JEE Main Examination 02-04-2017 Code-A - Career...

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[ CODE – A ]

JEE Main Exam 2017 (Paper & Solution)

Code – A Date : 02-04-2017

Part A – PHYSICS Q.1 A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density

remains same, the stress in the leg will change by a factor of-

(1) 9 (2) 91 (3) 81 (4)

811

Ans. [1] Sol. Let volume of man is Lbh As all dimension increases by a factor (K = 9) keeping the density constant

Stress on his legs = area

weight = A

gVρ

Initial stress = Stress1 = A

gVρ

Final stress = Stress2 = AK

gVK2

3 ρ

Stress2 = K Stress1 Where (K = 9) So stress is changed by a factor 9. Q.2 A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ?

(1) v

t

(2) v

t (3) v

t (4) v

t

Ans. [3] Sol. v = u – gt (straight line graph)

t

v

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Q.3 A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is

v0 = 10 ms–1. If, after 10 s, its energy is 20mv

81 , the value of k will be-

(1) 10–3 kg m–1 (2) 10–3 kg s–1 (3) 10–4 kg m–1 (4) 10–1 kg m–1 s–1 Ans. [3]

Sol. a = – m

kv2

dtdv = –

mkv2

∫v

102v

dv = – ∫10

0

dtmk

v

10v1

⎥⎦⎤

⎢⎣⎡− = –

mk × 10

– ⎥⎦⎤

⎢⎣⎡ −

101

v1 =

210k−

× 10

– 101

v1

+ = – k × 1000

According to question

KE = 21 mv2 = 2

0mv81

v = 2v0 =

210

– 101 × 2 +

101 = – k × 1000

101 = k × 1000

k = 10–4

Q.4 A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be-

(1) 4.5 J (2) 22 J (3) 9 J (4) 18 J Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Work-Power-Energy, Ex.2, Page No.25, Q. No.13]

Ans. [1]

Sol. a = 1t6 = 6 t

dtdv = 6t

v = 1

0

2

2t6

⎥⎦

⎤⎢⎣

⎡ = 3 × 12 = 3

KE = 21 × 1 × 32 = 4.5

W = ΔKE = 4.5 – 0 = 4.5 Joule

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Q.5 The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum ?

(1) 23 (2)

23 (3) 1 (4)

23

Ans. [1] Sol.

R

l

Moment of inertia of cylinder about perpendicular bisector is I

I = ⎥⎥⎦

⎤

⎢⎢⎣

⎡+

4R

12LM

22

For given mass and density

M = πR2Lρ

R2 = ρπL

M

I = M⎥⎥⎦

⎤

⎢⎢⎣

⎡

ρπ+

L4M

12L2

For maxima or minima of I

dLdI = 0

dLdI = M

⎥⎥⎦

⎤

⎢⎢⎣

⎡

ρπ− 2L4

M12

L2 = 0

6L =

ρπ 2L4M

6L =

ρπ

ρπ2

2

L4LR

2

2

RL =

23

23

RL

=

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Q.6 A slender uniform rod of mass M and length l is pivoted at one end so that is can rotate in vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is-

z

x

θ

(1) θsin2g3l

(2) θsin3g2l

(3) θcos2g3l

(4) θcos3g2l

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Rotational motion, Ex.3, Page No., 36, Q. No.11]

Ans. [1] Sol.

z

x

λ/2

α

θ mg

θsin2l

θ

τ = mg θsin2l

τ = Iα

mg θsin2l = α

3m 2l

α = θsing23

l

Q.7 The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by

(R = Earth's radius)-

(1)

g

O d (2)

g

O d R

(3)

g

O d R

(4)

g

O d R

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Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Gravitation, Ex.4, Page No., 42, Q. No.20]

Ans. [4] Sol.

R

g = 2rGM when r > R

g = 3RrGM when r < R

g

r R

g ∝ r g∝2r

1

Q.8 A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled

with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75ºC. T is given by- (Given : room temperature = 30ºC, specific heat of copper = 0.1 cal/gmºC)

(1) 800ºC (2) 885ºC (3) 1250ºC (4) 825ºC Ans. [2] Sol. Total heat gain = Total heat loss 100 × 0.1 (75 – 30) + 170 × 1 × (75 – 30) = 100 × 0.1 (T – 75) 10 × 45 + 170 × 45 = 10T – 750 1200 + 7650 = 10 T T = 885ºC Q.9 An external pressure P is applied on a cube at 0ºC so that it is equally compressed from all sides. K is the bulk

modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by-

(1) K3

Pα

(2) KP

α (3)

PK3α (4) 3PKα

Ans. [1]

Sol. B =

VdV

P−

V

dV = BP−

dV = BPV−

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By heating we have to increase the volume by B

PV

ΔV = VγΔT = V × 3α ΔT

V × 3αΔT = B

PV

ΔT = B3

Pα

Here B = K

∴ ΔT = B3

Pα

Q.10 Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that Cp – Cv = a for hydrogen gas Cp – Cv = b for nitrogen gas The correct relation between a and b is-

(1) a = b141 (2) a = b (3) a = 14 b (4) a = 28 b

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : KTG, Ex.1, Page No.17, Q. No.35] [JEE Advance, Chapter : KTG, Ex.1, Page No.25,Q. No.22]

Ans. [3] Sol. Cp – Cv = R If Cp and Cv are molar specific heat But if Cp and Cv are specific heat i.e. gram specific heat then C = MSg {Q = 1CΔT ⇒ Q = MSgΔT ⇒ C = MSg}

Sg = MC

MSgp – MSgv = R

Sgp –Sgv = MR

2R = a

28R = b

14 = ba

a = 14 b

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Q.11 The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1× 105 Pa. In ni and nf are the number of molecules in the room before and after heating, the nf – ni will be :

(1) – 1.61 × 1023 (2) 1.38 × 1023 (3) 2.5 × 1025 (4) – 2.5 × 1025

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : KTG, Ex.3, Page No.27, Q. No.5]

Ans. [4] Sol. PV = nRT

n = RTPV

Ti = 273 + 17 = 290 K Tf = 273 + 27 = 300 K

nf – ni = 314.8

30105 × ⎥⎦⎤

⎢⎣⎡

2901–

3001 × 6.023 × 1023

= 314.8103 6× ⎥⎦

⎤⎢⎣⎡

× 29030010– × 6.023 × 1023

= –

293314.8023.6103 27

××××

= –

29314.810023.6– 27

××

= – 0.025 × 1027 = – 2.5 ×1025 Q.12 A particle is executing simple harmonic motion with a time period T. At time t = 0 it is at its position of

equilibrium. The kinetic energy - time graph of the particle will look like :

(1)

0

KE

T T t 2T

(2) 0

KE

T t

(3)

0

KE

T t 2T

(4) 0

KE

t 4T

2T T

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : SHM, Ex.1, Page No.27, Q. No.33]

Ans. [4] Sol. x = A sin ωt

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v = dtdx = Aω cos ωt

KE =21 mA2ω2cos2ωt

=21 mA2ω2cos2

T2π t

0

KE

t 4T

2T T

Q.13 An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer ? (speed of light = 3 × 108 ms–1)

(1) 10.1 GHz (2) 12.1 GHz (3) 17.3 GHz (4) 15.3 GHz Ans. [3] Sol. According to theory of relativity

fapp = f

cv–1

cv1

2

2

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡+

(for approach)

= 2

c2/c–1

c2/c1

⎟⎠⎞

⎜⎝⎛

+ × 10 GHz

=

43

23

× 10 GHz

= 3 × 10 GHz fapp = 17.32 GHz Q.14 An electric dipole has fixed dipole moment p

r, which makes angle θ with respect to x-axis . When subjected

to an electric field 1Er

= E i , it experience a torque 1Tr

= kτ . When subjected to another electric field

2Er

= 3 E1 j it experiences a torque 2Tr

= – 1Tr

. The angle θ is. (1) 30° (2) 45° (3) 60° (4) 90°

Ans. [3] Sol.

