5

tangent to parabola y2 = 4 5 x

x2 + y2 = 25

circle

r from (0, 0) = 25



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1m

m/52

= 25

)1m(m5

22 = 25

m4 + m2 – 2 = 0

t2 + t – 2 = 0(t + 2) (t – 1) = 0t = – 2 or t = 1 m2 = 1m = ± 1equation of tangent

y = x + 5

y = – x – 5

58. A multiple choice examination has 5 questions.Each question has three alternative answers ofwhich exactly one is correct. The probability thata student will get 4 or more correct answersjust by guessing is :

(1) 5311

(2) 5310

(3) 5317

(4) 5313

Sol. 1RRRRW or RRRRR

!4!5 ×

4

31

. 3

2 + !5

!5 .

5

31

= 53110

= 5311

59. Statement - I : The value of the integral

3/

6/xtan1

dx is equal to

6

.

Statement - II : b

a

dx)x(f = b

a

dx)xba(f .

(1) If Statement-I is true but Statement - II is false.(2) If Statement-I is false but Statement-II is true.(3) If both Statement - I and Statement - II aretrue, and Statement - II is the correctexplanation of Statement- I.(4) If both Statement-I and Statement - II aretrue but Statement - II is not the correctexplanation of Statement-I.

Sol. 2

I =

3/

6/ xcosxsindxxcos

2I = 6dx

3/

6/

I =

12

60. 0xlim

x4tanx)xcos3)(x2cos1( is equal to :

(1) 1 (2) 2 (3) –41

(4) 21

Sol. 2

0xLim

(1 cos2x) (3 cos x)x tan 4x

= 0xLim

22

1 cos2x4x (3 cos x)(2x)

tan4x4x4x

= 21

. 4 = 2






Scholarship100%

61. Which of the following represents the correct orderof increasing first ionization enthalpy for Ca, Ba, S,Se and Ar ?(A) Ba < Ca < Se < S < Ar(B) Ca < Ba < S < Sr < Ar(C) Ca < S < Ba < Se < Ar(D) S < Se < Ca < Ba < Ar

Sol. AAr highest ionization energy because noble gas. BaLowest ionization energy because 6 period & moremetallic.

62. A gaseous hydrocarbon given upon combustion 0.72g of water and 3.08 g of CO2. The empirical formulaof the hydrocarbon is(A) C6H5 (B) C7H8 (C) C2H4 (D) C3H4

Sol. B

CX HY + O2 xCO2 + 2y

H2O

CO2 = 3.08 gm

CO2 (mole) = 4408.3

= 0.07 mole

H2O (moles) = 0.72 gm

H2O ( moles) = 1872.0

= 0.04 moles

moles of C : H0.07 : 0.08 (C7 H8)7 : 8

63. The rate of a reaction doubles when its temperaturechanges from 300 K to 310 K. Activation energy ofsuch a reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)(A) 58.5 kJ mol–1

(B) 60.5 kJ mol–1

(C) 53.6 kJ mol–1

(D) 48.6 kJ mol–1

Sol. C

log )TT(R303.2)TT(E

kk

21

12a

1

2

log 2 = )310)(300)(314.8(303.210Ea

Ea = 53.42 kJ /moleEa = 53.6 kJ /mole

PART - III [CHEMISTRY]

64. The gas leaked from a storage tank of the UnionCarbide plant in Bhopal gas tragedy was:(A) Ammonia(B) Phosgene(C) Methylisocynate(D) Methylamine

Sol. CMIC is poisonous gas.

65. An organic compound A upon reacting and NH3 givesB. On heating B gives C. in presence of KOH reactsis with Br2 to given CH3CH2NH2 A is

(A) CH3–CH

CH3

–COOH

(B) CH3CH2COOH(C) CH3COOH(D) CH3CH2CH2COOH

Sol. BCH — CH — C — OH3 2

O NH3

CH — CH — C — ONH3 2 4

O

CH — CH — C — NH 3 2 2

O

Hoffmann's Bromide Br2 + KOH

CH3 – CH2 – NH2

66. Compound (A), C8H9Br, gives a white precipitatewhen warmed with alcoholic AgNO3. Oxidation of(A) gives an acid (B), C8H6O4. (B) easily formsanhydride on heating. identify the compound (A).

