Solution Jee Main 2017

CAREER POINT

[ CODE – A ]

CAREER POINT JEE Main Exam 2017 2017 (Paper & Solution)

Part A – PHYSICS Q.1

A man man grows grows into a giant giant such that his linear dimensions increase by a factor of 9. Assuming that his density density remains same, the stress in the leg will change by a factor of(1) 9

1 9

(2)

(3) 81

(4)

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 =

1 81

weight Vρg = area A

Initial stress = Stress 1 =

Vρg A

K 3Vρg Final stress = Stress 2 = K 2 A Stress2 = K Stress 1 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 ? v

v

(1)

(2)

v t

(3)

v t

(4)

t

t

Ans.

[3]

Sol.

v = u – gt (straight line graph) v

t

CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200

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1

CAREER POINT

[ CODE – A ]

Q.3

A body of mass m = 10 –2 kg is moving in a medium and experiences a frictional force F = –kv 2. Its initial speed is 1 v0 = 10 ms –1. If, after 10 s, its energy is mv02 , the value of k will be8 –3 –1 –3 (1) 10 kg m (2) 10 kg s –1 (3) 10 –4 kg m –1 (4) 10 –1 kg m –1 s –1

Ans.

[3]

Sol.

a=–

kv 2 m dv kv 2 = – m dt v

10

dv k = – dt 2 m v 10 0

∫

∫

v

⎡− 1 ⎤ = – k × 10 ⎢⎣ v ⎥⎦ m 10 k ⎡1 1 ⎤ – ⎢ − ⎥ = − 2 × 10 ⎣ v 10 ⎦ 10 1 1 + = – k × 1000 v 10 According to question 1 1 KE = mv2 = mv02 2 8 v 10 v = 0 = 2 2 1 1 – × 2 + = – k × 1000 10 10 1 = k × 1000 10 k = 10 –4 –

Q.4

Ans.

A time time dependent force F = 6t acts acts on a particle particle of mass mass 1 kg. If the particle starts from rest, the work 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 Stud Students nts may find fi nd similar similar quest uestion ion in CP exe exercise rcise she sheet : [ JE J E E A dvance nce, Chap Chapter : Work Work-Pow -Powe er-E nerg nergy, y, E x.2, Page No.25, No.25, Q. No.13 No.13] [1]

a=

Sol.

6t = 6 t 1

dv = 6t dt 1

⎡ 6t 2 ⎤ v = ⎢ ⎥ = 3 × 12 = 3 ⎣ 2 ⎦0 KE =

1 × 1 × 3 2 = 4.5 2

W = ΔKE = 4.5 – 0 = 4.5 Joule CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


2

CAREER POINT Q.5

The moment of inertia of a uniform cylinder of length l and and radius R about about its perpendicular perpendicular bisector is I. What is the ratio l /R /R such that the moment of inertia is minimum ? (1)

Ans.

[ CODE – A ]

3 2

(2)

3 2

(3) 1

(4)

3 2

[1]

Sol.

R

l

Moment of inertia of cylinder about perpendicular bisector is I

⎡ L2

I = M⎢

⎢⎣ 12

+

R 2 ⎤ ⎥ 4 ⎥⎦

For given mass and density M = πR 2Lρ R 2 =

M πLρ

⎡ L2

I = M⎢

⎣⎢ 12

+

M ⎤ ⎥ 4πLρ ⎦⎥

For maxima or minima of I dI = 0 dL

⎡ 2L dI M ⎤ = M ⎢ − = 0 2 ⎥ dL ⎣⎢ 12 4πL ρ ⎦⎥ L M = 6 4πL2ρ L πR 2 Lρ = 6 4πL2ρ 3 L2 = 2 2 R L 3 = R 2



3

CAREER POINT Q.6

[ CODE – A ]

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

θ

x

(1)

3g sin θ 2l

(2)

2g sin θ 3l

(3)

3g cos θ 2l

(4)

2g cos θ 3l

Stud Students nts may find fi nd similar similar quest uestion ion in CP exe exercise rcise she sheet : [ JE J E E A dvance nce, Chap Chapter : R otationa ional motion, ion, Ex.3, Ex .3, Page No., 36, 36, Q. Q. No.11 No.11] Ans.

[1]

Sol.

z

α λ/2

θ

θ mg

l

2

x

sin θ l

τ = mg sin θ 2

τ = Iα l

mg sin θ = 2

α =

Q.7

ml 2 α 3

3g sin θ 2 l

The variation variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)g

g

(1)

d

O

(2)

g

(3)

O

d

O

R g

d R


(4)

O

d R


4

CAREER POINT

[ CODE – A ]

Students may find similar question in CP exercise sheet : [ JE E M ain, Chapter : Gravitation, Ex.4, Page No., 42, Q. No.20] Ans.

[4]

Sol.

R

GM when r > R r 2 GM r g= when r < R R 3 g=

g g∝

g∝ r

R

1 r 2 r

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 byP P 3α (1) (2) (3) (4) 3PK α αK 3αK PK

Ans.

[1]

Sol.

B=

−P dV V

−P dV = V B dV =

−PV B



5

CAREER POINT

[ CODE – A ]

By heating we have to increase the volume by

PV B

ΔV = VγΔT = V × 3 α ΔT V × 3αΔT =

ΔT =

PV B

P 3αB

Here B = K

∴ ΔT =

Q.10

P 3αB

C p and Cv are specific heats at constant pressure and constant volume respectively. It is observed that C p – Cv = a for hydrogen gas C p – Cv = b for nitrogen gas The correct relation between a and b is(1) a =

1 b 14

(2) a = b

(3) a = 14 b

(4) a = 28 b

Students may find similar question in CP exercise sheet : [ JE E M ain, Chapter : KTG , E x.1, Page No.17, Q. No.35] [ JE E A dvance, Chapter : KTG , E x.1, Page No.25,Q. No.22] Ans.

[3]

Sol.