E1 = E p

x θ

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τ = 1Tr

= PE sinθ k

p

x θ

E2 = jE3 1

τ = 2T

r = )k(–cosE3P θ

PE sin θ ( k ) = )k(–cosE3P– θ

PE sin θ = θcosE3

tan θ = 3 θ = 60° Q.15 A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large

number of 1μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :

(1) 2 (2) 16 (3) 24 (4) 32 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Capacitance, Ex.3, Page No.47, Q. No.12]

Ans. [4] Sol.

n capacitor in a row

m rows

Potential on each capacitor V = n

1000

n

1000 = 300

n = 3

10 ≈ 4

Ceq = mnC

×

mn1

× = 2

m = n × 2 = 4 × 2 = 8 Minimum number of capacitor = 8 × 4 = 32

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Q.16 In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:

E r

C

r1

r2

(1) CE (2) CE )rr(

r

2

1

+ (3) CE

)rr(r

2

2

+ (4) CE

)rr(r

1

1

+

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Capacitance, Ex.2, Page No.44, Q. No.25]

Ans. [3] Sol.

E r

C

r1

r2

i

At steady state current through capacitor branch become zero.

i = 2rr

E+

Potential difference across capacitor ΔV ΔV = ir2

ΔV = ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2rrE r2

charge on capacitor = C ΔV

= CE ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

rrr

Q.17 In the above circuit the current in each resistance is :

2V

2V

1Ω 1Ω

2V

2V

2V

2V

1Ω

(1) 1A (2) 0.25 A (3) 0.5 A (4) 0 A

Ans. [4]

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Sol. 2V

2V

1Ω 1Ω

2V

2V

2V

2V

1Ω

6 4 2 0

024

Potential difference across each resistor is zero so current in each resistor also zero. Q.18 A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is performing

simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is : (1) 6.65 s (2) 8.89 s (3) 6.98 s (4) 8.76 s

Ans. [1]

Sol. M = 6.7 × 10–2 Am2

I = 7.5 × 10–6 kg m2, B = 0.01 T τ = – MB sin θ

Ια = – MB θ (for small oscillations )

α = ⎟⎠⎞

⎜⎝⎛

IMB θ ⇒ ω =

IMB ⇒ T = 2π

MBI

T = 2π 01.0107.6

105.72–

6–

×××

⇒ T = 0.6644 sec Time for 10 oscillation Δt = 10 T ⇒ Δt = 6.65 sec Q.19 When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω, it shows full

scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 – 10 V is

(1) 1.985 × 103 Ω (2) 2.045 × 103 Ω (3) 2.535 × 103 Ω (4) 4.005 × 103 Ω Ans. [1] Sol. Ig max = 5 mA , Rg = 15 Ω Range of voltmeter = 10 volt

RH Rg

ΔV

Ig

ΔV = Ig (Rg + RH) Range ΔVmax = Ig max (Rg + RH) 10 = 5 × 10–3 (15 + RH) RH = 1985 Ω

RH = 1.985 × 103 Ω

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Q.20 In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is :

Time 0.5 sec

Current(amp.)

10

(1) 200 Wb (2) 225 Wb (3) 250 Wb (4) 275 Wb

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : EMI, Ex.1, Page No.31, Q. No.1]

Ans. [3] Sol.

Time 0.5 sec

Current(amp.)

10

Emf = iR

– dtdφ = iR

( )∫ φd– = ( )∫ idt R

(– Δφ) = ∫5.0

0

idtR

( )φΔ– = R (area of i – t curve)

Δφ = R (21 × 0.5 × 10)

Δφ = 100 (2.5) Δφ = 250 Wb Q.21 An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-ray. It

produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log λmin with log V is correctly represented in -

(1)

log λmin

log V (2)

log λmin

log V

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(3)

log λmin

log V (4)

log λmin

log V

Ans. [1] Sol.

log λmin

log V

K.E. = eV (K.E. = kinetic energy of electron) EPmax = K.E.

min

hcλ

= eV ⇒ λmin V = ⎟⎠⎞

⎜⎝⎛

ehc = constant

ln λmin + ln V = ln constant ln λmin = – ln V + ln (constant) Straight line of – ve slope

Q.22 A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed as -

(1) real and at a distance of 40 cm from convergent lens (2) virtual and at a distance of 40 cm from convergent lens (3) real and at a distance of 40 cm from the divergent lens (4) real and at a distance of 6 cm from the convergent lens

Ans. [1] Sol.

15 cm

ƒ = 25 cm ƒ = 20 cm Image form by diverging is at the focus of diverging lens. Now image form by diverging act as a source for converging lens. For converging lens object real at a distance 40 cm from it which is at (2ƒ).

u = – 2 ƒ, ƒ = + ƒ, v = + 2ƒ ⎟⎟⎠

⎞⎜⎜⎝

⎛=−

ƒ1

u1

v1

Final image real at a distance 2ƒ = 40 cm from converging lens.

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[ CODE – A ]

Q.23 In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is -

(1) 1.56 mm (2) 7.8 mm (3) 9.75 mm (4) 15.6 mm

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Wave nature of light : Interference, Ex.6, Page No.43, Q. No.23]

Ans. [2] Sol. d = 0.5 mm, D = 1.5 m, λ1 = 650 nm, λ2 = 520 nm

β1 = dD1λ = 3–

9–

105.05.110650

×

×× = 1.95 mm

β2 = dD2λ = 3–

9–

105.05.110520

×

×× = 1.56 mm

Least distance where their maxima again coincides from central maxima is = LCM of β1 & β2 = 7.8 mm

Q.24 A particle A of mass m and initial velocity v collides with a particle B of mass 2m which is at rest. The

collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is -

(1) 31

B

A =λλ (2)

B

A

λλ = 2 (3)

32

B

A =λλ (4)

21

B

A =λλ

Ans. [2] Sol.

m v

m/2 mv1

m/2v2⇒

v1 = 2m/m2m/m

+− v + 0

v1 = 2m/32m/ v

v1 = 3v

Similarly

v2 = 0 + 2m/m

m2+

× v

v2 = 2m/3

m2 v

v2 = 34 v

Q λ1 = 11vm

h , λ2 = 22vm

h

2

1

λλ =

11

22

vmvm =

m2/m .

3/v3/v4 = 2

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[ CODE – A ]

Q.25 Some energy levels of a molecule are shown in the figure. The ratio of the wavelength r = 2

1

λλ , is given by -

– E

– 34 E

– 2 E

– 3 E

λ2

λ1

(1) r = 34 (2) r =

32 (3) r =

43 (4) r =

31

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Atomic Structure, Ex.5, Q. No.27]

Ans. [4]

Sol. 1

hcλ

= (– E) – (– 2E)

1

hcλ

= E ….(i)

2

hcλ

= (– E) – ⎟⎠⎞

⎜⎝⎛−

3E4 = – E +

3E4 =

3E4E3 +− =

3E ….(ii)

By (i) & (ii)

r = 2

1

λλ =

3/EhcEhc

= 31

Q.26 A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At

sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by -

(1) t = 2T

3.1log2log (2) t = T

2log3.1log (3) t = T log (1.3) (4) t =

)3.1log(T

Ans. [2] Sol. A ⎯⎯→ B

A = A0e–λt ; B = A0 (1 – e–λt) ; AB = t

0

t0

eA)e1(A

λ−

λ−−

0.3 = eλt – 1 eλt = 1.3

λt = ln (1.3)

T

)2ln( t = ln (1.3)

t = T )2ln()3.1ln(

⇒ t = )2log(

)3.1log(T

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[ CODE – A ]

Q.27 In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be -

(1) 45º (2) 90º (3) 135º (4) 180º Ans. [4] Sol. In C-E amplifier phase difference between input-output voltage is 180º. Q.28 In amplitude modulation, sinusoidal carrier frequency used is denoted by ωc and the signal frequency is

denoted ωm. The bandwidth (Δωm) of the signal is such that Δωm << ωc. Which of the following frequencies is not contained in the modulated wave?

