(1)

Time : 3 hrs. Max. Marks: 360

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Code - P

Answers & Solutionsforforforforfor

JEE (MAIN)-2013

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(Physics, Chemistry & Mathematics)

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1. A uniform cylinder of length L and mass M havingcross-sectional area A is suspended, with its lengthvertical, from a fixed point by a massless spring,such that it is half submerged in a liquid of densityσ at equilibrium position. The extension x0 of thespring when it is in equilibrium is

(1)Mg

k (2)

1

Mg LAk M

(3)

1

2Mg LAk M (4)

1

Mg LAk M

(Here k is spring constant)

Answer (3)

Sol. For equilibrium,

02LMg A g kx

2L A g kx0

Mg⇒

0 12

Mg LAxk M

2. A metallic rod of length l is tied to a string of length2l and made to rotate with angular speed ω on ahorizontal table with one end of the string fixed. Ifthere is a vertical magnetic field B in the region, thee.m.f. induced across the ends of the rod is

(1) 222

B l (2) 232

B l

(3) 242

B l (4) 252

B l

Answer (4)

Sol.

l2l

32 23

2 2

5( )2 2

ll

l l

x B lB x dx B

PART–A : PHYSICS

3. This question has Statement-I and Statement-II. Of thefour choices given after the Statements, choose the one thatbest describes the two Statements.

Statement-I: A point particle of mass m movingwith speed v collides with stationary point particleof mass M. If the maximum energy loss possible is

given as

212

f mv then

mfM m

.

Statement-II: Maximum energy loss occurs whenthe particles get stuck together as a result of thecollision.

(1) Statement-I is true, Statement-II is true,Statement-II is a correct explanation ofStatement-I

(2) Statement-I is true, Statement-II is true,Statement-II is not a correct explanation ofStatement-I

(3) Statement-I is true, Statement-II is false.

(4) Statement-I is false, Statement-II is true.

Answer (4)

Sol. Maximum energy loss is

21 ( 0)2

Mm vM m =

212

M mvM m

So, Statement-1 is wrong.

4. Let [ε0] denote the dimensional formula of thepermittivity of vacuum. If M = mass, L = length,T = time and A = electric current, then

(1) [ε0] = M–1 L–3 T2 A]

(2) [ε0] = M–1 L–3 T4 A2]

(3) [ε0] = M–1 L2 T–1 A–2]

(4) [ε0] = M–1 L2 T–1 A]

Answer (2)

Sol. ε0 = 8.85 × 10–12 C2N–1m–2

[ε0] = L–2A2T2 (MLT–2)–1

= L–3 A+2 M–1 T4 = [M–1L–3T4A2]

(3)

5. A projectile is given an initial velocity of

ˆ ˆ( 2 )m/si j where i is along the ground and j is

along the vertical. If g = 10 m/s2, the equation of itstrajectory is

(1) y = x – 5x2 (2) y = 2x – 5x2

(3) 4y = 2x – 5x2 (4) 4y = 2x – 25x2

Answer (2)

Sol.

2

2 2tan2 cos

gxy x

u

Here tan 2y

x

v

v also

1cos5

, 5u

⇒

22102 2 512 5

5

xy x x x

6. The amplitude of a damped oscillator decreases to0.9 times its original magnitude in 5 s. In another10 s it will decrease to α times its originalmagnitude, where α equals

(1) 0.7 (2) 0.81

(3) 0.729 (4) 0.6

Answer (3)

Sol. In 10 seconds it will become (0.9)3 = 0.729 times.

7. Two capacitors C1 and C2 are charged to 120 V and200 V respectively. It is found that by connectingthem together the potential on each one can be madezero. Then

(1) 5C1 = 3C2 (2) 3C1 = 5C2

(3) 3C1 + 5C2 = 0 (4) 9C1 = 4C2

Answer (2)

Sol. For this, charge must be same Q = C1V1 = C2V2

⇒ 120C1 = 200C2

⇒ 3C1 = 5C2

8. A sonometer wire of length 1.5 m is made of steel.The tension in it produces an elastic strain of 1%.What is the fundamental frequency of steel ifdensity and elasticity of steel are 7.7 × 103 kg/m3

and 2.2 × 1011 N/m2 respectively?

(1) 188.5 Hz (2) 178.2 Hz

(3) 200.5 Hz (4) 770 Hz

Answer (2)

Sol.

1 1 Stress 1 strain2 2 Density 2 Density

Tfl l l

⇒

11

3

12.2 101 1002 1.5 7.7 10

f

= 61 2 103 7

= 1000 2 178.2 Hz

3 7

9. 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 the biggerloop. The distance between their centres is 15 cm. Ifa current of 2.0 A flows through the smaller loop,then the flux linked with bigger loop is

(1) 9.1 × 10–11 weber

(2) 6 × 10–11 weber

(3) 3.3 × 10–11 weber

(4) 6.6 × 10–9 weber

Answer (1)

Sol. By the principle of reversibility, we can take thesame current through the bigger coil and calculatethe flux through smaller coil.

20

2 2 3/22

4 ( )i RB

R x

=

7 2 2

2 2 2 2 3/210 2 2 (20 10 )

[(20 10 ) (15 10 ) ]

=

7 2

2 310 2 2 4 10

(25 10 )

= 1.024π × 10–6 T

φ = 1.024π × 10–6 × π(0.3 × 10–2)2

= 9.1 × 10–11 weber

(4)

10. Diameter of a plano-convex lens is 6 cm andthickness at the centre is 3 mm. If speed of light inmaterial of lens is 2 × 108 m/s, the focal length ofthe lens is(1) 15 cm (2) 20 cm(3) 30 cm (4) 10 cm

Answer (3)Sol.