(A)

CH Br2

CH3

(B)

CH2Br

CH3

(c)

CH2Br

CH3

(D)

C H2 5

Br

Sol. B



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67. A compound with molecular mass 180 its acrylatedwith CH3COCl to get a compound with molecularmass 390. the number of amino groups present permolecular of the former compound is(A) 4 (B) 6 (C) 2 (D) 5

Sol. D

– NH – H – NH – C – CH3Acetylation

O

M.w = 42

No. of –NH2 group = 42.wM

42180390

= 5

68. When one of the following molecules is expected toexhibit diamagnetic behaviour ?(A) O2 (B) S2 (C) C2 (D) N2

Sol. C,DS2 Paramagnetic two unpaired electronsC2 Diamagnetic zero unpaired electronsN2 Diamagnetic zero unpaired electronsO2 Paramagnetic two unpaired electrons

69. In which of the following pairs of molecules/ions,both the species are not likely to exist?(A) H2

2+, He2

(B) H2–, He2

2+

(C) H2+

, He22–

(D) H2–, He2

2–

Sol. ABond order zero mean molecule does not exist

H2 H2 H2 H22– H2

2 He2

1 .5 .5 0 0 0bond order

70. Which of the following complex species is notexpected to exhibit optical isomerism?(A) [Co(NH3)3Cl3](B) [Co(en)(NH3)2Cl2]

+

(C) [Co(en)3]3+

(D) [Co(en)2Cl2]+

Sol. AIt is a type of Ma3b3 octahedral complex so do notshow optical isomerism therefore answer is A.

71. The coagulating power of electrolytes having ionsNa+, Al3+ and Ba2+ for arsenic sulphide sol increasesin the order:(A) Ba2+ , Na+ < Al3+ (B) Al3+ < Na+ < Ba2+

(C) Al3+ < Ba2+ < Na+ (D) Na+ < Ba2+ < Al3+

Sol. DAs2 S3

is negative colloideCoagulation power charge (+)Na+ < Ba+2 < Al+3

72. Consider the following reactionx MnO4

– + yC2O42– + zH+ xMn2+ + 2yCO2 + z/2H2O

The value of x,y and z in the reaction arerespectively.(A) 2, 5 and 16(B) 5, 2 and 8(C) 5, 2 and 16(D) 2, 5 and 8

Sol. A2MnO4

– + 5C2O4–2 + 16H+ 2Mn+2 + 10CO2 + 8H2O

73. Which of the following exists as covalent crystalsin the solid state ?(A) Sulphur (B) Phosphorus(C) Iodine (D) Silicon

Sol. D

74. A solution of (–) –1–chloro–1-phenylethane intoluene racemizes slowly in the presence of a smallamount of SbCl5. due to formation of :(A) carbocation (B) free radical(C) carbanion (D) carbene

Sol. A

CH – CH3 CH – CH3

Cl

SbCl5Lewis acid

Carbocation

75. An unknown alcohol is treated with the "Lucasreagent" to determine whether the alcohol isprimary, secondary or tertiary. Which alcohol reactsfastest and by what mechanism(A) secondary alcohol by SN2

(B) tertiary alcohol by SN2

(C) Secondary alcohol by SN1

(D) tertiary alcohol by SN1

Sol. DReactivity of alcohol towards Luca's reagent= 3º > 2º > 1º > CH3OH [SN1 Rxn]Intermediate is carbocation






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76. How many litres of water must be added to litre ofan aqueous solution of HCl with a pH of 1 to createan aqueous solution with pH of 2?(A) 2.0 L (B) 9.0 L (C) 0.1 L (D) 0.9 L

Sol. BM1V1 = M2V2

10–1 × 1 = 10–2 V2

V2 = 10 Lwater added = 10L – 1 L = 9 L

77. The molarity of a solution obtained by mixing 750mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be(A) 1.75 M (B) 0.975 M(C) 0.875 M (D) 1 M

Sol. CM1V1 + M2V2 = M3V3

10007505.02502

= M3

M3 = 0.875

78. A piston filled with 0.04 mol of an ideal gas expandsreversibly from 50.0 mL to 375 mL at a constanttemperature of 37.0ºC. As it does so, it absorbs208 J of heat. The values of q and w for the processwill be: (R = 3.314 J/mol K) (Ln 7.5 = 2.01)(A) q = – 208 J, w = + 208 J(B) q = + 208 J, w = + 208 J(C) q = + 208 J, w = – 208 J(D) q = – 208 J, w = – 208 J

Sol. C

Wrev, exp. = – | 2.303 mRT log 1

2

vv

= –2.303 (0.04) (8.3) (310) log 50375

= – 208 J(E = 0)E = Q + w., Q = –w = + 208 J

79. Experimentally it was found that a metal oxide hasformula M0.98O. Metal M, is present as M2+ and M3+

in its oxide. Fraction of the metal which exists asM3+ would be:(A) 6.05 % (B) 5.08 %(C) 7.01 % (D) 4.08 %

Sol. DM0.98O0.98(x) – 2 = 0

x = 98200

= Average O.N.