(1) ωm (2) ωc (3) ωm + ωc (4) ωc – ωm Ans. [1] Sol.

ωc – ωm ωc ωc + ωm

LSB USB

because Δωm << ωc

∴ ωm is not present in modulated wave. Q.29 Which of the following statements is false ?

(1) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude (2) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is

disturbed (3) A rheostat can be used as a potential divider (4) Kirchhoff’s second law represents energy conservation

Ans. [2] Sol.

G A B

P Q

S R

(I)

P

G

Q

SR

(II)

In Ist case for balance

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[ CODE – A ]

RP =

SQ

QP =

SR ….(i)

In IInd case for balance

QP =

SR ….(ii)

In both case Null point is same. Q.30 The following observations were taken for determining surface tension T of water by capillary method : diameter of capillary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m.

Using g = 9.80 m/s2 and the simplified relation T = 2

rhg × 103 N/m, the possible error in surface tension is

closest to - (1) 0.15% (2) 1.5% (3) 2.4% (4) 10%

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Practical Physics, Ex.2, Page No.70, Q. No.7]

Ans. [2] Sol. D = 1.25 × 10–2 m, ΔD = 0.01 × 10–2 m, h = 1.45 × 10–2 m, Δh = 0.01 ×10–2 m g = 9.80 m/s2

T = 2

rhg × 103 N/m

Tr–1h–1 = 2g ×103

Tr–1h–1 = constant

hh

rr

TT Δ

−Δ

−Δ = 0

⎟⎠⎞

⎜⎝⎛ Δ

+Δ

=Δ

hh

rr

TT

TTΔ = ⎟

⎟⎠

⎞⎜⎜⎝

⎛

×

×2–

2–

1025.11001.0 + ⎟

⎟⎠

⎞⎜⎜⎝

⎛

×

×2–

2–

1045.11001.0

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

Δ=

ΔΔ=Δ

=

rr

DD

r2Dr2D

% error = TTΔ × 100

≈ 1.5 %

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[ CODE – A ]

Part B – CHEMISTRY Q.31 Given

)g(CO)g(OC 22)graphite( ⎯→⎯+ ;

ΔrH° = – 393.5 kJ mol–1

H2(g) + 21 O2(g) ⎯→ H2O(1);

ΔrH° = – 285.8 kJ mol–1

CO2(g) + 2H2O(1) ⎯→ CH4(g) + 2O2(g); ΔrH° = + 890.3 kJ mol–1

Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction C(graphite) + 2H2(g) ⎯→ CH4(g) will be ;

(1) –74.8 kJ mol–1 (2) –144.0 kJ mol–1 (3) +74.8 kJ mol–1 (4) +144.0 kJ mol–1

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Energetic ,Exercise-1 Q. No.12] [JEE Advance, Chapter : Chemical Energetic ,Exercise-4 Q. No.11]

Ans. [1]

Sol. )g(CO)g(OC 22)graphite( ⎯→⎯+ ; ΔrH° = – 393.5 kJ .... (i)

H2(g) + 21 O2(g) ⎯→ H2O(1); ΔrH° = – 285.8 kJ .... (ii)

CO2(g) + 2H2O(1) ⎯→ CH4(g) + 2O2(g); ΔrH° = + 890.3 ..... (iii)

)g(CH)g(H2C 42)graphite( ⎯→⎯+ ; ΔrH° = ? .... (iv)

eq. (iv) = eq. (i) + 2 × eq (ii) + eq. (iii) = –393. 5 + 2 (–285.8) + 890.3 = –74.8 kJ/mol Q.32 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of

M2CO3 in g mol–1 is - (1) 118.6 (2) 11.86 (3) 1186 (4) 84.3

Ans. [4]

Sol. 01186.

21gm

32 CO HCl COM ⎯→⎯+

POAC on carbon

1

01186.x1

= ⇒ x = 84.3

Q.33 ΔU is equal to -

(1) Adiabatic work (2) Isothermal work (3) Isochoric work (4) Isobaric work

CAREER POINT


[ CODE – A ]

Ans. [1]

Sol. FLOT (According to first law of thermodynamics)

ΔE = q + w

If q = 0, ΔE = w ∴ adiabatic process

Q.34 The Tyndall effect is observed only when following conditions are satisfied -

(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used

(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

(1) (a) and (c) (2) (b) and (c) (3) (a) and (d) (4) (b) and (d)

Students may find similar question in CP exercise sheet :

[JEE Advance, Chapter : Surface Chemistry, Key Concept, Optical Properties]

Ans. [4]

Sol. Facts

Q.35 A metal crystallises in the face centred cubic structure. if the edge length of its unit cell is 'a', the closest approach

between two atoms in metallic crystal will be :

(1) a2 (2) 2

a (3) 2a (4) a22


[JEE Main, Chapter : Solid State, Solved Example, Q. No.21]

[JEE Advance, Chapter : Solid State, Solved Exercise-4, Q. No.4]

Ans. [2]

Sol. ∴ nearest distance = 2

a

2

a

fcc

CAREER POINT


[ CODE – A ]

Q.36 Given

V36.1E Cl/Cl2=°

− , V74.0E Cr/Cr3 −=°+

V33.1E 3272 Cr/OCr =°

+− , V51.1E 24 Mn/MnO =°

+− ,

Among the following, the strongest reducing agent is - (1) Cr3+ (2) Cl– (3) Cr (4) Mn2+

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Electro Chemistry, Exercise-4, Q. No.26] [JEE Advance, Chapter : Electro Chemistry, Exercise-5, Section [A], Q. No.11]

Ans. [3] Sol. Less is the SRP, more is the reducing power & strongest is the reducing agent. Q.37 The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic

acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (Kf for benzene = 5.12 K kg mol–1)

(1) 74.6% (2) 94.6% (3) 64.6% (4) 80.4%

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Solution, Solved Example Q. No.40] [JEE Advance, Chapter : Solution, Exercise-5, Section [B], Q. No.2]

Ans. [2] Sol. ΔTf = i × Kf × m

0.45 = 2060

10002.0 5.12 i×

×××

i = .527

n/11

i−

−1=β = 946.

2/11527.

=−−1 or 94.6 %

Q.38 The radius of the second Bohr orbit for hydrogen atom is – (Planck's Const. h = 6.6262 × 10–34 Js; mass of electron = 9.1091 × 10–31 kg; charge of electron

e = 1.60210 ×10–19 C; permittivity of vaccum ∈0 = 8.854185 ×10–12 kg–1m–3A2) (1) 0.529 Å (2) 2.12 Å (3) 1.65 Å (4) 7.76 Å

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Atomic Structure, Exercise-1, Q. No.14] [JEE Advance, Chapter : Atomic Structure, Exercise-1, Q. No.21]

Ans. [2]

Sol. r = Åz

n529.02

×

= 1)2(529.0

2× = 2.116 Å ≅ 2.12 Å

CAREER POINT


[ CODE – A ]

Q.39 Two reactions, R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1 . If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to. (R = 8.314 J mol–1 K–1).