R

6 cm0.3 cmR

1.5cv

R2 = 32 + (R – 0.3)2

⇒ R2 = 9 + R2 + 0.09 – 0.6R⇒ 0.6R = 9.09

⇒ R = 9.090.6 ≈ 15 cm.

Now,

151 0.5

Rf = 30 cm.

11. What is the minimum energy required to launch asatellite of mass m from the surface of a planet ofmass M and radius R in a circular orbit at analtitude of 2R?

(1)5

6GmM

R(2)

23

GmMR

(3) 2GmM

R (4) 3GmM

RAnswer (1)

Sol. At surface, GMmE

R

In orbit, 2(3 ) 6GMm GMmE

R R

⇒ Required energy = 5

6GMm

R12. A diode detector is used to detect an amplitude

modulated 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) 10.62 MHz (2) 10.62 kHz(3) 5.31 MHz (4) 5.31 kHz

Answer (2)

Sol.

12

fRCm

13. A beam of unpolarized light of intensity I0 is passedthrough a polaroid A and then through anotherpolaroid B which is oriented so that its principalplane makes an angle of 45° relative to that of A.The intensity of the emergent light is

(1) I0 (2) 0

2I

(3) 0

4I

(4) 0

8I

Answer (3)

Sol. 01 2

II , 2 02 1 cos 45

4II I

14. The supply voltage to a room is 120 V. Theresistance of the lead wires is 6 Ω. A 60 W bulb isalready switched on. What is the decrease of voltageacross the bulb, when a 240 W heater is switchedon in parallel to the bulb?

(1) Zero volt (2) 2.9 volt

(3) 13.3 volt (4) 10.04 volt

Answer (4)

Sol.

6

120 V 60 W Heater (240 W)

Assuming that both bulb and the heater has a ratingof 120 V

bulb

120 120 24060

R

Rheater = 60 Ω

Initial current is

120 120240 6 246

i

New current is

120 60 120 1 2448 6 60 240 54 5 54

i

⇒ Decrease in voltage = 120 24240 10.4246 54

V

Nearest answer is (4)

(5)

15.

2p0

v0

p0

p

2v0 vThe above p-v diagram represents thethermodynamic cycle of an engine, operating withan ideal monoatomic gas. The amount of heat,extracted from the source in a single cycle is

(1) p0v0 (2) 0 0

132

p v

(3) 0 0

112

p v (4) 4p0v0

Answer (2)

Sol.

2p0

v0

p0

p

2v0 v

A B

D C

Heat is extracted from source in AB & DA

In AB, Q = nCPΔT

5 ( )2RQ n T

0 0 0 0 0 05 (4 2 ) 52

p v p v p v

In DA, Q = nCVΔT

0 03 3( ) ( )2 2Rn T p v

Total = 0 0132

p v

16. A hoop of radius r and mass m rotating with anangular velocity ω0 is placed on a rough horizontalsurface. The initial velocity of the centre of the hoopis zero. What will be the velocity of the centre of thehoop when it ceases to slip?

(1)0

4r

(2)0

3r

(3)0

2r

(4) rω0

Answer (3)

Sol. By conservation of angular momentum about apoint on ground

mr2ωr = mr2w + mvr

⇒ mr2ω0 = 2 mvr

⇒

0

2r

v

17. 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 is P0.The piston is slightly displaced from the equilibriumposition and released. Assuming that the system iscompletely isolated from its surrounding, the pistonexecutes a simple harmonic motion with frequency

(1)

0

0

12

A PV M

(2) 0 0

21

2V MP

A

(3)

20

0

12

A PMV

(4) 0

0

12

MVA P

Answer (3)

Sol. PVγ = constant

Differentiating after taking log

0dP dVP V

P dVdP

V

P AxdP

V

2

restoringP A xF

V Take P = P0, V = V0

20

0

1 12 2

P AkfM V M

(6)

18. If a piece of metal is heated to temperature θ andthen allowed to cool in a room which is attemperature θ0 , the graph between the temperatureT of the metal and time t will be closest to :

(1)T

t o(2)

T

t o

(3)

T

t o

(4)

T

t o

Answer (3)

Sol. The temperature will decrease exponentially withtime.

19. This question has Statement I and Statement II. Of thefour choices given after the Statements, choose the one thatbest describes the two Statements.

Statement - I : Higher the range, greater is theresistance of ammeter.

Statement - II : To increase the range of ammeter,additional shunt needs to be used across it.

(1) Statement - I is true, Statement - II is true,Statement - II is the correct explanation ofStatement-I.

(2) Statement - I is true, Statement - II is true,Statement - II is not the correct explanation ofStatement-I.

(3) Statement - I is true, Statement - II is false.

(4) Statement - I is false, Statement - II is true.

Answer (4)

Sol. Statement - I is false as shunt is added in parallel.

20. In an LCR circuit as shown below both switches areopen initially. Now switch S1 is closed, S2 kept open(q is charge on the capacitor and τ = RC isCapacitive time constant). Which of the followingstatement is correct ?

S1

S2

R

V

C

L

(1) Work done by the battery is half of the energydissipated in the resistor

(2) At t = τ , q = CV/2

(3) At t = 2τ , q = CV(1 – e–2)

(4) At t = 2

, q = CV(1 – e–1)

Answer (3)

Sol. It is a simple RC charging circuit.

q = CV (1 – e–t/τ)

⇒ q = CV (1 – e–2), when t = 2τ

21. Two coherent point sources S1 and S2 are separatedby a small distance 'd' as shown. The fringesobtained on the screen will be :

D

S1 S2

d

Screen

(1) Points

(2) Straight lines

(3) Semi - circles

(4) Concentric Circles

Answer (4)

Sol. On any circle with centre at O, path difference willremain constant.