C p – Cv = R If C p and Cv are molar specific heat But if C p and Cv are specific heat i.e. gram specific heat then {Q = 1CΔT ⇒ Q = MS gΔT ⇒ C = MS g}

C = MSg C M MSgp – MSgv = R Sg =

Sgp –Sgv =

R M

R = a 2 R = b 28 a b a = 14 b 14 =



6

CAREER POINT Q.11

[ CODE – A ]

The temperature of an open room of volume 30 m 3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1× 10 5 Pa. In ni and nf are the number of molecules in the room before and after heating, the n f – ni will be : (1) – 1.61 × 10 23 (2) 1.38 × 1023 (3) 2.5 × 1025 (4) – 2.5 × 1025 Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : KTG , Ex.3, Page No.27, Q. No.5]

Ans.

[4]

Sol.

PV = nRT PV RT Ti = 273 + 17 = 290 K Tf = 273 + 27 = 300 K n =

105 × 30 ⎡ 1 1 ⎤ nf – ni = – × 6.023 × 10 23 ⎢ ⎥ 8.314 ⎣ 300 290 ⎦ 3 × 106 ⎡ – 10 ⎤ = × 6.023 × 10 23 ⎢ ⎥ 8.314 ⎣ 300 × 290 ⎦ 3 × 10 27 × 6.023 = – 8.314 × 3 × 29 – 6.023 × 1027 = – 8.314 × 29 = – 0.025 × 10 27 = – 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 : KE

(1)

0

KE T

T 2

T

t

(2)

KE

(3)

0

0

T

t

KE T 2

T

t

(4)

0

T 4

T 2

T

t

Students may find similar question in CP exercise sheet : [ JE E M ain, Chapter : SH M, E x.1, Page No.27, Q. No.33] Ans.

[4]

Sol.

x = A sin ωt



7

CAREER POINT

v=

[ CODE – A ]

dx = Aω cos ωt dt

1 KE = mA2ω2cos2ωt 2 1 2π = mA2ω2cos2 t 2 T KE 0

T 4

T 2

t

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

⎡ ⎤ ⎢ 1+ v ⎥ c ⎥ f (for approach) f app = ⎢⎢ 2 ⎥ ⎢ 1 – v ⎥ ⎢⎣ c 2 ⎥⎦ 1+

=

c/2 c

⎛ c / 2 ⎞ 1 – ⎜ ⎟ ⎝ c ⎠ =

2

× 10 GHz

3 2 × 10 GHz 3 4

= 3 × 10 GHz f app = 17.32 GHz Q.14

r

An electric dipole has fixed dipole moment p , which makes angle θ with respect to x-axis . When subjected r r ˆ . When subjected to another electric field to an electric field E = E iˆ , it experience a torque T = τk 1

r

r

r

1

E 2 = 3 E1 jˆ it experiences a torque T2 = – T1 . The angle θ is.

(1) 30° Ans.

(2) 45°

(3) 60°

(4) 90°

[3]

Sol.

E1 = E p θ

x



8

CAREER POINT

[ CODE – A ]

r ˆ τ = T1 = PE sin θ k

E2 =

3E1 jˆ

p θ

x

r

ˆ) τ = T2 = P 3E cos θ (– k ˆ ) = – P 3E cos θ (– k ˆ) PE sin θ ( k PE sin θ =

3E cos θ

tan θ = 3 θ = 60° Q.15

Ans.

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 : [ JE E M ain, Chapter : Capacitance, Ex.3, Page No.47, Q. No.12] [4]

Sol.

n capacitor in a row

m rows

Potential on each capacitor V =

1000 n

1000 = 300 n 10 ≈ 4 n= 3 C Ceq = × m n 1 × m = 2 n m=n×2=4×2=8 Minimum number of capacitor = 8 × 4 = 32



9

CAREER POINT Q.16

[ CODE – A ]

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 r 1

C r 2

(1) CE

(2) CE

r 1

(3) CE

(r 2 + r )

r 2 (r + r 2 )

(4) CE

r 1 (r 1 + r )

Students may find similar question in CP exercise sheet : [ JE E M ain, Chapter : Capacitance, Ex.2, Page No.44, Q. No.25] Ans.

[3]

Sol. E

i

r r 1

C r 2

At steady state current through capacitor branch become zero. i=

E r + r 2

Potential difference across capacitor ΔV

ΔV = ir 2 ⎛ E ⎞ ⎟⎟ r 2 ΔV = ⎜⎜ ⎝ r + r 2 ⎠ charge on capacitor = C ΔV ⎛ r 2 ⎞ ⎟⎟ r r + ⎝ 2 ⎠

= CE ⎜⎜

Q.17

In the above circuit the current in each resistance is : 2V

2V

1Ω

2V

(1) 1A Ans.

2V

2V

1Ω

1Ω

2V

(2) 0.25 A

(3) 0.5 A

(4) 0 A

[4]



10

CAREER POINT

[ CODE – A ]

Sol. 2V

4

2V

1Ω 6 2V

4

2V

2V

2

0

1Ω

2

2V

1Ω

0

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 ) ⎛ MB ⎞ α = ⎜ ⎟ θ ⎝ I ⎠ T = 2π

⇒ ω =

MB ⇒ T = 2π I

I MB

7.5 × 10 – 6 6.7 × 10 – 2 × 0.01

⇒ 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 15 Ω, 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

Ans.

[1]

Sol.

Ig max = 5 mA , R g = 15 Ω Range of voltmeter = 10 volt Ig

(3) 2.535 × 103 Ω

(4) 4.005 × 103 Ω

R H R g ΔV

ΔV = Ig (R g + R H) Range ΔVmax = Ig max (R g + R H) 10 = 5 × 10 –3 (15 + R H) R H = 1985 Ω R H = 1.985 × 103 Ω CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


11

CAREER POINT Q.20

[ CODE – A ]

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 : 10 Current (amp.)

Time 0.5 sec

(1) 200 Wb

(2) 225 Wb

(3) 250 Wb

(4) 275 Wb

Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : E MI , Ex.1, Page No.31, Q. No.1] Ans.

[3]

Sol.