(1) 6 (2) 4 (3) 8 (4) 12 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Kinetics, Exercise-5, Section [B], Q. No.27] [JEE Advance, Chapter : Chemical Kinetics, Exercise-4, Q. No.25]

Ans. [2] Sol. R1 R2

A A

Ea +10 Ea k1 k2

k1 = A e–(Ea + 10)/RT k2 = A e–Ea /RT

RT/)10EaEa(

1

2 ekk ++−=

RT/10

1

2 ekk

= = 300 8.314/ 1010 3e ××

= 4 2494.2 / 10000 e e =

4kkln

1

2 =

Q.40 pka of a weak acid (HA) and pkb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB)

solution is- (1) 7.0 (2) 1.0 (3) 7.2 (4) 6.9

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Ionic Equilibrium, Exercise-4, Q. No.8] [JEE Advance, Chapter : Ionic Equilibrium, Exercise-1, Q. No.28]

Ans. [4]

Sol. pH = baw pK21pK

21pK

21

−+

= 4.3212.3

2114

21

×−×+×

= 7 + 1.6 – 1.7 = 6.9 Q.41 Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one

which is incorrect, is : (1) Both form nitrides (2) Nitrates of both Li and Mg yield NO2 and O2 on heating (3) Both form basic carbonates (4) Both form soluble bicarbonates

Ans. [3] Sol. It is the best possible option but, it should be bonus because Li2CO3 is less basic and MgCO3 is basic

CAREER POINT


[ CODE – A ]

Q.42 Which of the following species is not paramagnetic ? (1) O2 (2) B2 (3) NO (4) CO

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical bonding, Exercise # 1, Q. No.92] [JEE Advance, Chapter : Chemical bonding, Exercise # 2, Q. No.43]

Ans. [4] Sol. CO is diamagnetic because all electrons are paired in it. Q.43 Which of the following reactions is an example of a redox reaction ?

(1) XeF6 + H2O XeOF4 + 2HF (2) XeF6 + 2H2O XeO2F2 + 4HF (3) XeF4 + O2F2 XeF6 + O2 (4) XeF2 + PF5 [XeF]+ PF6

– Ans. [3]

Sol. +4 XeF4 + O2F2 → XeF6 + O2

+1 +6 0

This reaction is a redox reaction Q.44 A water sample has ppm level concentration of following anions

F– = 10 ; −24SO = 100 ; −

3NO = 50

The anion / anions that make / makes the water sample unsuitable for drinking is / are

(1) Only F– (2) Only −24SO

(3) Only −3NO (4) Both −2

4SO and −3NO

Ans. [1] Sol. F– ion concentration above 2 ppm causes brown mottling of teeth Q.45 The group having isoelectronic species is

(1) O2–, F–, Na, Mg2+ (2) O–, F–, Na+, Mg2+ (3) O2–, F–, Na+, Mg2+ (4) O–, F–, Na, Mg+

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Periodic Table, Exercise # 4, Q. No.5] [JEE Advance, Chapter : Periodic Table, Exercise # 2, Q. No.3]

Ans. [3] Sol. O2–, F–, Na+, Mg2+ are isoelectronic species. They all have 10e– Q.46 The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are

(1) Cl– and ClO– (2) Cl– and −2ClO (3) ClO– and −

3ClO (4) −3ClO and −

3ClO

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : p-Block elements, Exercise # 2, Q. No. 39] [JEE Advance, Chapter : p-Block elements (Halogens), Exercise # 2, Q. No. 9]

Ans. [1] Sol. Cl2 + NaOH(cold/dilute) → NaCl + NaOCl + H2O

CAREER POINT


[ CODE – A ]

Q.47 In the following reactions, ZnO is respectively acting as a / an

(a) Zn + Na2O Na2ZnO2

(b) Zn + CO2 ZnCO3

(1) acid and acid (2) acid and base (3) base and acid (4) base and base

Ans. [2]

Sol. ZnO + Na2O Na2ZnO2

acid

ZnO + CO2 ZnCO3

Base

ZnO is an amphoteric oxide

Q.48 Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO4. 'X' reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. 'X' is -

(1) CH3COONa (2) Na2C2O4 (3) C6H5COONa (4) HCOONa


[JEE Advance, Chapter : Salt analysis, Exercise # 1, Q. No. 19]

Ans. [2]

[X] Na2C2O4

(Oxalate-ion)

CaCl2

CaC2O4 ↓ + 2NaClWhite ppt

Conc. H2SO4 Na2CO3 + H2C2O4

CO2 ↑Effervesce

Sol.

KMnO4 + 242OC − H+ CO2↑ + Mn+2

Pink Colour

Oxalate ion

Colourless

Q.49 The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%), Carbon (22.9 %),

Hydrogen (10.0%) and Nitrogen (2.6%). The weight which a 75kg person would gain if all 1H atoms are replaced by 2H atoms is

(1) 7.5 kg (2) 10 kg (3) 15 kg (4) 37.5 kg

Ans. [1]

Sol. Weight of H11 in person = 75 ×

10010 = 7.5 kg

Now if 11H is replaced by 2

1H

Than person will gain weight by = 7.5 kg

CAREER POINT


[ CODE – A ]

Q.50 On treatment of 100 mL of 0.1 M solution of CoCl3.6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is :

(1) [Co(H2O)6]Cl3 (2) [Co(H2O)5Cl]Cl2.H2O

(3) [Co(H2O)4Cl2]Cl.2H2O (4) [Co(H2O)3Cl3].3H2O Ans. [2]

CoCl3.6H2O + AgNO3 (excess) 100 ml, 0.1 M No. of moles = 0.01 mol

AgClSol.1.2×1022 ion1.2×1022

6×1023 = 0.02 mol

Q 0.01 mol of CoCl3.6H2O produce 0.02 mol of AgCl ∴ 1 mol of CoCl3.6H2O produce 2 mol of AgCl ∴ Correct complex is [Co(H2O)5Cl]Cl2.H2O Q.51 Which of the following compounds will form significant amount of meta product during mono-nitration reaction?

(1)

NH2

(2)

NHCOCH3

(3)

OH

(4)

OCOCH3

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Amines and Nitrogen Compounds, Example-4]

Ans. [1]

Sol.

NH2

HNO3

NH2

NO2

+

[47%]

NH2

NO2

[2%]

+

NH2

[51%]

NO2

In aniline protonation of NH2 Group make it deactivating and meta directing. So meta product is significant. Q.52 Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to

decolourize the colour of bromine?

(1)

Br

O

(2) Br

O

(3)

Br

O

(4) Br

C6H5

CAREER POINT


[ CODE – A ]

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Halogen Derivatives, Exercise # 2, Q. No.6]

Ans. [3]

Sol.

Br

O

Elimination reaction not possible

CH3 – C – OΘ

CH3

CH3

Non Nucleophillic base

CH3 – C – CH3

O

It do not give unsaturation test

CH2

CH3

O

Q.53 The formation of which of the following polymers involves hydrolysis reaction?

(1) Nylon 6, 6 (2) Terylene (3) Nylon 6 (4) Bakelite

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Polymer, Exercise # 5, Q. No.1] [JEE Advance, Chapter : Carbohydrate, Amino acid, Protein & Polymer, Example-21]

Ans. [3] Sol. Nylon-6 is formed by monomer which is obtain by hydrolysis of CAPROLACTAM

NH C

O

H2O/Δ

Caprolactam

NH2–(CH2)5–COOHAmino caproic acid

Polymerisation ––NH–(CH2)5–C––––– O n

Nylon-6 Q.54 Which of the following molecules is least resonance stabilized?

(1)

N (2)

O

(3) (4)

O

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : GOC, Page 51, Q. No.32] [JEE Advance, Chapter :GOC, Exercise # 1, Q. No.54]

Ans. [2]

Sol.

N

Aromatic

O Non Aromatic Aromatic

O

Aromatic

..

..