S1 S2

O

22. The magnetic field in a travelling electromagneticwave has a peak value of 20 nT. The peak value ofelectric field strength is :

(1) 3 V/m

(2) 6 V/m

(3) 9 V/m

(4) 12 V/m

Answer (2)

Sol. E cB

⇒ E = cB = 3 × 108 × 20 × 10–9 = 6 V/m

(7)

23. The anode voltage of a photocell is kept fixed. Thewavelength λ of the light falling on the cathode isgradually changed. The plate current I of thephotocell varies as follows :

(1)

I

O

(2)

I

O

(3)

I

O

(4)

I

O

Answer (4)

Sol. When wavelength exceeds a certain wavelength,photoelectric effect ceases to exist.

24. The I - V characteristic of an LED is

(1)

Red

Yello

wG

reen

Blue

(Y)(G)(B)

I

O V

(R)

(2) YGB

I

O V

R

(3)I

O V

(4) YGB

I

OV

R

Answer (1)

Sol. The LED emitting red colour will have a smallerenergy gap between conduction band and valanceband. Accordingly its knee voltage would be less.

25. Assume that a drop of liquid evaporates by decreasein its surface energy, so that its temperature remainsunchanged. What should be the minimum radius ofthe drop for this to be possible? The surface tensionis T, density of liquid is ρ and L is its latent heat ofvaporization.

(1) ρL/T

(2) / T L

(3) T/ρL

(4) 2 T/ρL

Answer (4)

Sol. This could happen if dm × L = dA × T

4πR2dR × ρ × L = 8 πRdR × T

2TR

L

26. In a hydrogen like atom electron makes transitionfrom an energy level with quantum number n toanother with quantum number (n – 1). If n >> 1, thefrequency of radiation emitted is proportional to

(1) n1

(2) n21

(3) n3/21

(4) n31

Answer (4)

Sol.

2 21 1

( 1) ( )f

n n

2 2

2 2( 1)

( 1)n n

fn n

2 2

2 21 2

( 1)n n nf

n n

2 22 1( 1)nf

n n

n >> 1 ⇒ 31f

n

(8)

27. The graph between angle of deviation (δ) and angleof incidence (i) for a triangular prism is representedby

(1)

O

i

(2)

O

i

(3)

O

i

(4)

O

iAnswer (3)

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

0 2q

q is placed at the origin. If charge q0 is given

a small displacement (y << a) along the y-axis, thenet force acting on the particle is proportional to

(1) y (2) – y

(3)1y (4)

1y

Answer (1)

Sol. Fnet = 2F cosθ

=

2

2 2 2 2 1/22

2( ) ( )kq y

a y a y

2

3kq yF

a

F F

q qo

( y

a aF y

29. Two short bar magnets of length 1 cm each havemagnetic moments 1.20 Am2 and 1.00 Am2

respectively. They are placed on a horizontal tableparallel to each other with their N poles pointingtowards the South. They have a common magneticequator and are separated by a distance of 20.0 cm.The value of the resultant horizontal magneticinduction at the mid-point O of the line joiningtheir centres is close to

(Horizontal component of earth's magneticinduction is 3.6 × 10–5 Wb/m2)

(1) 3.6 × 10–5 Wb/m2 (2) 2.56 × 10–4 Wb/m2

(3) 3.50 × 10–4 Wb/m2 (4) 5.80 × 10–4 Wb/m2

Answer (2)

Sol. Bnet = B1 + B2 + BH

N S

N S

B B B1 2 H + + South

0

1 34MBr =

7

31.201010

= 1.2 × 10–4 T

B2 = 1 × 10–4 T, BH = 3.6 × 10–5 T

Bnet = 2.56 × 10–4 T

30. A charge Q is uniformly distributed over a long rodAB of length L as shown in the figure. The electricpotential at the point O lying at a distance L fromthe end A is

OL L

A B

(1) 08Q

L (2) 0

34

QL

(3) 04 ln 2QL (4) 0

ln 24Q

L

Answer (4)

Sol.

kdqV

L x

L

O

kQ dLVL L x ln 2kQ

L

0

ln 24

QVL

(9)

31. Which of the following complex species is notexpected to exhibit optical isomerism?

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

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

(3) [Co(NH3)3Cl3]

(4) [Co(en)(NH3)2Cl2]+

Answer (3)

Sol. [Co(NH3)3Cl3) has two geometrical isomers, namelyfac. and mer. Both of them have a plane ofsymmetry. So, they are not expected to show opticalisomerism.

Co

Cl

NH3

Cl

Cl NH3

NH3

fac.

Co

NH3

Cl

Cl Cl

NH3

mer.

NH3

32. Which one of the following molecules is expected toexhibit diamagnetic behaviour?

(1) C2 (2) N2

(3) O2 (4) S2

Answer (1 & 2)

Sol. The electronic configuration of the given diatomicmolecules is

C2: x y2 2 2 2 2 21s 1s 2s 2s 2p 2p* *

N2: zx y

2 2 2 2 2 2 21s 1s 2s 2s 2p 2p 2p* *

O2: z yx y x

2 2 2 2 2 2 2 1 11s 1s 2s 2s 2p 2p 2p 2p 2p* * * *

S2: z yx y x

z z x y x y

2 2 2 2 2 2 2 2 21s 1s 2s 2s 2p 2p 2p 2p 2p

2 2 2 2 2 2 1 12p 3s 3s 3p 3p 3p 3p 3p

* * * *

* * * *

So, C2 and N2 are diamagnetic.