10 Current (amp.)

Time 0.5 sec

Emf = iR dφ – = iR dt

∫ ( – dφ) = ∫ (idt ) R 0.5

∫

(– Δφ) = R idt 0

( – Δφ) = R (area of i – t curve) 1 2 Δφ = 100 (2.5) Δφ = 250 Wb

Δφ = R ( × 0.5 × 10)

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 -

log λmin

(1)

log λmin

(2)

log V


log V


12

CAREER POINT

[ CODE – A ]

log λmin

log λmin

(3)

(4)

log V Ans.

log V

[1]

Sol.

log λmin log V

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

⎛ hc ⎞ = constant ⎟ ⎝ e ⎠

= eV ⇒ λmin V = ⎜

λ min 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

⎛ 1 1 1 ⎞ ⎜⎜ − = ⎟⎟ ⎝ v u ƒ ⎠

Final image real at a distance 2ƒ = 40 cm from converging lens. CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


13

CAREER POINT Q.23

[ CODE – A ]

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 : [ JE E A dvance, Chapter : Wave nature of light : I nterference, 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 =

λ1D d

=

650 × 10 – 9 × 1.5 = 1.95 mm 0.5 × 10 – 3

520 × 10 – 9 × 1.5 β2 = = = 1.56 mm d 0.5 × 10 – 3 Least distance where their maxima again coincides from central maxima is = LCM of β1 & β2 = 7.8 mm

λ2D

m which is at rest. The 2 collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is λ λ 1 2 1 λ λ (1) A = (2) A = 2 (3) A = (4) A = λB 3 λB λB 3 λB 2

Q.24

A particle A of mass m and initial velocity v collides with a particle B of mass

Ans.

[2]

Sol.

m

m

m/2 v

v1 =

m − m/ 2 v+0 m + m/ 2

v1 =

m/2 v 3m/ 2

⇒

m/2 v1

v2

v 3 Similarly v1 =

2m ×v m + m/2 2m v2 = v 3m/ 2 4 v2 = v 3 h h Q λ1 = , λ2 = m 1 v1 m 2 v2 v2 = 0 +

λ1 m 2 v 2 m / 2 4v / 3 = = . = 2 m v/3 λ 2 m1v1 CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


14

CAREER POINT Q.25

[ CODE – A ]

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

λ1 , is given by λ2

– E

4 – E 3

λ2 λ1

– 2 E

– 3 E

4 2 (2) r = 3 3 Students may find similar question in CP exercise sheet : [ JE E M ain, Chapter : Atomic Structure, E x.5, Q. No.27 ] (1) r =

Ans. Sol.

(3) r =

3 4

(4) r =

1 3

[4]

hc

λ1 hc

λ1 hc

λ2

= (– E) – (– 2E) =E

….(i)

⎛ 4E ⎞ = – E + 4E = −3E + 4E = E ⎟ 3 3 3 ⎝ 3 ⎠

= (– E) – ⎜ −

….(ii)

By (i) & (ii) hc λ 1 r = 1 = E = hc 3 λ2 E/3 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 T log 2 log1.3 T (1) t = (2) t = T (3) t = T log (1.3) (4) t = 2 log1.3 log 2 log(1.3)

Ans.

[2]

Sol.

A ⎯⎯→ B A = A0e – λt ; B = A0 (1 – e – λt) ;

A (1 − e − λt ) B = 0 − λt A A 0e

0.3 = eλt – 1 eλt = 1.3 λt = ln (1.3) ln(2) t = ln (1.3) T ln(1.3) T log(1.3) ⇒ t= t=T ln(2) log(2) CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


15

CAREER POINT

[ 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

Ans.

(2) ωc

(3) ωm + ωc

(4) ωc – ωm

[1]

Sol.

LSB

USB

ωc – ωm

ωc + ωm

ωc

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.

(I)

Q

P G

A R

(II) B

Q

P G

S R

S

In Ist case for balance CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


16

CAREER POINT

[ CODE – A ]

P Q = R S P R = S Q

….(i)

In IInd case for balance R P = S Q

….(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/s 2 and the simplified relation T = closest to (1) 0.15%

rhg × 10 3 N/m, the possible error in surface tension is 2

(2) 1.5%

(3) 2.4%

(4) 10%

Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : Pr actical Physics, E x.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=

rhg × 103 N/m 2

Tr –1h –1 =

g ×103 2

Tr –1h –1 = constant

ΔT T

−

Δr Δh − = 0 r

h

ΔT ⎛ Δr Δh ⎞ =⎜ + ⎟ T ⎝ r h ⎠ ⎛ 0.01× 10 – 2 ⎞ ⎟ =⎜ ⎜ 1.25 × 10 – 2 ⎟ + T ⎝ ⎠

ΔT

% error =

ΔT T

⎛ 0.01 × 10 – 2 ⎞ ⎜ ⎟ ⎜ 1.45 × 10 – 2 ⎟ ⎝ ⎠

⎧D = 2r ⎫ ⎪ ⎪ ⎪ΔD = 2Δr ⎪ ⎨ ⎬ ⎪ ΔD = Δr ⎪ ⎪D r ⎪ ⎩ ⎭

× 100

≈ 1.5 % CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


17

CAREER POINT

[ CODE – A ]

Part B – CHEMISTRY Q.31

Given C (graphite) + O 2 (g) ⎯ ⎯→ CO 2 (g) ;

Δr H° = – 393.5 kJ mol –1 H2(g) +

1 O2(g) ⎯→ H2O(1); 2

Δr H° = – 285.8 kJ mol –1 CO2(g) + 2H2O(1) ⎯→ CH4(g) + 2O2(g); Δr H° = + 890.3 kJ mol –1 Based on the above thermochemical equations, the value of Δr H° 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 : [ JE E Main, Chapter : Chemical E nergetic ,E xercise-1 Q. No.12] [ JE E A dvance, Chapter : Chemical E nergetic ,E xercise-4 Q. No.11] Ans.

[1]

Sol.

C (graphite) + O 2 (g) ⎯ ⎯→ CO 2 (g) ; Δr H° = – 393.5 kJ

H2(g) +

1 O2(g) ⎯→ H2O(1); Δr H° = – 285.8 kJ 2

.... (i) .... (ii)

CO2(g) + 2H2O(1) ⎯→ CH4(g) + 2O2(g); Δr H° = + 890.3 ..... (iii) C(graphite ) + 2H 2 (g) ⎯ ⎯→ CH 4 (g) ; Δr H° = ?