2 is least stable as other are aromatic and 2 is non aromatic

CAREER POINT


[ CODE – A ]

Q.55 The increasing order of the reactivity of the following halides for the SN1 reaction is :

CH3CHCH2CH3

Cl

(I)

CH3CH2CH2Cl

(II)

p – H3CO – C6H4 – CH2Cl

(III)

(1) (I) < (III) < (II) (2) (II) < (III) < (I) (3) (III) < (II) < (I) (4) (II) < (I) < (III) Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Haloalkane, page 28, Q. No.17] [JEE Advance, Chapter : Halogen Derivatives, Exercise # 1, Q. No.14]

Ans. [4] Sol. Rate of SN1 reaction ∝ stability of carbocation

CH3CHCH2CH3

Cl

(I)

CH3 – CH – CH2 – CH3 + Cl

2º Carbocation

⊕ Θ

CH3 – CH2 – CH2 – Cl CH3 – CH2 – CH2

1º Carbocation

⊕

CH3O

Resonance Stabilize

⊕CH2 – Cl CH3O CH2

Order of SN1 = II < I < III Q.56 The major product obtained in the following reaction is :

C6H5

Br

H

C6H5

( + )

Δ⎯⎯⎯ →⎯ BuOKT

(1) ( + )C6H5CH(OtBu)CH2C6H5

(2) ( – )C6H5CH(OtBu)CH2C6H5 (3) ( ± )C6H5CH(OtBu)CH2C6H5

(4) C6H5CH = CHC6H5 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Hydrocarbon, page 37, Q. No. 31] [JEE Advance, Chapter : Hydrocarbon, Exercise # 2, Q. No. 19]

Ans. [4]

Sol.

C6H5

Br

H

C6H5 tBuOK

Δ C6H5 – CH = CH – C6H5

It is example of E2 elimination as t butoxide is stronger base and heating is also used.

CAREER POINT


[ CODE – A ]

Q.57 Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?

(1)

HOH2C

OH

HO

O CH2OH

OCH3

(2)

HOH2C

OH

O CH2OCH3

OH

OH

(3)

HOH2C

OH

HO

O CH2OH

OCOCH3 (4)

HOH2C

OH

HO

O CH2OH


[JEE Main, Chapter : Carbohydrate, page 74, Q. No. 9]

[JEE Advance, Chapter : Carbohydrate, Protein & Polymer, Exercise # 5, Q. No. 2]

Ans. [3]

Sol.

HOCH2

OH

HO

O CH2OH

O – C – CH3 O

KOHAq.

Hydrolysis

HOCH2

OH

HO

O CH2OH

OH

It is hemiacetal that is why it can behave as reducing sugar.

Q.58 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is :

(1) Two (2) Four (3) Six (4) Zero


[JEE Main, Chapter : Hydrocarbon, page 30, Q. No.50]

[JEE Advance, Chapter : Hydrocarbon, Example-14, Q. No.50]

Ans. [2]

Sol.

H3C – CH2 – C = CH – CH3

CH3

Peroxide

HBr⎯⎯ →⎯ CH3 – CH2 – CH – CH – CH3

CH3

Br

Major Product

*

*

Two new chiral center are formed so total 4 stereoisomer

This HBr/Peroxide gives antimarkownikov product as major product

CAREER POINT


[ CODE – A ]

Q.59 The correct sequence of reagents for the following conversion will be : O

CHO

HO CH3

HO CH3

CH3 (1) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH (2) [Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH (3) [Ag(NH3)2]+OH–, H+/CH3OH, CH3MgBr (4) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH–

Ans. [3]

Sol.

O

C – H

HO CH3

HO CH3CH3

O

[Ag(NH3)2]+ OHΘ

Tollen’s reagent oxidise Aldehyde in carboxylic acid

O

C – OΘ

O

H+/CH3OH CH3MgBr

O

C – OCH3

O

Q.60 The major product obtained in the following reaction is :

COOH

OO

⎯⎯⎯ →⎯ H–DIBAL

(1) CHO

COOH

(2) CHO

CHO

(3) CHO

COOH

OH

(4) CHO

CHO

OH

Ans. [4]

Sol.

COOH

O O

DiBAL–H

CHO

HO

CHO

DiBAL-H is selective reducing agent which reduce carboxylic acid and it’s derivative up to aldehyde only

[further reduction not possible]

CAREER POINT


[ CODE – A ]

Part C – MATHS

Q.61 The function f : R → ⎥⎦⎤

⎢⎣⎡−

21,

21 defined as f(x) = 2x1

x+

, is

(1) injective but not surjective (2) surjective but not injective (3) neither injective nor surjective (4) invertible

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Function, Page 55, Ex. 5A, Q. No. 23] [JEE Advance, Chapter : Function, Page 20, Ex. 1, ]

Ans. [2] Sol.

–1

min. –1/2

1

max. 1/2

O

y-axis

x-axis

f(x) = 2x1x

+ (odd)

(symmetry about origin)

dxdy = 22

2

)x1()x20(x1·)x1(

++−+

= 22

2

)x1(x1

+− = 0 ⇒ x = 1, –1

+ – –

–1 1

max. min.

Line parallel to x-axis cuts the graph more than one points hence function is many one.

Range = ⎥⎦⎤

⎢⎣⎡−

21,

21 = codomain hence function is onto

Q.62 If, for a positive integer n, the quadratic equation, x (x + 1) + (x + 1) (x + 2) + .... + (x + 1n − ) (x + n) = 10n has two consecutive integral solutions, then n is equal to

(1) 9 (2) 10 (3) 11 (4) 12 Ans. [3]

CAREER POINT


[ CODE – A ]

Sol. x (x + 1) + (x + 1) (x + 2) + .... + (x + (n – 1)) (x + n) = 10n After simplify nx2 + (1 + 3 + 5 + 7 + ...... + (2n –1)x + (0·1 + 1·2 + 2·3 + ... + (n –1)n) = 10n

nx2 + n2x + 3

)1n(n 2 − – 10n = 0

x2 + nx + 3

301n2 −− = 0

x2 + nx + 3

31n2 − = 0

Put n = 11 (where n ∈ I+)

x2 + 11x + 3

31121− = 0

x2 + 11x + 30 = 0 (x + 6) (x + 5) = 0 i.e. x = –5, –6 (Two consecutive integral solutions) So, n = 11

Q.63 Let ω be a complex number such that 2ω + 1 = z where z = 3− . If 72

22

111

111

ωωω−ω− = 3k, then k is equal to

(1) z (2) –1 (3) 1 (4) –z

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Complex Number, Page 29, Ex. 3 ]

Ans. [4] Sol. Apply operation C1 = C1 + C2 + C3

ωω

ωω+−2

22

0)1(0

113 = 3k

ωω

ωω2

2

00

113 = 3k (Because 1 + ω + ω2 = 0)

open by C1 3(ω2 – ω4) = 3k 3(ω2 – ω) = 3k 3(–1 – ω – ω) = 3k –3(1 + 2ω) = 3k Given that 2ω + 1 = z –3z = 3k k = – z

CAREER POINT


[ CODE – A ]

Q.64 If A = ⎥⎦

⎤⎢⎣

⎡−

−1432

, then adj (3A2 + 12A) is equal to

(1) ⎥⎦

⎤⎢⎣

⎡72846351

(2) ⎥⎦

⎤⎢⎣

⎡72638451

(3) ⎥⎦

⎤⎢⎣

⎡−

−51846372

(4) ⎥⎦

⎤⎢⎣

⎡−

−51638472

Ans. [1]

Sol. A = ⎥⎦

⎤⎢⎣

⎡−

−1432

A2 = ⎥⎦

⎤⎢⎣

⎡−

−1432

⎥⎦

⎤⎢⎣

⎡−

−1432

= ⎥⎦

⎤⎢⎣

⎡−

−1312

916

3A2 + 12A = ⎥⎦

⎤⎢⎣

⎡−

−39362748

+ ⎥⎦

⎤⎢⎣

⎡−

−12483624

= ⎥⎦

⎤⎢⎣

⎡−

−51846372

adj (3A2 + 12A) = ⎥⎦

⎤⎢⎣

⎡72846351

Q.65 If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is

(1) an infinite set (2) a finite set containing two or more elements (3) a singleton (4) an empty set

Ans. [3] Sol. Δ = 0 and at the one of Δ1 or Δ2 or Δ3 ≠ 0

Δ = 1ba1a1111

= 0

1[a – b] –1[1 – a] + 1[b – a2] = 0 2a – b – 1 + b – a2 = 0 a2 – 2a + 1 = 0 a = 1 x + y + z = 1 x + y + z = 1 x + by + z = 0 only one value of b, S is singleton set

CAREER POINT


[ CODE – A ]

Q.66 A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is

(1) 468 (2) 469 (3) 484 (4) 485

Students may find similar question in CP exercise sheet : [JEE Main, Chapter :P & C, Page 16, Ex. 1, Q. No. 29]

Ans. [4] Sol.