33. A solution of (–) – 1 – chloro – 1 – phenylethane intoluene racemises slowly in the presene of a smallamount of SbCl5, due to the formation of

(1) Carbanion (2) Carbene

(3) Carbocation (4) Free radical

Answer (3)

Sol.

CH3

H — C — Cl

C H6 5

(–)

SbCl5 C+

H

CH3

C H6 5

+ SbCl6

–

H — C Cl + Cl C H— — —

CH3 CH3

C H6 5 C H6 5

(Racemic mix.)34. Given

3+ – 2+4

0 0Cr /Cr MnO /Mn

E = – 0.74 V;E = 1.51 V

2– 3 –2 7

0 0Cr O /Cr Cl/Cl

E = 1.33 V;E = 1.36 V

Based on the data given above, strongest oxidisingagent will be

(1) Cl– (2) Cr3+

(3) Mn2+ (4) –4MnO

Answer (4)Sol. Of the given values of standard reduction potential,

the value of 24MnO /Mn

E is highest. Therefore,

4MnO is the strongest reducing agent.

35. A piston filled with 0.04 mol of an ideal gasexpands reversibly from 50.0 mL to 375 mL at aconstant temperature of 37.0°C. As it does so, itabsorbs 208 J of heat. The values of q and w for theprocess will be

(R = 8.314 J/mol K) (ln 7.5 = 2.01)(1) q = + 208 J, w = – 208 J(2) q = – 208 J, w = – 208 J

(3) q = – 208 J, w = + 208 J(4) q = + 208 J, w = + 208 J

Answer (1)Sol. For isothermal reversible expansion of an ideal gas,

the work done is given by

rev375W 0.04 8.314 310 ln50

–207.22 208 J

revq –W 208 J

PART–B : CHEMISTRY

(10)

36. The molarity of a solution obtained by mixing750 mL of 0.5 (M) HCl with 250 mL of 2 (M) HClwill be

(1) 0.875 M

(2) 1.00 M

(3) 1.75 M

(4) 0.975 M

Answer (1)

Sol. The molarity of a solution obtained by mixing750 mL of 0.5 M HCl with 250 mL of 2 M HCl isgiven by

750 0.5 250 2Molarity 0.875 M1000

37. Arrange the following compounds in order ofdecreasing acidity

OH

Cl(I)

OH

CH3(II)

OH

NO2(III)

; ; ;

OH

OCH3(IV)

(1) II > IV > I > III

(2) I > II > III > IV

(3) III > I > II > IV

(4) IV > III > I > II

Answer (3)

Sol. The acidic strength of the given derivatives ofphenol is decided by the stability of their conjugatebases. p-nitrophenol is most acidic as –I and –Reffects of NO2 group stabilises its conjugate basemost effectively. This is followed by p-chlorophenoldue to –I effect of Cl group, p-cresol due to +I and+H effects of CH3 group which destabilises itsconjugate base. p-methoxphenol is least acidic dueto +R effect of OCH3 group.

38. For gaseous state, if most probable speed is denotedby C*, average speed by C and mean square speedby C, then for a large number of molecules the ratiosof these speeds are

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

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

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

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

Answer (3)*

Sol. The ratio of most probable speed (C*), average speed

(C) and root mean square speed (C) is given by

* 8C : C : C 2 : : 3

1 : 1.128 : 1.225

*(Note: root mean square speed has been wronglywritten in the question as mean square speed).

39. 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 log2 = 0.301)

(1) 53.6 kJ mol–1

(2) 48.6 kJ mol–1

(3) 58.5 kJ mol–1

(4) 60.5 kJ mol–1

Answer (1)

Sol.

a2

1 1 2

Ek 1 1logk 2.303R T T

aE 1 1log 2

2.303 8.314 300 310

a

0.301 2.303 8.314 300 310E10

1 153600 J mol or 53.6 kJ mol

40. A compound with molecular mass 180 is acylatedwith CH3COCl to get a compound with molecularmass 390. The number of amino groups present permolecule of the former compound is

(1) 2 (2) 5

(3) 4 (4) 6

Answer (2)

Sol.

2 3 3M M 42

R NH CH COCl RNHCOCH HCl

If the increase in molecular mass is 42, the reactanthas one NH2 group.

Actual increase in molecular mass = 390 –180 = 210

Number of NH2 groups 210 542

(11)

41. Which of the following arrangements does notrepresent the correct order of the property statedagainst it?

(1) V2+ < Cr2+ < Mn2+ < Fe2+ : paramagneticbehaviour

(2) Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size

(3) Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueoussolution

(4) Sc < Ti < Cr < Mn : number of oxidation states

Answer (1)

Sol. Number of unpaired electrons of V2+, Cr2+, Mn2+ andFe2+ are 3, 4, 5 and 4 respectively. Hence the givenorder of paramagnetic behaviour is incorrect.

42. The order of stability of the following carbocations

CH = CH – CH ; CH – CH – CH ; 2 2 3 2 2

I II

CH2

III

is

(1) III > II > I (2) II > III > I

(3) I > II > III (4) III > I > II

Answer (4)

Sol. Benzyl carbocation (III) is more stable than allylcarbocation (I) as it has more number of resonatingstructures. n-propyl carbocation (II) is least stable asit is stabilised by +I and +H effects of ethyl group.

43. Consider the following reaction :

– 2–4 2 4xMnO yC O zH

22 2

zxMn 2yCO H O2

The values of x, y and z in the reaction are,respectively

(1) 5, 2 and 16 (2) 2, 5 and 8

(3) 2, 5 and 16 (4) 5, 2 and 8

Answer (3)

Sol. 2 24 2 4 2 22MnO 5C O 16H 2Mn 10CO 8H O

44. Which of the following is the wrong statement ?

(1) ONCl and ONO– are not isoelectronic.