Let M+2 M+3

(100 – y) % y %

98200

100y3)y100(2

y = 4.08 %

80. For gaseous state if most probable speed is denotedC*, average speed by C and mean square speedby C, then for a large number of molecules theratios of these speeds are :

(A) C* : C : C = 1 : 128 : 1.225

(B) C* : C = 1 : 1.225 : 1.128

(C) C* : C : C = 1.225 : 1.128 : 1

(D) C* : C : C = 1.128 : 1.225 : 1Sol. A

C* : C : C = 1 : 1.128 : 1.225

Cmps = MRT2

CAS = MRT8

Crms = MRT3

C* : C : C = 2 : 3:8

= 1 : 1.128 : 1.225

81 . Arrange the following compound in order ofdecreasing acidity :

OH

Cl(I)

OH

CH3

(II)

OH

NO2

(III)

OH

OCH3

(IV)

(A) III > I > II > IV(B) IV > III > I > II(C) II > IV > I > III(D) I > II > III > IV

Sol. A

Order of Acidity I/H/MI/H/M

–Cl –I > + M –CH3 + I < +H–NO2 –I < – M–OCH3 –I < + MIII > I > II > IV



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82. The order of stability of the following carbocations:

(I) CH2= CH – CH2 (II) CH3–CH2– CH2

(III)

CH2

is :(A) I > II > III (B) III > I > II(C) III > II > I (D) II > III > I

Sol. BBenzylic > allylic > 1º alkyl

83. The first ionisation potential of Na is 5.1 eV. Thevalue of electron gain enthalpy of Na+ will be:(A) – 10.2 eV (B) + 2.55 eV(C) – 2.55 eV (D) – 5.1 eV

Sol. DFirst I.P.Na + I IP Na+ + e– H = +5.1 eVNa+ + 1e– Na H = –5.1 eV

(e– gain enthalpy)

84. Four successive member of the first now transitionelements are listed below with atoms number. which

one of them is expected to have the highest 0M/M 23E

value ?(A) Fe(Z = 26) (B) Co(Z = 27)(C) Cr(Z = 24) (D) Mn(Z = 25)

Sol. BCo (Z = 27) + 1.80Cr –0.407Mn + 1.54Fe + 0.771

85. Stability of the species Li2, Li2– and Li2

+ increases inthe order of(A) Li2

< Li2– < Li2

+

(B) Li2–< Li2 < Li2

+

(C) Li2 < Li2

+ < Li2–

(D) Li2–< Li2

+ < Li2Sol. D

Li < Li < Libond order 1 .5 .5

more bond order more stability, Li is less stablethan Li because it contain more antibondingelectron.

86. Energy of an electron is given by

E = –2.178 × 10–18

2

2

nZ

.Wavelength of ligh required

to excite an electron in an hydrogen atom fromlevel n = 1 to n = 2 will be(h = 6.62 × 10–34 Js and c = 3.0 × 108 ms–1)(A) 6.500 × 10–7 m (B) 8.500 × 10–7 m(C) 1.214 × 10–7 m (D) 2.816 × 10–7 m

Sol. C

E2 – E1 = hc

+2.178 × 10–8

834 103106.6

411

= 178.2

46.6 × 10–8

= 12.12 × 10–8

= 1.212 × 10–7 m

87. Synthesis of each molecule of glucose inphotosynthesis involves:(A) 8 molecules of ATP(B) 6 molecules of ATP(C) 18 molecules of ATP(D) 10 molecules of ATP

Sol. CPer CO2 require 3 molecule of ATP.1 molecule of Glucosegives 6 CO2 so Ans is 18 ATP.

88. Which of the following is the wrong statement ?(A) Ozone is violet black in solid state(B) Ozone is diamagnetic gas(C) ONCl and ONO– are not isoelectron(D) O3 molecule is bent

Sol. C

OO

:

O sp2 bent molecule

diamagnetic* ONCl & ONO are isoelectronic because totalvalence electron is same6 + 5 + 7 = 18 6 + 5 + 6 + 1 = 18

89. Which of the following arrangements does notrepresent the correct order of the property statedagainst it?(A) Co3+ < Fe3+ < Cr3+ < Sc3+ stability in aqueoussolution.(B) Sc < Ti < Cr < Mn : number of oxidation states.(C) V2+ < Cr2+ < Mn2+ < Fe2+ : Paramagneticbehaviour.(D) Ni2+< Co2+< Fe2+ < Mn2+ : ionic size






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Sol. CMore the unpaired electron more paramagnetic

magnetic moment = )2n(n

n = no of unpaired electronunpaired electron

v+2 3Cr+2 4Mn+2 5Fe+2 4

90. Given

V51.1E;V74.0E 0Mn/MnO

0Cr/Cr 2

43

V36.1E;V33.1E 0Cl/Cl

0Cr/OCr 32

72

Based on the data given above, strongest oxidistingagent will be(A) Mn2+ (B) MnO4

–

(C) Cl– (D) Cr3+

Sol. BOxidising power × EºR self reductionMnO4

–/Mn+2 = 1.51MnO4

–

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