X

7 friends

3M 4L

Y

7 friends

4M 3L Case I : 3L from X side and 3M from Y side 4C3 × 4C3 = 4 × 4 = 16 Case II : 3M from X side and 3L from Y side 3C3 × 3C3 = 1 × 1 = 1 Case III : 2L and 1M from X side and 2M and 1L from Y side (4C2 × 3C1) × (4C2 × 3C1) = (6 × 3) × (6 × 3) = 18 × 18 = 324 Case IV : 2M and 1L from X side and 1M and 2L from Y side (3C2 × 4C1) × (4C1 × 3C2) = (3 × 4) × (4 × 3) = 12 × 12 = 144 Total number of ways = Case I + Case II + Case III + Case IV = 16 + 1 + 324 + 144 = 485 Q.67 The value of (21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4) + ... + (21C10 – 10C10) is

(1) 221 – 210 (2) 220 – 29 (3) 220– 210 (4) 221 – 211

Ans. [3]

Sol. (21C1 + 21C2 + .... + 21C10) – (10C1 + 10C2 + ..... + 10C10)

= [ ]1021

221

121 C2...C2C2

21

×++×+× – (210 – 1)

= [ ])CC(–CC...CC...CCC21

2121

021

2121

2021

1121

1021

221

121

021 +++++++++ – (210 – 1)

= 21 (221 – 2) – (210 – 1)

= 220 – 1 – 210 + 1 = 220 – 210

CAREER POINT


[ CODE – A ]

Q.68 For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then (1) b, c and a are in A.P. (2) a, b and c are in A.P. (3) a, b and c are in G.P. (4) b, c and a are in G.P.

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Progression, Page 27, Ex. 4, Q. No. 19] [JEE Advance, Chapter : Progression, Page 20, Ex. 2, Q. No. 26 ]

Ans. [1] Sol. 225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0 450a2 + 18b2 + 50c2 – 150ac – 10ab – 30bc = 0 (15a – 3b)2 + (3b – 5c)2 + (15a – 5c)2 = 0 (15a – 3b)2 = 0, (3b – 5c)2 = 0, (15a – 5c)2 = 0 15a = 3b, 3b = 5c 15a = 3b = 5c

3c

5b

1a

== = k (let)

a = k, b = 5k, c = 3k Then a, c and b are in A.P. Q.69 Let a, b, c ∈ R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, ∀ x, y ∈ R, then

∑=

10

1n

)n(f is equal to

(1) 165 (2) 190 (3) 255 (4) 330 Ans. [4] Sol. As a + b + c = 3 So, f(1) = 3 f(x + y) = f(x) + f(y) + xy Put x = 1, y = 1 f(2) = 2f(1) + 1 = 7 Put x = 2, y = 1 f(3) = f(2) + f(1) + 2 = 12 Put x = 2, y = 2 f(4) = 2f(2) + 4 = 18

So, ∑=

10

1n

)n(f = f(1) + f(2) + f(3) + .... + f(10)

Let S = 3 + 7 + 12 + 18 + .... + f(n) S = 3 + 7 + 12 + ........ + f(n) 0 = 3 + 4 + 5 + 6 + ....... – f(n)

f(n) = 3 + 4 + 5 + 6 + ....... = 2n [6 + (n – 1)]

CAREER POINT


[ CODE – A ]

f(n) = 2

)5n(n +

∴ ∑=

10

1n

)n(f = ∑=

10

1n

2n21 + ∑

=

10

1n

n25

= 21 ×

6211110 ×× +

25 ×

21110×

= 330

Q.70 3

2x )x2(

xcosxcotlim−π−

π→

equals :

(1) 161 (2)

81 (3)

41 (4)

241

Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Limit, Page 20, Ex. 2, Q. No. 30] [JEE Advance, Chapter :Limit, Page 16, Ex. 1, Q. No. 18]

Ans. [1]

Sol. Put x = 2π + h

30hh

22

)h2

cosh2

cotlim

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +

π−π

⎟⎠⎞

⎜⎝⎛ +

π−⎟

⎠⎞

⎜⎝⎛ +

π

→

30h h8hsinhtanlim

−+−

→

30h h8hsinhtanlim −

→

by expansion method

3

535

3

0h h8

...!5

h!3

hh....h152

3hh

lim⎟⎟⎠

⎞⎜⎜⎝

⎛+−−⎟⎟

⎠

⎞⎜⎜⎝

⎛++

→

3

53

0h h8

...!5

1152h

!31

31h

lim+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

→

= 8

061

31

++ =

161

CAREER POINT


[ CODE – A ]

Q.71 If for ⎟⎠⎞

⎜⎝⎛∈

41,0x , the derivative of ⎟

⎟⎠

⎞⎜⎜⎝

⎛

−−

31

x91xx6tan is )x(gx ⋅ , then g(x) equals :

(1) 3x91xx3

− (2) 3x91

x3−

(3) 3x913

+ (4) 3x91

9+

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Differentiation, Page 25, Ex. 2, Q. No. 26]

Ans. [4]

Sol. ⎟⎟⎠

⎞⎜⎜⎝

⎛

−= −

31

x91xx6tany

⎟⎟⎠

⎞⎜⎜⎝

⎛

−⋅

= −2

1

)xx3(1xx32tan

)xx3(tan2 1−=

)1xx2

1x(3)xx3(1

2dxdy

2 ⋅+⋅⋅+

= (differentiating w.r.t. x)

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+= x

2x

x916

3

3x91x9

+=

3x919x

+⋅=

Q.72 The normal to the curve y(x – 2) (x – 3) = x + 6 at the point where the curve intersects the y-axis passes

through the point :

(1) ⎟⎠⎞

⎜⎝⎛

21,

21 (2) ⎟

⎠⎞

⎜⎝⎛ −

31,

21 (3) ⎟

⎠⎞

⎜⎝⎛

31,

21 (4) ⎟

⎠⎞

⎜⎝⎛ −−

21,

21

Ans. [1] Sol. y(x – 2) (x – 3) = x + 6 at y axis, x = 0 y(–2) (–3) = 0 + 6 y = 1 Now y(x2 – 5x + 6) = x + 6

6x5x

6xy 2 +−+

=

22

2

)6x5x()5x2()6x(1)6x5x(

dxdy

+−−+−⋅+−

=

at x = 0, y = 1 16

)5()6(62 =

−−=

CAREER POINT


[ CODE – A ]

equation of normal y – 1 = – 1 (x – 0) x + y = 1

passes ⎟⎠⎞

⎜⎝⎛

21,

21 (by option)