(2) O3 molecule is bent.

(3) Ozone is violet-black in solid state.

(4) Ozone is diamagnetic gas.

Answer (No correct answer)

Sol. All the four given options are correct.

45. A gaseous hydrocarbon gives upon combustion0.72 g of water and 3.08 g of CO2. The empiricalformula of the hydrocarbon is

(1) C2H4 (2) C3H4

(3) C6H5 (4) C7H8

Answer (4)

Sol.

x y 2 2 2y y

C H x O xCO H O4 2

Number of moles of 23.08CO 0.07; x 0.0744

Number of moles of 20.72H O 0.04; y 0.0818

x : y = 7 : 8.

Therefore, empirical formula of hydrocarbon is C7H8

46. In which of the following pairs of molecules/ions,both the species are not likely to exist?

(1) 2–2 2H , He (2) – 2–

2 2H , He

(3) 22 2H , He (4) – 2

2 2H , He

Answer (3)

Sol. Bond order of each of 22H and He2 is zero.

47. Which of the following exists as covalent crystals inthe solid state ?

(1) Iodine (2) Silicon

(3) Sulphur (4) Phosphorus

Answer (2)

Sol. Silicon is a covalent crystal in which silicon atomsare covalently bonded to give three dimensionalnetwork.

48. Synthesis of each molecule of glucose inphotosynthesis involves

(1) 18 molecules of ATP

(2) 10 molecules of ATP

(3) 8 molecules of ATP

(4) 6 molecules of ATP

Answer (1)

Sol. 3 ATP molecules are needed per molecule of CO2.Since one molecule of glucose has 6 C-atoms, thenumber of ATP molecules required is 18.

(12)

49. The coagulating power of electrolytes having ionsNa+, Al3+ and Ba2+ for arsenic sulphide sol increasesin the order

(1) Al3+ < Ba2+ < Na+

(2) Na+ < Ba2+ < Al3+

(3) Ba2+ < Na+ < Al3+

(4) Al3+ < Na+ < Ba2+

Answer (2)

Sol. The coagulating power of cation to causecoagulation of arsenic sulphide sol increases withthe increase of valency of the cation.

50. Which of the following represents the correct orderof increasing first ionization enthalpy for Ca, Ba, S,Se and Ar?

(1) Ca < S < Ba < Se < Ar

(2) S < Se < Ca < Ba < Ar

(3) Ba < Ca < Se < S < Ar

(4) Ca < Ba < S < Se < Ar

Answer (3)

Sol. The increasing order of first ionisation enthalphy ofthe given elements is Ba < Ca < Se < S < Ar .

51. Energy of an electron is given by

218

2ZE 2.178 10 J .n Wavelength of light

required to excite an electron in an hydrogen atomform level n = 1 to n = 2 will be

(h = 6.62 × 10–34 Js and c = 3.0 × 108 ms–1)

(1) 1.214 × 10–7 m

(2) 2.816 × 10–7 m

(3) 6.500 × 10–7 m

(4) 8.500 × 10–7 m

Answer (1)

Sol. 181E 2.178 10 J

182

1E 2.178 10 J4

182 1

hc 3E E 2.178 104

34 87

186.62 10 3 10 4 1.214 10 m

2.178 3 10

52. 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).

(1)

CH Br2

CH3

(2)

C H2 5

Br

(3)

CH3

CH Br2

(4)

CH Br2

CH3

Answer (4)

Sol.

CH Br2

CH3

alc. AgNO3

CH NO2 3

CH3

+ AgBr(Light yellow ppt.)(A)

OxidationCOOH

COOH

CO

COO + H O2

(B)

Note: Compound (A) reacts with alc. AgNO3 to givelight yellow ppt. of AgBr and not white ppt. asgiven in the question.

53. Four successive members of the first row transitionelements are listed below with atomic numbers.Which one of them is expected to have the highest

3 20M /M

E value?

(1) Cr (Z = 24)

(2) Mn (Z = 25)

(3) Fe (Z = 26)

(4) Co (Z = 27)

Answer (4)

Sol. Cobalt has the highest value of 3 2Co /CoE 1.97 V.

Other values of standard reduction potential are

3 2Mn /MnE 1.57 V, 3 2Fe /Fe

E 0 0.77 V and

3 2Cr /Cr

E 0.41 V .

(13)

54. How many litres of water must be added to 1 litreof an aqueous solution of HCl with a pH of 1 tocreate an aqueous solution with pH of 2?

(1) 0.1 L (2) 0.9 L

(3) 2.0 L (4) 9.0 L

Answer (4)

Sol. [H+] of the given HCl solution = 0.1 M

Let x litre of water be added to 1 litre of the givenHCl solution to get pH = 2 or [H+] = 10–2 M

0.1 0.01 x 9 litre1 x

55. The first ionisation potential of Na is 5.1 eV. Thevalue of electron gain enthalpy of Na+ will be

(1) –2.55 eV (2) –5.1 eV

(3) –10.2 eV (4) +2.55 eV

Answer (2)

Sol. IENa(g) Na (g) e, H 5.1 eV

egNa (g) e Na(g), H 5.1 eV

56. An organic compound A upon reacting with NH3gives B. On heating, B gives C. C in presence ofKOH reacts with Br2 to give CH3CH2NH2. A is

(1) CH3COOH

(2) CH3CH2CH2COOH

(3) CH3 CH COOH

CH3

(4) CH3CH2COOH

Answer (4)

Sol. CH CH COOH + NH3 2 3 CH CH CO NH3 2 2 4

CH CH CONH3 2 2 2 + H OCH CH NH3 2 2

(A) (B)

(C)

KOH + Br2

57. Stability of the species Li2, Li2– and Li2+ increases in

the order of

(1) Li2 < Li2+ < Li2

–

(2) Li2– < Li2

+ < Li2(3) Li2 < Li2– < Li2

+

(4) Li2– < Li2 < Li2

+

Answer (2)

Sol. 2 2 2*2 1s 1s 1sLi : Bond order = 1

2 2 1*2 1s 1s 2sLi : Bond order = 0.5

2 2 2 1* *2 1s 1s 2s 2sLi : Bond order = 0.5

2Li is less stable than 2Li because the incomingelectron goes to antibonding molecular orbital.