.... (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 CO 2. The molar mass of M2CO3 in g mol –1 is (1) 118.6 (2) 11.86 (3) 1186 (4) 84.3

Ans.

[4]

Sol.

M 2 CO3 + HCl ⎯ ⎯→ CO 2 1gm

.01186

POAC on carbon 1 .01186 = ⇒ x = 84.3 x 1 Q.33

ΔU is equal to (1) Adiabatic work

(2) Isothermal work


(3) Isochoric work

(4) Isobaric work


18

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 : [ JE E

A dvance, 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)

2a

(2)

a 2

(3) 2a

(4) 2 2a

Students may find similar question in CP exercise sheet : [ JE E

Main, Chapter : Solid State, Solved E xample, Q. No.21]

[ JE E

A dvance, Chapter : Solid State, Solved E xercise-4, Q. No.4]

Ans.

[2]

Sol.

∴ nearest distance =

a 2 a 2

fcc



19

CAREER POINT Q.36

[ CODE – A ]

Given E°Cl

2

/ Cl −

= 1.36V , E °Cr 3+ / Cr = −0.74V

E °Cr O2− / Cr 3+ = 1.33V , E °MnO− / Mn 2+ = 1.51V , 2

7

4

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

(4) Mn2+

(3) Cr

Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : E lectro Chemistry, E xercise-4, Q. No.26] [ JE E A dvance, Chapter : E lectro Chemistry, E xercise-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 (K f 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 : [ JE E Main, Chapter : Solution, Solved E xample Q. No.40] [ JE E A dvance, Chapter : Solution, E xercise-5, Section [B ], Q. No.2]

Ans.

[2]

Sol.

ΔTf = i × K f × m i × 5.12 × 0.2 × 1000 0.45 = 60 × 20 i = .527

β=

Q.38

1− i 1 − .527 = .946 or 94.6 % = 1 − 1/ n 1 − 1/ 2

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 : [ JE E Main, Chapter : Atomic Structure, E xercise-1, Q. No.14] [ JE E A dvance, Chapter : Atomic Structure, Exercise-1, Q. No.21]

Ans.

[2]

Sol.

r = 0.529 ×

n2 Å z

(2) 2 = 2.116 Å ≅ 2.12 Å = 0.529 × 1 CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


20

CAREER POINT Q.39

[ CODE – A ]

Two reactions, R 1 and R 2 have identical pre-exponential factors. Activation energy of R 1 exceeds that of R 2 by 10 kJ mol –1 . If k 1 and k 2 are rate constants for reactions R 1 and R 2 respectively at 300 K, then ln(k 2/k 1) 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 : [ JE E M ain, Chapter : Chemical Kinetics, E xercise-5, Section [B ], Q. No.27 ] [ JE E A dvance, Chapter : Chemical Kinetics, E xercise-4, Q. No.25]

Ans.

[2]

Sol.

R 1 R 2 A A Ea +10 Ea k 1 k 2 –(Ea + 10)/RT k 1 = A e k 2 = A e –Ea /RT k 2 = e( − Ea + Ea +10) / RT k 1 3 k 2 = e10 / RT = e10×10 / 8.314 × 300 k 1

= e10000 / 2494.2 = e 4 k ln 2 = 4 k 1 Q.40

Ans. Sol.

pk a of a weak acid (HA) and pk b 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 : [ JE E Main, Chapter : I onic E quilibrium, E xercise-4, Q. No.8] [ JE E A dvance, Chapter : I onic E quilibrium, Exercise-1, Q. No.28] [4]

1 1 1 pH = pK w + pK a − pK b 2 2 2 1 1 1 = × 14 + × 3.2 − × 3.4 2 2 2 = 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 NO 2 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 Li 2CO3 is less basic and MgCO3 is basic



21

CAREER POINT Q.42

[ CODE – A ]

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 : [ JE E Main, Chapter : Chemical bonding, E xercise # 1, Q. No.92] [ JE E A dvance, Chapter : Chemical bonding, E xercise # 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) XeF 6 + 2H2O  XeO2F2 + 4HF (3) XeF4 + O2F2  XeF6 + O2 (4) XeF2 + PF5  [XeF]+ PF6 –

Ans.

[3]

Sol.

+4

+1

+6

0

XeF4 + O2F2 → XeF6 + O2

This reaction is a redox reaction Q.44

A water sample has ppm level concentration of following anions F – = 10 ; SO 24− = 100 ; NO3− = 50 The anion / anions that make / makes the water sample unsuitable for drinking is / are (1) Only F –

(2) Only SO 24−

(3) Only NO 3−

(4) Both SO 24− and NO3−

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, Mg 2+ (2) O – , F – , Na+, Mg2+

(3) O2– , F – , Na+, Mg2+

(4) O – , F – , Na, Mg +

Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : Periodic Table, E xercise # 4, Q. No.5] [ JE E A dvance, 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 ClO −2

(3) ClO – and ClO3−

(4) ClO3− and ClO3−

Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : p-B lock elements, E xercise # 2, Q. No. 39] [ JE E Advance, Chapter : p-Block elements (H alogens), E xercise # 2, Q. No. 9] Ans.

[1]

Sol.

Cl2 + NaOH(cold/dilute) → NaCl + NaOCl + H 2O



22

CAREER POINT Q.47

[ CODE – A ]

In the following reactions, ZnO is respectively acting as a / an (a) Zn + Na 2O  Na2ZnO2 (b) Zn + CO 2  ZnCO3 (1) acid and acid

(2) acid and base

Ans.

[2]

Sol.

ZnO + Na2O  Na2ZnO2

(3) base and acid

(4) base and base

acid ZnO + CO2  ZnCO3 Base ZnO is an amphoteric oxide Q.48

Sodium salt of an organic acid 'X' produces effervescence with conc. H 2SO4. 'X' reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO 4. 'X' is (1) CH3COONa

(2) Na 2C2O4

(3) C6H5COONa

(4) HCOONa

Students may find similar question in CP exercise sheet : [ JE E Ans. Sol.