Q.73 Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the

maximum area (in sq. m) of the flower-bed, is : (1) 10 (2) 25 (3) 30 (4) 12.5

Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Maxima & Minima, Page 96, Ex. 2, Q. No. 13]

Ans. [2] Sol. Given r + r + rθ = 20

r

r220 −=θ

θ

r r

rθ

Area θ= 2r21

⎟⎠⎞

⎜⎝⎛ −

⋅=r

r220r21 2

)r2r20(21z 2−=

0)r420(21

drdz

=−= ⇒ r = 5

at r = 5, θ = 2, 0dr

zd2

2< (hence maxima)

maximum area

θ= 2r21z

22 m252521

=××=

CAREER POINT


[ CODE – A ]

Q.74 Let ∫ >= ).1n(,dxxtanI nn If I4 + I6 = a tan5 x + bx5 + C, where C is a constant of integration, then the

ordered pair (a, b) is equal to :

(1) ⎟⎠⎞

⎜⎝⎛ 0,

51 (2) ⎟

⎠⎞

⎜⎝⎛ −1,

51 (3) ⎟

⎠⎞

⎜⎝⎛− 0,

51 (4) ⎟

⎠⎞

⎜⎝⎛− 1,

51

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Indefinite Integration, Page 33, Ex. 3, Q. No. 2] [JEE Advance, Chapter : Indefinite Integration, Page 35, Ex. 4, Q. No. 1]

Ans. [1]

Sol. In ∫= dxxtann

I4 + I6 ∫ += dx)xtanx(tan 64

∫ += dx)xtan1(xtan 24

∫ ⋅= dxxsecxtan 24 put t = tan x

∫ ⋅= dtt4

C5t5

+=

C5

xtan5+=

On comparison, we get

0b,51a ==

Q.75 The integral ∫π

π +

43

4

xcos1dx is equal to :

(1) 2 (2) 4 (3) – 1 (4) – 2

Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Definite Integration, Page 42, Ex. 5A, Q. No. 17]

Ans. [1]

Sol. ∫π

π+

=4/3

4/xcos1

dxI

∫π

π−π+

=4/3

4/)x(cos1

dxI by using ∫ ∫ −+=b

a

b

a

dx)xba(fdx)x(f

CAREER POINT


[ CODE – A ]

∫π

π

⎟⎠⎞

⎜⎝⎛

−+

+=

4/3

4/

dxxcos1

1xcos1

1I2 ∫π

π

⎟⎠⎞

⎜⎝⎛

−=

4/3

4/2 dx

xcos12

∫π

π

=4/3

4/

2 dxxeccos2I2

4/34/)x(cotI π

π−=

– (– 1 – 1) = 2

Q.76 The area (in sq. units) of the region {(x, y) : x ≥ 0, x + y ≤ 3, x2 ≤ 4y and y ≤ 1 + x } is :

(1) 23 (2)

37 (3)

25 (4)

1259

Ans. [3] Sol.

(0, 3)

(1, 2)

(2, 1)

(3, 0)

(0, 1)

y

xO

2

1

Required Area = ∫ ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛−−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

1

0

2

1

22dx

4xx3dx

4xx1

2

1

321

0

32/3

12x

2xx3

12x

2/3xx ⎥

⎦

⎤⎢⎣

⎡−−+⎥

⎦

⎤⎢⎣

⎡−+=

25

1211

1219

=+=

Q.77 If (2 + sin x) )1y(dxdy

++ cos x = 0 and y(0) = 1, then y ⎟⎠⎞

⎜⎝⎛ π

2 is equal to :

(1) 32

− (2) 31

− (3) 34 (4)

31

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Differential Equation, Page 70, Ex. 5A, Q. No. 6]

Ans. [4]

CAREER POINT


[ CODE – A ]

Sol. 0xcos)1y(dxdy)xsin2( =+++

xsin2

xcos)1y(dxdy

++−

=

⇒ dxxsin2

xcos1y

dy∫ ∫ ⎟

⎠⎞

⎜⎝⎛

+−=

+

⇒ log(y + 1) = – log(2 + sin x) + log c

⇒ y + 1 = xsin2

c+

…(1)

Given that y(0) = 1

∴ 1 + 1 = 2c ⇒ c = 4

∴ Equation of curve

xsin2

41y+

=+

at 2

x π= ⇒

1241y+

=+

⇒ 134y −=

31y =

Q.78 Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then

the orthocentre of this triangle is at the point :

(1) ⎟⎠⎞

⎜⎝⎛

43,1 (2) ⎟

⎠⎞

⎜⎝⎛ −

43,1 (3) ⎟

⎠⎞

⎜⎝⎛

21,2 (4) ⎟

⎠⎞

⎜⎝⎛ −

21,2

Ans. [3] Sol. Area

2812k1k51k3k

21

±=−

−

k(k – 2) + 3k(5 + k) + 1(10 + k2) = ± 56 k2 – 2k + 15k + 3k2 + 10 + k2 = ± 56 5k2 + 13k + 10 = ± 56 5k2 + 13k – 46 = 0 5k2 + 13k + 66 = 0 5k2 + 23k – 10k – 46 = 0 D = 169 – 4 × 5 × 66 < 0 k(5k + 23) – 2(5k + 23) = 0 No solution (5k + 23) (k – 2) = 0 k = 2 (k is integer) Hence co-ordinate (2, –6) (5, 2) (–2, 2)

CAREER POINT


[ CODE – A ]

E(2, β) AD ⊥ BC

138

222

−=×+−β

83

42 −

=−β

232 −=−β

232 −=β

21

=

⎟⎠⎞

⎜⎝⎛

21,2

(–2, 2) A

C (5, 2)

(2, β)

D

B (2, –6)

x = 2

x

E

y

O

Q.79 The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is :

(1) )12(2 − (2) )12(4 − (3) )12(4 + (4) )12(2 +

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Area under the curve, Ex. 3]

Ans. [2]

Sol.

(0, 4) y = x

r (0, k)

y = –x

y

xO

CAREER POINT


[ CODE – A ]

Circle touches the line

By graph radius = 4 – k

Perpendicular distance from centre = radius

⇒ 2k0k4 −

=−

⇒ 16 + k2 – 8k = 2k2

⇒ k2 – 16k + 32 = 0

⇒ 2

)32(425616k

−±=

⇒ 2

12816k ±=

⇒ 2

2816k ±=

⇒ 248k ±=

⇒ 248k −= (k should be 0 < k < 4) Radius = 4 – k

= )248(4 −−

= )12(4 −

Q.80 The eccentricity of an ellipse whose centre is at the origin is .21 If one of its directrices is x = – 4, then the

equation of the normal to it at ⎟⎠⎞

⎜⎝⎛

23,1 is :

(1) 4x – 2y = 1 (2) 4x + 2y = 7 (3) x + 2y = 4 (4) 2y – x = 2

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Ellipse, Page 58, Ex. 3, Q. No. 3] [JEE Advance, Chapter : Ellipse, Page 24, Ex. 3, Q. No. 13]

Ans. [1]

Sol. Let the equation of ellipse 2

2

2

2

by

ax

+ = 1

given that e = 21

and directrix ⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−=

−=

4xor

e/ax

CAREER POINT


[ CODE – A ]

⇒ ea = 4

⇒ a = 2 Now, b2 = a2 (1 – e2)

b2 = 4 ⎟⎠⎞

⎜⎝⎛ −

411 = 3

Equation of ellipse

3y

4x 22

+ = 1

diff. w.r.t. x

dxdy

3y2

4x2

+ = 0

21

dxdy

)2/3,1(−=⎟

⎠⎞

⎜⎝⎛

∴ Equation of normal at ⎟⎠⎞

⎜⎝⎛

23,1 is

y – y1 = )xx(

dxdy1

1−⎟⎠⎞

⎜⎝⎛

−

⇒ y – 23 = 2(x – 1)