58. An unknown alcohol is treated with the "Lucasreagent" to determine whether the alcohol isprimary, secondary or tertiary. Which alcohol reactsfastest and by what mechanism?

(1) Secondary alcohol by SN1

(2) Tertiary alcohol by SN1

(3) Secondary alcohol by SN2

(4) Tertiary alcohol by SN2

Answer (2)

Sol. Tertiary alcohol gives instant turbidity with Lucasreagent and the reaction follows SN1 mechanismdue to stability of tertiary carbocation.

59. The gas leaked from a storage tank of the UnionCarbide plant in Bhopal gas tragedy was

(1) Methylisocyanate (2) Methylamine

(3) Ammonia (4) Phosgene

Answer (1)

Sol. Methylisocynate (CH3NCO) a deadly poisonous gasleaked from a storage tank of the Union CarbidePlant in Bhopal gas tragedy.

60. 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

(1) 7.01% (2) 4.08%

(3) 6.05% (4) 5.08%

Answer (2)

Sol. Let the number of moles of M3+ be x and that of M2+

is (0.98 – x)

Number of moles of oxide ions = 3x0.98 x 12

or, x = 0.04

Percentage of M3+ = 4.08%

(14)

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

(1)32

(2)52

(3)72

(4)92

Answer (3)

Sol. Distance between the planes

5 8 72 units3 2

62. At present, a firm is manufacturing 2000 items. It isestimated that the rate of change of production Pw.r.t. additional number of workers x is given by

100 12dP xdx

. If the firm employs 25 more

workers, then the new level of production of itemsis

(1) 2500 (2) 3000

(3) 3500 (4) 45000

Answer (3)

Sol. 25

2000 0

100 – 12P

dp x dx

⇒ P – 2000 = 2500 – 212 × × 1253

= 1500

⇒ P = 3500

63. Let A and B be two sets containing 2 elements and4 elements respectively. The number of subsets ofA × B having 3 or more elements is

(1) 256 (2) 220

(3) 219 (4) 211

Answer (3)

Sol. n(A × B) = 2 × 4 = 8

The number of subsets of A × B having 3 or moreelements.

= 8C3 + 8C4 + ... + 8C8

= 28 – 8C0 – 8C1 – 8C2

= 256 – 1 – 8 – 28

= 219

64. If the lines

32 41 1

yx zk and

41 52 1

yx zk are coplanar, then k can have

(1) Any value (2) Exactly one value

(3) Exactly two values (4) Exactly three values

Answer (3)

Sol. Given lines are coplanar if 1 1 –

2 11 –1 –1

kk = 0

⇒ 1 (– 2 + 1) – 1 (– k – 1) – k (– k – 2) = 0

– 1 + k + 1 + k2 + 2k = 0

⇒ k = 0 or – 3

∴ Exactly two values of k.

65. If the vectors ˆˆ3 4AB i k and

ˆˆ ˆ5 2 4AC i j k arethe sides of a triangle ABC, then the length of themedian through A is

(1) 18 (2) 72

(3) 33 (4) 45

Answer (3)

Sol.AD =

2AB + AC

A B

D

C

5 – 2 +

4

ij

k

3 + 4i k

= ˆ ˆ ˆˆ ˆ – 2 j 3 + 4 5 4

2

i k i k

= ˆ ˆˆ – j 4 4i k

∴

| |AD = 16 + 1 + 16 = 33

PART–C : MATHEMATICS

(15)

66. The real number k for which the equation 32 3 0x x k has two distinct real roots in [0, 1]

(1) Lies between 1 and 2

(2) Lies between 2 and 3

(3) Lies between –1 and 0

(4) Does not exist

Answer (4)

Sol. Let f(x) = 2x3 + 3x + k

f ′(x) = 6x2 + 3 > 0, x∈ R

∴ f(x) is strictly increasing function for all real values of k.

∴ No real k exists such that equation has two distinct roots in [0, 1].

67. The sum of first 20 terms of the sequence

0.7, 0.77, 0.777, ......, is

(1) 207(179 10 )

81(2) 207

(99 10 )9

(3) 207 (179 10 )81 (4) 207 (99 10 )

9

Answer (3)

Sol. S = 0.7 + 0.77 + 0.777 + ... upto 20 terms

= .9 .99 .999 .... 79

= 79

1 – 0.1 + 1 – 0.01 + 1 – 0.001+ ...upto 20 terms

=

2 3 20

7 1 1 1 120 – + + + ... +9 10 10 10 10

= 20

1 11–10 10

11–10

7 20 –9

=

20

7 1 120 – 1 –9 9 10

= 201

10

7 179 +81

–207 179 10

81

68. A ray of light along 3 3x y gets reflectedupon reaching x-axis, the equation of the reflectedray is

(1) 3y x (2) 3 3y x

(3) 3 3y x (4) 3 1y xAnswer (2)Sol.