A dvance, Chapter : Salt analysis, E xercise # 1, Q. No. 19]

[2]

[X] Na2C2O4

Conc. H2SO4

Na2CO3 + H2C2O4

(Oxalate-ion)

CO2 ↑ Effervesce

CaCl2

CaC2O4 ↓ + 2NaCl White ppt −2

KMnO4 + C 2 O 4 Pink Colour

Q.49

H+

+2

CO2↑ + Mn

Oxalate ion

Colourless

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

Ans.

[1]

Sol.

Weight of 11 H in person = 75 ×

(3) 15 kg

(4) 37.5 kg

10 = 7.5 kg 100

Now if 1 H1 is replaced by 1 H 2 Than person will gain weight by = 7.5 kg CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


23

CAREER POINT

[ CODE – A ]

Q.50

On treatment of 100 mL of 0.1 M solution of CoCl 3.6H2O 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]

Sol.

CoCl3.6H2O + AgNO3 (excess)

AgCl

100 ml, 0.1 M

1.2×10 ion

No. of moles = 0.01 mol

1.2×10

22 22

6×10 Q 0.01

23

= 0.02 mol

mol of CoCl3.6H2O produce 0.02 mol of AgCl

∴ 1 mol of CoCl3.6H2O produce 2 mol of AgCl ∴ Correct complex is [Co(H 2O)5Cl]Cl2.H2O Q.51

Which of the following compounds will form significant amount of meta product during mono-nitration reaction? NH2

NHCOCH3

(1)

(2)

OH

(3)

OCOCH3

(4)

Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : Amines and Nitrogen Compounds, E xample-4 ] Ans.

[1]

NH2

NH2

NH2

NH2 NO2

HNO3

Sol.

+

+

NO2 NO2 [47%]

[2%]

[51%]

In aniline protonation of NH 2 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? O O

(1)

(2) Br

Br C6H5

O

(3)

(4) Br


Br www.careerpoint.ac.in

24

CAREER POINT

[ CODE – A ]

Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : Halogen Deri vatives, E xercise # 2, Q. No.6] Ans.

[3]

CH3 CH3 – C – O

O

Θ

O

CH2

CH3 Non Nucleophillic base

Br Sol.

O CH3 – C – CH 3

Elimination reaction not possible

CH3 It do not give unsaturation test 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 : [ JE E Main, Chapter : Polymer, E xercise # 5, Q. No.1] [ JE E A dvance, Chapter : Carbohydrate, Amino acid, Protein & Polymer, E xample-21] Ans. Sol.

[3]

Nylon-6 is formed by monomer which is obtain by hydrolysis of CAPROLACTAM O NH

C H2O/Δ

NH2 –(CH2)5 –COOH Amino caproic acid

Polymerisation

O

Caprolactam

Q.54

––NH–(CH2)5 –C––––– n

Nylon-6

Which of the following molecules is least resonance stabilized?

(1)

N

(2)

(3)

(4)

O

O

Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : GOC, Page 51, Q. No.32] [ JE E A dvance, Chapter :GOC, E xercise # 1, Q. No.54] Ans.

Sol.

[2]

.. O ..

N Aromatic

O Non Aromatic

Aromatic

Aromatic

2 is least stable as other are aromatic and 2 is non aromatic CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


25

CAREER POINT Q.55

[ CODE – A ]

The increasing order of the reactivity of the following halides for the S N1 reaction is : CH3CHCH2CH3

CH3CH2CH2Cl

p – H3CO – C6H4 – CH2Cl

Cl (II)

(I)

(III)

(1) (I) < (III) < (II) (2) (II) < (III) < (I) (3) (III) < (II) < (I) Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : H aloalkane, page 28, Q. No.17 ] [ JE E A dvance, Chapter : Halogen Deri vatives, E xercise # 1, Q. No.14] Ans.

[4]

Sol.

Rate of SN1 reaction ∝ stability of carbocation ⊕

CH3CHCH2CH3

(4) (II) < (I) < (III)

Θ

CH3 – CH – CH2 – CH3 + Cl

Cl

2º Carbocation (I) ⊕

CH3 – CH2 – CH2 – Cl

CH3 – CH2 – CH2 1º Carbocation ⊕

CH3O

CH2 – Cl

CH3O

CH2

Resonance Stabilize

Order of SN 1 = II < I < III Q.56

The major product obtained in the following reaction is : Br H T

⎯BuOK ⎯ ⎯ → C6H5 ⎯ Δ

C6H5 (+) t

(1) ( + )C6H5CH(O Bu)CH2C6H5 (2) ( – )C6H5CH(OtBu)CH2C6H5 (3) ( ± )C6H5CH(OtBu)CH2C6H5 (4) C6H5CH = CHC6H5 Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : H ydrocarbon, page 37, Q. No. 31] [ JE E Advance, Chapter : H ydrocarbon, E xercise # 2, Q. No. 19] Ans.

[4]

Br H

Sol.

C6 H 5

tBuOK Δ

C6H5 – CH = CH – C6H5

C6 H 5

It is example of E 2 elimination as t butoxide is stronger base and heating is also used. CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


26

CAREER POINT Q.57

[ CODE – A ]

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

HOH2C

(1)

HOH2C

CH2OH

O CH2OCH3

(2)

HO OCH3

OH OH OH

OH O

HOH2C

(3)

OH2C

CH2OH

O

(4)

HO OCOCH3

CH2OH

HO

OH

OH

Students may find similar question in CP exercise sheet :

Ans.

[ JE E

M ain, Chapter : Carbohydrate, page 74, Q. No. 9]

[ JE E

A dvance, Chapter : Carbohydrate, Pr otein & Polymer, E xercise # 5, Q. No. 2]

[3]

HOCH2

O

Sol.

CH2OH

HO O – C – CH3

Aq. KOH Hydrolysis

HOCH2

O

CH2OH

HO OH

O

OH

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

Students may find similar question in CP exercise sheet :

Ans.