⇒ 2y – 3 = 4x – 4 ⇒ 4x – 2y = 1

Q.81 A hyperbola passes through the point P( 2 , 3 ) and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point :

(1) (2 2 , 3 3 ) (2) ( 3 , 2 ) (3) (– 2 , – 3 ) (4) ( 23 , 32 )

Ans. [1] Sol. Let the equation of hyperbola

2

2

ax – 2

2

by = 1

given that foci (±ae, 0) = (±2, 0) ⇒ ae = 2

Hyperbola passes through P( 2 , 3 )

∴ 2a2 – 2b

3 = 1

⇒ 2a2 –

)1–e(a322 = 1

CAREER POINT


[ CODE – A ]

⇒ 2a2 – 2a–4

3 = 1

⇒ 8 – 2a2 – 3a2 = 4a2 – a4

⇒ a4 – 9a2 + 8 = 0

⇒ (a2 – 8) (a2 – 1) = 0 ⇒ a2 = 8 and a2 = 1 b2 = a2(e2 – 1) b2 = a2(e2 – 1) b2 = 4 – 8 b2 = 4 – 1 = 3 b2 = – 4 (not possible) ∴ Equation of hyperbola

1

x2–

3y2

= 1

tangent at P( 2 , 3 )

T = 0

x2 – 3

y = 1

By option it passes through ( 22 , 33 )

Q.82 The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal

perpendicular to both the lines 1

1–x =2–2y + =

34–z and

22–x =

1–1y + =

1–7z + , is :

(1) 83

10 (2) 835 (3)

7410 (4)

7420

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : 3D, Page 110, Ex. 2, Q. No. 17]

Ans. [1]

Sol. n = 1–1–2

32–1kji

= k3)7(–j–i5 +

Equation of plane 5(x – 1) + 7(y + 1) + 3(z + 1) = 0

5x + 7y + 3z + 5 = 0

Perpendicular distance of the plane from (1, 3, –7) is = 94925

|521–215|++++ =

8310

CAREER POINT


[ CODE – A ]

Q.83 If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,

1x =

4y =

5z is Q, then PQ is equal to :

(1) 422 (2) 42 (3) 56 (4) 53

Ans. [1] Sol.

R (2, 2, 8)

P(1, –2, 3)given line

Q The equation of line PR

1

1–x = 4

2y + = 5

3–z = k

let R(k + 1, 4k – 2, 5k + 3) it lies on the plane 2x + 3y – 4z + 22 = 0 ∴ 2(k + 1) + 3(4k – 2) – 4(5k + 3) + 22 = 0 ⇒ – 6k + 6 = 0 ⇒ k = 1 ∴ R(2, 2, 8) Image of P in the plane is (R is the mid-point of PQ) ∴ Q(3, 6, 13)

PQ = 100644 ++ = 168 = 422

Q.84 Let →a = k2–ji2 + and

→b = ji + . Let

→c be a vector such that |

→c –

→a | = 3, |(

→a ×

→b ) ×

→c | = 3 and the angle

between →c and

→a ×

→b be 30º. Then

→a ·

→c is equal to :

(1) 2 (2) 5 (3) 81 (4)

825

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Vector, Page 55, Ex. 5A, Q. No. 10] [JEE Advance, Chapter : Vector, Page 38, Ex. 4, Q. No. 35]

Ans. [1]

Sol. |→a ×

→b | =

0112–12

kji

CAREER POINT


[ CODE – A ]

|→a ×

→b | = kj2–i2 +

|→a ×

→b | = 3

given | (→a ×

→b ) ×

→c | = 3

|→a ×

→b | |

→c | sin 30º = 3

⇒ 3|→c | ·

21 = 3

⇒ |→c | = 2

Now |→c –

→a | = 3

|→c |2 + |

→a |2 – 2

→a .

→c = 9

⇒ 4 + 9 – 2→a .

→c = 9

⇒ →a .

→c = 2

Q.85 A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement,

then the variance of the number of green balls drawn is :

(1) 6 (2) 4 (3) 256 (4)

512

Ans. [4]

Sol. Total no. of balls = 25

(15 green and 10 yellow balls)

(Varience) σ2 = npq

where n → No. of Trial

p → Probability of happening of that event

q → Probability of not happening of that event

n = 10, p =2515 =

53 , q =

2510 =

52

So, σ2 = 10 × 53 ×

52

σ2 = 2560 =

512

CAREER POINT


[ CODE – A ]

Q.86 For three events A, B and C,

P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = 41

and P(All the three events occur simultaneously) = 161 .

Then the probability that at least one of the events occurs, is :

(1) 167 (2)

647 (3)

163 (4)

327

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Probability, Page 35, Ex. 3, Q. No. 6]

Ans. [1]

Sol. P(Exactly one of A or B occurs) = P(A) + P(B) – 2P(A ∩ B) = 41 ... (1)

P(Exactly one of B or C occurs) = P(B) + P(C) – 2P(B ∩ C) = 41 ... (2)

P(Exactly one of C or A occurs) = P(C) + P(A) – 2P(C ∩ A) = 41 ... (3)

Adding (1), (2) and (3)

2[P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A)] = 43

P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) = 83

P(All the three events occurs simultaneously) = P(A ∩ B ∩ C) =161

P(Atleast one of the events occurs) = P(A ∪ B ∪ C) = 83 +

161 =

167

Q.87 If two different numbers are taken from the set {0, 1, 2, 3, ……, 10}, then the probability that their sum as

well as absolute difference are both multiple of 4, is :

(1) 5512 (2)

4514 (3)

557 (4)

556

Ans. [4] Sol. n(S) = 11C2 = 55

(0, 4) (0, 8) (2, 6), (2, 10)(4, 8), (6, 10)

Favorable events

So, required probability = EventsTotalEvents.Fav =

556

CAREER POINT


[ CODE – A ]

Q.88 If 5(tan2x – cos2x) = 2cos2x + 9, then the value of cos4x is :

(1) 31 (2)

92 (3) –

97 (4) –

53

Ans. [3] Sol. 5(tan2x – cos2x) = 2cos2x + 9

⇒ 5 ⎥⎦

⎤⎢⎣

⎡xcos–

xcosxsin 2

2

2= 2(2cos2x – 1) + 9

⇒ 5[(1 – cos2x) – cos4x] = 4cos4x – 2cos2x + 9cos2x ⇒ 9cos4x + 12cos2x – 5 = 0 ⇒ 9cos4x + 15cos2x – 3cos2x – 5 = 0 ⇒ 3cos2x (3cos2x + 5) – (3cos2x + 5) = 0

⇒ cos2x = 31

⇒ cos2x = 2cos2x – 1

⇒ cos2x =32 – 1 =

31–

Now cos4x = 2cos22x – 1

cos4x = 2 ⎟⎠⎞

⎜⎝⎛

91 – 1

cos4x = 97–

Q.89 Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on

the ground such that AP = 2AB. If ∠BPC = β, then tan β is equal to :

(1) 41 (2)

92 (3)

94 (4)

76

Ans. [2] Sol.

B

x

C

A P4x

x βγ α

β = α – γ

tanβ = γα+

γαtan.tan1

tan–tan =

811

41–

21

+=

92

CAREER POINT


[ CODE – A ]

Q.90 The following statement (p → q) → [(~p → q) → q] is :

(1) equivalent to ~p → q (2) equivalent to p → ~q (3) a fallacy (4) a tautology Ans. [4] Sol.

q p ~p (p → q) (~p → q) (~p → q) → q (p → q) → [(~p → q) → q] T F T F

T T F F

F F T T

T F T T

T T T F

T F T T

T T T T

It is a tautology.

JEE Main Examination 02-04-2017 Code-A - Career...

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