30°0 ( 3, 0)

y

x

Reflected ray

Slope of incident ray = 13

⇒ α = 150°∴ Slope of reflected ray = tan30°

= 13

∴ Reflected ray is y = 1 ( 3)3

x

⇒ 3 3y x69. The number of values of k, for which the system of

equations(k + 1)x + 8y = 4kkx + (k + 3)y = 3k – 1has no solution, is(1) Infinite (2) 1(3) 2 (4) 3

Answer (2)

Sol.

1 80

3k

k k⇒ k2 + 4k + 3 – 8k = 0⇒ k = 1, 3When k = 1, equation change to

2x + 8y = 4 ⇒ x + 4y = 2and x + 4y = 2 ⇒ x + 4y = 2⇒ Infinitely many solutionsWhen k = 3

4x + 8y = 12 ⇒ k + 2y = 3

and 3x + 6y = 8 and x + 2y = 83

⇒ No solution∴ One value of k exists for which system of

equation has no solution.

(16)

70. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0,a, b, c ∈ R, have a common root, then a : b : c is

(1) 1 : 2 : 3 (2) 3 : 2 : 1

(3) 1 : 3 : 2 (4) 3 : 1 : 2

Answer (1)

Sol. The equation x2 + 2x + 3 = 0 has complex rootsand coefficients of both equations are real.

∴ Both roots are common.

∴ 1 2 3a b c

71. The circle passing through (1, – 2) and touching theaxis of x at (3, 0) also passes through the point

(1) (–5, 2) (2) (2, –5)

(3) (5, –2) (4) (–2, 5)

Answer (3)

Sol. Let the circle be (x – 3)2 + (y + k)2 = k2

(3, – )k

It passes through (1, –2)

4 + (4 + k2 –4k) = k2

⇒ k = 2

∴ The circle is (x –3)2 + (y + 2)2 = 4

Clearly the point (5, –2) lies on it.

72. If x, y, z are in A.P. and tan–1x, tan–1y and tan–1z arealso in A.P., then

(1) x = y = z (2) 2x = 3y = 6z

(3) 6x = 3y = 2z (4) 6x = 4y = 3z

Answer (1)

Sol. 2y = x + z and 2tan–1y =

–1tan

1x z

xz

⇒

22

11y x z

xyy

⇒ 2y xz

⇒ x, y, z are in GP

∴ x = y = z

73. Consider :

Statement - I : (p ∧ ∼ q) ∧ (∼ p ∧ q) is a fallacy.

Statement - II : (p q) (∼ q ∼ p) is atautology.

(1) Statement - I is true; Statement-II is true;Statement - II is a correct explanation forStatement - I.

(2) Statement - I is true; Statement - II is true;Statement-II is not a correct explanation forStatement-I.

(3) Statement-I is true; Statement - II is false.

(4) Statement - I is false; Statement - II is true.

Answer (2)

Sol.

( ) (~ ~ )T T F F T TT F F T F FF T T F T TF F T T T T

p q a p q b q p~ p ~ q

( ~ ) (~ )T FT FT FT F

a b p q p q

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

(1) 3 3 2 31 ( ) ( )3

x x x x dx C

(2) 3 3 3 31 3 ( )3

x x x x dx C

(3) 3 3 2 31 ( )3

x x x x dx C

(4) 3 3 3 31 ( )3

x x x x dx C

Answer (3)

Sol. 5 3( )x f x dx

3 3 21 ( ) 33

x f x x dx

3 3 2 2 3 21 1( ) 3 3 ( ) 33 3

x f x x dx x f x x dx dx

3 3 2 31 ( ) ( )3

x x x x dx C

(17)

75.

0

(1 cos 2 )(3 cos )tan 4x

x xlimx x is equal to

(1) 14

(2)12

(3) 1 (4) 2Answer (4)

Sol.

2

0 2

2sin (3 cos )lim tan 444

x

x xxx

x

2 4 2

476. Statement - I :

The value of the integral

3

6

1 tandx

x is equal to 6

Statement - II :

( ) ( ) .b b

a a

f x dx f a b x dx

(1) Statement - I is true; Statement - II is true;Statement - II is a correct explanation forStatement - I.

(2) Statement - I is true; Statement - II is true;Statement - II is not a correct explanation forStatement - I.

(3) Statement - I is true; Statement - II is false.(4) Statement - I is false; Statement - II is true.

Answer (4)Sol. Statement (1)

3

61 tan

dxIx

⇒3

61 cot

dxIx

⇒3

6

2

I dx

⇒ 26

I

12

I

Statement (1) is false, Statement (2) is true.

77. The equation of the circle passing through the foci

of the ellipse 22

116 9

yx, and having centre at

(0, 3) is

(1) x2 + y2 – 6y – 7 = 0

(2) x2 + y2 – 6y + 7 = 0

(3) x2 + y2 – 6y – 5 = 0

(4) x2 + y2 – 6y + 5 = 0

Answer (1)

Sol.22

116 9

yx

9 = 16 (1 – e2)

2 716

e

74

e

( 7 , 0) foci

Equation of required circle is

(x – 0)2 + (y – 3)2 = 7 + 9

⇒ x2 + y2 – 6y – 7 = 0

78. A multiple choice examination has 5 questions.Each question has three alternative answers ofwhich exactly one is correct. The probability that astudent will get 4 or more correct answers just byguessing is

(1) 5173

(2) 5133

(3) 5113

(4) 5103

Answer (3)

Sol. Required probability

4 55 5

4 51 2 13 3 3

C C

51 2 15

81 3 3

5 5 5

10 1 113 3 3

(18)

79. The x-coordinate of the incentre of the triangle thathas the coordinates of mid points of its sides as(0, 1), (1, 1) and (1, 0) is

(1) 2 2 (2) 2 2

(3) 1 2 (4) 1 2

Answer (2)

Sol.