[ JE E

Main, Chapter : H ydrocarbon, page 30, Q. No.50]

[ JE E

A dvance, Chapter : H ydrocarbon, E xample-14, Q. No.50]

[2]

CH3 Sol.

H3C – CH2 – C = CH – CH3

CH3 HBr ⎯ ⎯ ⎯ → Peroxide

* CH3 – CH2 – CH – CH – CH3 * Br Major Product

Two new chiral center are formed so total 4 stereoisomer This HBr/Peroxide gives antimarkownikov product as major product



27

CAREER POINT Q.59

[ CODE – A ]

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

CHO +

–

HO

CH3

HO

CH 3 CH3

+

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

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

[3] O

+

Sol.

[Ag(NH3)2] OH

C–H

H+/CH3OH

Θ

Tollen’s reagent oxidise Aldehyde in carboxylic acid

O

Q.60

O

O

C–O

HO

CH3

HO

CH3 CH3

CH3MgBr

Θ

C – OCH 3 O

O

The major product obtained in the following reaction is : O O DIBAL – H ⎯ ⎯ ⎯ ⎯ →

COOH

(1)

(2)

CHO

CHO

COOH

CHO

OH

OH

(3)

(4)

CHO COOH

Ans.

CHO CHO

[4]

O O HO CHO

DiBAL–H

Sol.

COOH

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 Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


28

CAREER POINT

[ CODE – A ]

Part C – MATHS Q.61

x ⎡ 1 1⎤ The function f : R → ⎢− , ⎥ defined as f(x) = , is 1+ x2 ⎣ 2 2⎦ (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 : [ JE E Main, Chapter : F unction, Page 55, Ex. 5A, Q. No. 23] [ JE E A dvance, Chapter : F unction, Page 20, Ex. 1, ]

Ans.

[2]

Sol.

y-axis max.

1/2 –1 O min.

1

x-axis

–1/2

x (odd) 1+ x2 (symmetry about origin) f(x) =

dy (1 + x 2 ) · 1 − x (0 + 2x ) = dx (1 + x 2 ) 2 1− x2 = = 0 ⇒ x = 1, –1 (1 + x 2 ) 2 min. –

max. –

+ –1

1

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

⎡ 1 1⎤ Range = ⎢− , ⎥ = codomain hence function is onto ⎣ 2 2⎦ Q.62

If, for a positive integer n, the quadratic equation, x (x + 1) + (x + 1) (x + 2) + .... + (x + n −1 ) (x + n) = 10n has two consecutive integral solutions, then n is equal to (1) 9 (2) 10 (3) 11 (4) 12

Ans.

[3]



29

CAREER POINT Sol.

[ CODE – A ]

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 +

n (n 2 − 1) – 10n = 0 3

n 2 − 1 − 30 x + nx + = 0 3 2

x2 + nx +

n 2 − 31 = 0 3

Put n = 11 (where n ∈ I+) 121 − 31 = 0 3 x2 + 11x + 30 = 0 (x + 6) (x + 5) = 0 i.e. x = –5, –6 (Two consecutive integral solutions) So, n = 11 x2 + 11x +

Q.63

Let ω be a complex number such that 2 ω + 1 = z where z = (1) z

1 1 1 − 3 . If 1 − ω2 − 1 ω2 = 3k, then k is equal to 1 ω2 ω7

(2) –1

(3) 1

(4) –z

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

[4]

Sol.

Apply operation C1 = C1 + C2 + C3 3 1 1 2 0 − (1 + ω ) ω2 = 3k 0 ω2 ω 3 1 1 0 ω ω2 = 3k (Because 1 + ω + ω2 = 0) 0 ω2 ω open by C1 3(ω2 – ω4) = 3k 3(ω2 – ω) = 3k 3(–1 – ω – ω) = 3k –3(1 + 2ω) = 3k –3z = 3k k=–z

Given that 2 ω + 1 = z



30

CAREER POINT Q.64

[ CODE – A ]

⎡ 2 − 3⎤ If A = ⎢ , then adj (3A2 + 12A) is equal to ⎥ ⎣− 4 1 ⎦ ⎡51 63⎤ (1) ⎢ ⎥ ⎣84 72⎦

Ans.

[1]

Sol.

⎡ 2 − 3⎤ A= ⎢ ⎥ ⎣− 4 1 ⎦

⎡51 84⎤ (2) ⎢ ⎥ ⎣63 72⎦

⎡ 2 − 3⎤ ⎡ 2 − 3⎤ A2 = ⎢ ⎥ ⎢− 4 1 ⎥ = 4 1 − ⎣ ⎦⎣ ⎦

⎡ 16 − 9⎤ ⎢− 12 13 ⎥ ⎣ ⎦

⎡ 48 − 27⎤ 3A2 + 12A = ⎢ ⎥ + ⎣− 36 39 ⎦

⎡ 24 − 36⎤ ⎢− 48 12 ⎥ ⎣ ⎦

⎡ 72 − 63⎤ (3) ⎢ ⎥ ⎣− 84 51 ⎦

⎡ 72 − 84⎤ (4) ⎢ ⎥ ⎣− 63 51 ⎦

⎡ 72 − 63⎤ = ⎢ ⎥ ⎣− 84 51 ⎦ ⎡51 63⎤ adj (3A2 + 12A) = ⎢ ⎥ ⎣84 72⎦ 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 1 1 1 Δ = 1 a 1 = 0 a b 1 1[a – b] –1[1 – a] + 1[b – a 2] = 0 2a – b – 1 + b – a 2 = 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



31

CAREER POINT Q.66

[ CODE – A ]

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 : [ JE E Main, Chapter :P & C, Page 16, E x. 1, Q. No. 29]

Ans.

[4]

Sol.

X

Y

7 friends

7 friends

3M

4L

4M

3L

Case I : 3L from X side and 3M from Y side 4

C3 × 4C3 = 4 × 4 = 16 Case II : 3M from X side and 3L from Y side 3 C3 × 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

Ans.

[3]

Sol.