(1, 1)

(2, 0)(1, 0)

(0, 2)

(0, 1)

A

B

O (0, 0)

Required triangle is ΔOAB

So, x co-ordinate of incentre

2 0 2 2 2 2 02 2 2 2

44 2 2

22 2

2 2

80. The term independent of x in expansion of

10

2 1 13 3 2

1 1

1

x x

x xx x is

(1) 4 (2) 120

(3) 210 (4) 310

Answer (3)

Sol. Given expression can be written as

101/2 101/3 1/3 1/2

1/211

xx x xx

General term = 1010 1/3 1/2 r r

rC x x

From question,

10 03 3 2

r r

⇒ r = 4

i.e., constant term 104 210 C

81. The area (in square units) bounded by the curves

y = , 2 3 0x y x , x-axis, and lying in the firstquadrant is

(1) 9 (2) 36

(3) 18 (4)274

Answer (1)

Sol.

(3, 0)

(9, 3)

Required area9

0

1 6 32

x dx

= 18 – 9 = 9

82. Let Tn be the number of all possible triangles formedby joining vertices of an n-sided regular polygon. IfTn+1 – Tn = 10, then the value of n is :

(1) 7 (2) 5

(3) 10 (4) 8

Answer (2)

Sol. 13 3 10 n nC C

⇒ 2 10nC

⇒ n = 5

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

argument θ, then arg

11

zz

equals

(1) – θ (2)

2

(3) θ (4) π – θ

Answer (3)

Sol.1arg1

zz

arg1

zz zz

= arg (z)

= θ

(19)

84. ABCD is a trapezium such that AB and CD areparallel and BC ⊥ CD. If ∠ADB = θ, BC = p andCD = q, then AB is equal to

(1)

2 2( )sincos sinp q

p q

(2)

2 2 coscos sinp q

p q

(3)

2 2

2 2cos sinp q

p q

(4)

2 2

2( )sin

( cos sin )p q

p q

Answer (1)

Sol.

qC

p

BA

D

pq

2

2

+

sin sin( )

AB BD

⇒

2 2· sinsin · cos cos · sin

p qAB

2 2

2 2 2 2

· sin

sin · cos ·

p qq p

p q p q

2( )sincos sinp q

p q

85. If

1 31 3 32 4 4

P is the adjoint of a 3 × 3 matrix A

and |A| = 4, then α is equal to

(1) 4 (2) 11

(3) 5 (4) 0

Answer (2)

Sol.

1 31 3 32 4 4

|P| = 1(12 – 12) – α(4 – 6) + 3(4 – 6)

= 2α – 6

Also, |P| = |A|2 = 16

2α = 22

α = 11

86. The intercepts on x-axis made by tangents to the

curve, 0

| | ,x

y t dt x R , which are parallel to the

line y = 2x, are equal to

(1) ± 1 (2) ± 2

(3) ± 3 (4) ± 4

Answer (1)

Sol. 0

xy t dt

| | 2dy

xdx

x = ±2

Case-1, x = 2

2

0

2y t dt

Equation of tangent is y – 2 = 2(x – 2)

⇒

12 1

y x

x-intercept = 1

When x = –2

22 2

0 02ty t dt

= –2

y + 2 = 2(x + y)

⇒ y = 2x + 2

Hence, here x-intercept is –1

∴ x-intercepts = ±1

(20)

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

2 4 5y x

Statement-I : An equation of a common tangent to

these curves is 5y x .

Statement-II : If the line, 5y mx

m (m ≠ 0) is their

common tangent, then m satisfies m4 – 3m2 + 2 = 0.

(1) Statement-I is true; statement-II is true;statement-II is a correct explanation forstatement-I

(2) Statement-I is true; statement-II is true;statement-II is not a correct explanation forstatement-I

(3) Statement-I is true; statement-II is false

(4) Statement-I is false; statement-II, is true

Answer (2)

Sol. 2 2 52

x y

2 4 5y x

Equation of tangent to parabola is

5y mx

m…(i)

For circle,

25 12

y mx m …(ii)

(i) and (ii) are identical,

22

5 5 (1 )2

mm2 = m4 + m2

⇒ m4 + m2 – 2 = 0

⇒ m = ±1

which satisfy given equation

Statement (1) is true and statement (2) is true.

88. If 1sec(tan ),y x then dydx

at x = 1 is equal to

(1)12 (2)

12

(3) 1 (4) 2

Answer (1)

Sol. –1 2 2secsec 1 1y x x

21

dy xdx x

1

12x

dydx

89. The expression tan cot

1 cot 1 tanA A

A A can be written

as

(1) sinA cosA + 1 (2) secA cosecA + 1

(3) tanA + cotA (4) secA + cosecA

Answer (2)

Sol. tan cot

1 cot 1 tanA A

A A

2tan cottan 1 1 tan

A AA A

2tan cottan 1

A AA

3tan 1tan (tan 1)

AA A

2tan tan 1tan

A AA

= tanA + 1 + cotA

sin cos 1cos sin

A AA A

1 sin cossin · cos

A AA A

= 1 + secA· cosecA

90. All the students of a class performed poorly inMathematics. The teacher decided to give gracemarks of 10 to each of the students. Which of thefollowing statistical measures will not change evenafter the grace marks were given?

(1) Mean (2) Median

(3) Mode (4) Variance

Answer (4)

Sol. With increase in data, mean will also increase bythe same, hence variance will remain unchanged.