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

1 [2 × 21C1 + 2 × 21C 2 + ... + 2 × 21C10 ] – (210 – 1) 2

=

1 21 [ C 0 + 21C1 + 21C 2 + ... + 21C10 + 21C11 + ... + 21C 20 + 2

(3) 220 – 210

(4) 221 – 211

C 21 – ( 21C 0 + 21C 21 )] – (210 – 1)

21

1 21 (2 – 2) – (2 10 – 1) 2 = 220 – 1 – 210 + 1 = 220 – 210 =



32

CAREER POINT Q.68

[ CODE – A ]

For any three positive real numbers a, b and c, 9(25a 2 + 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 : [ JE E Main, Chapter : Pr ogression, Page 27, E x. 4, Q. No. 19] [ JE E Advance, Chapter : Pr ogression, 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 a b c = = = k (let) 1 5 3 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

∑ f (n) is equal to n =1

(1) 165 Ans.

[4]

Sol.

As So,

(2) 190

(3) 255

(4) 330

a+b+c=3 f(1) = 3 f(x + y) = f(x) + f(y) + xy x = 1, y = 1 f(2) = 2f(1) + 1 = 7 x = 2, y = 1 f(3) = f(2) + f(1) + 2 = 12 x = 2, y = 2 f(4) = 2f(2) + 4 = 18

Put Put Put 10

So,

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

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 + ....... =

n [6 + (n – 1)] 2



33

CAREER POINT

f(n) =

n (n + 5) 2

10

∑

∴

[ CODE – A ]

f (n ) =

n =1

1 10 2 5 10 n + n 2 n =1 2 n =1

∑

∑

1 10 × 11 × 21 5 10 × 11 × + × 2 6 2 2 = 330 =

Q.70

cot x − cos x equals : π (π − 2x ) 3 x→ lim

2

(1)

1 16

(2)

1 8

(3)

1 4

(4)

1 24

Students may find similar question in CP exercise sheet : [ JE E M ain, Chapter :Limit, Page 20, E x. 2, Q. No. 30] [ JE E A dvance, Chapter :Limit, Page 16, E x. 1, Q. No. 18] Ans.

[1]

Sol.

Put x =

π 2

⎛ π

lim

+ h

⎛ π

⎞

⎞

cot⎜ + h ⎟ − cos⎜ + h ) ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

h →0

⎛ π ⎞ ⎞ ⎜⎜ π − 2⎛ ⎜ + h ⎟ ⎟⎟ ⎝ 2 ⎠ ⎠ ⎝ − tan h + sin h lim h →0 − 8h 3 tan h − sin h

3

lim

8h 3 by expansion method h →0

⎛ h 3 2 5 ⎞ ⎛ h 3 h 5 ⎞ ⎜⎜ h + + h .... ⎟⎟ − ⎜⎜ h − + ...⎟⎟ 3 15 ⎠ ⎝ 3! 5! ⎠ lim ⎝ 8h 3

h →0

lim

h →0

⎛ 1 1 ⎞ ⎛ 2 1 ⎞ h 3 ⎜⎜ + ⎟⎟ + h 5 ⎜⎜ − ⎟⎟ + ... ⎝ 3 3! ⎠ ⎝ 15 5! ⎠ 8h 3

1 1 + +0 1 3 6 = = 8 16



34

CAREER POINT Q.71

[ CODE – A ]

⎛ 6x x ⎞ ⎛ 1 ⎞ ⎟ is If for x ∈ ⎜ 0, ⎟ , the derivative of tan −1 ⎜⎜ 3⎟ 4 ⎝ ⎠ ⎝ 1 − 9x ⎠ (1)

3x x 1 − 9x 3

(2)

x ⋅ g( x ) , then g(x) equals :

3x 1 − 9x 3

(3)

3 1 + 9x 3

(4)

9 1 + 9x 3

Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : Differentiation, Page 25, Ex. 2, Q. No. 26] Ans. Sol.

[4]

⎛ 6x x ⎞ ⎟ 3⎟ 1 9 x − ⎝ ⎠

y = tan −1 ⎜⎜

⎛ 2 ⋅ 3x x ⎞ ⎟ = tan −1 ⎜⎜ 2⎟ 1 ( 3 x x ) − ⎝ ⎠ = 2 tan −1 (3x x ) dy 2 1 3 ( x = ⋅ ⋅ + x ⋅1) (differentiating w.r.t. x) dx 1 + (3x x ) 2 2 x

=

6 1 + 9x 3

=

9 x 1 + 9x 3

= x⋅

Q.72

⎛ x ⎞ ⎜ ⎟ x + ⎜ 2 ⎟ ⎝ ⎠

9 1 + 9x 3

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 1 ⎞ (1) ⎜ , ⎟ ⎝ 2 2 ⎠ Ans. Sol.

⎛ 1 1 ⎞ (2) ⎜ , − ⎟ ⎝ 2 3 ⎠

⎛ 1 1 ⎞ (3) ⎜ , ⎟ ⎝ 2 3 ⎠

⎛ 1 1 ⎞ (4) ⎜ − , − ⎟ ⎝ 2 2 ⎠

[1]

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 y=

x+6 x − 5x + 6 2

dy (x 2 − 5x + 6) ⋅1 − ( x + 6) (2x − 5) = dx ( x 2 − 5 x + 6) 2

at x = 0, y = 1

=

6 − (6) (−5) =1 62



35

CAREER POINT

[ CODE – A ]

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

⎛ 1 passes ⎜ , ⎝ 2 Q.73

1 ⎞ ⎟ 2 ⎠

(by option)

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 : [ JE E Main, Chapter :Maxima & Mi nima, Page 96, Ex. 2, Q. No. 13]

Ans.

[2]

Sol.

Given r + r + r θ = 20

θ=

r

20 − 2r r

θ

r

r θ

1 Area = r 2θ 2 1 20 − 2r ⎞ = r 2 ⋅ ⎛ ⎜ ⎟ 2 ⎝ r ⎠ 1 z = (20r − 2r 2 ) 2 dz 1 = (20 − 4r ) = 0 ⇒ r = 5 dr 2 at r = 5, θ = 2,

d2z < 0 (hence maxima) dr 2

maximum area 1 z = r 2θ 2 1 2

= × 52 × 2 = 25m 2



36

CAREER POINT Q.74

[ CODE – A ]

∫

Let I n = tan n x dx , (n > 1). If I4 + I6 = a tan 5 x + bx5 + C, where C is a constant of integration, then the ordered pair (a, b) is equal to :

⎛ 1 ⎞ (1) ⎜ , 0 ⎟ ⎝ 5 ⎠

⎛ 1 ⎞ (2) ⎜ , − 1⎟ ⎝ 5 ⎠

⎛ 1 ⎞ (3) ⎜ − , 0 ⎟ ⎝ 5 ⎠

⎛ 1 ⎞ (4) ⎜ − , 1⎟ ⎝ 5 ⎠

Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : I ndefinite I ntegration, Page 33, E x. 3, Q. No. 2] [ JE E Advance, Chapter : I ndefinite I ntegration, Page 35, E x. 4, Q. No. 1] Ans.

[1]

∫ = ∫ (tan x + tan x ) dx = ∫ tan x (1 + tan x ) dx = ∫ tan x ⋅ sec x dx = ∫ t ⋅ dt

In = tan n x dx

Sol.

I4 + I6

4

6

4

2

4

2

put t = tan x

4

t5 = +C 5 tan 5 x = +C 5 On comparison, we get 1 a = , b = 0 5 3π 4

Q.75

The integral

dx

∫ 1 + cos x is equal to : π 4

(1) 2

(2) 4

(3) – 1

(4) – 2

Students may find similar question in CP exercise sheet : [ JE E M ain, Chapter :Definite I ntegration, Page 42, Ex. 5A, Q. No. 17 ] Ans.

[1] 3π / 4

Sol.

I=

dx 1 + cos x π/ 4

∫

3π / 4

dx I= 1 + cos (π − x ) π/ 4

∫

b

∫

b

∫

by using f ( x ) dx = f (a + b − x ) dx a


a


37

CAREER POINT

[ CODE – A ]

3π / 4

⎛ 1 + 1 ⎞ dx = 2I = ⎜ ⎟ + − 1 cos x 1 cos x ⎝ ⎠ π/ 4

∫

3π / 4

⎛

⎞

2

∫ ⎜⎝ 1 − cos x ⎠⎟ dx 2

π/ 4

3π / 4

∫ cosec x dx 2

2I = 2

π/ 4

I = − (cot x )3ππ/ 4/ 4 – (– 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)

Ans.

3 2

(2)

7 3

(3)

5 2

(4)

59 12

(4)

1 3

[3]

Sol.

y (0, 3)

(1, 2) 2 (2, 1)

(0, 1)

O

1

x

(3, 0)

1

2 ⎛ ⎛ x 2 ⎞⎟ x 2 ⎞⎟ ⎜ ⎜ Required Area = ⎜1 + x − ⎟ dx + ⎜ 3 − x − ⎟ dx 4 ⎠ 4 ⎠ 0 ⎝ 1 ⎝

∫

∫

1

2

⎡ x3/ 2 x3 ⎤ ⎡ x 2 x3 ⎤ = ⎢x + − ⎥ + ⎢3x − − ⎥ 2 12 ⎦1 ⎣ 3 / 2 12 ⎦ 0 ⎣ =

Q.77

If (2 + sin x) (1) −

2 3

19 11 5 + = 12 12 2

dy π ⎞ + ( y + 1) cos x = 0 and y(0) = 1, then y ⎛ ⎜ ⎟ is equal to : dx ⎝ 2 ⎠ 1 (2) − 3

(3)

4 3

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

[4]



38

CAREER POINT Sol.

(2 + sin x )

[ CODE – A ]

dy + ( y + 1) cos x = 0 dx

dy −( y + 1) cos x = dx 2 + sin x

⇒

⎛ cos x ⎞

dy

∫ y + 1 = −∫ ⎜⎝ 2 + sin x ⎠⎟ dx

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

c 2 + sin x

…(1)

Given that y(0) = 1 c ∴ 1+1= ⇒ c=4 2

∴ Equation of curve y +1 = at

x=

π 2

4 2 + sin x

⇒ y +1 =

4 2 +1

4 3

⇒ y = −1 y =

Q.78

1 3

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 :

⎛ 3 ⎞ (1) ⎜1, ⎟ ⎝ 4 ⎠ Ans.

[3]

Sol.

Area 1 2

⎛ 3 ⎞ (2) ⎜1, − ⎟ ⎝ 4 ⎠

⎛ 1 ⎞ (3) ⎜ 2, ⎟ ⎝ 2 ⎠

⎛ 1 ⎞ (4) ⎜ 2, − ⎟ ⎝ 2 ⎠

k −3k 1 5 k 1 = ± 28 − k 2 1

k(k – 2) + 3k(5 + k) + 1(10 + k 2) = ± 56 k 2 – 2k + 15k + 3k 2 + 10 + k 2 = ± 56 5k 2 + 13k + 10 = ± 56 5k 2 + 13k + 66 = 0 D = 169 – 4 × 5 × 66 < 0 No solution

5k 2 + 13k – 46 = 0 5k 2 + 23k – 10k – 46 = 0 k(5k + 23) – 2(5k + 23) = 0 (5k + 23) (k – 2) = 0 k = 2 (k is integer)

Hence co-ordinate (2, –6) (5, 2) (–2, 2) CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


39

CAREER POINT

[ CODE – A ]

E(2, β) AD ⊥ BC β−2 8 × = −1 2+2 3 β − 2 −3 = 4 8 3 β−2 = − 2 3 1 β= 2− = 2 2 ⎛ 2, 1 ⎞ ⎜ ⎟ ⎝ 2 ⎠ y

(–2, 2) A

C (5, 2) E (2, β) D

x

O

B (2, –6) x=2 Q.79

The radius of a circle, having minimum area, which touches the curve y = 4 – x 2 and the lines, y = |x| is : (1) 2 ( 2 − 1)

(2) 4 ( 2 − 1)

(3) 4 ( 2 + 1)

(4) 2 ( 2 + 1)

Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : Ar ea under the curve, E x. 3] Ans.

[2]

y Sol.

(0, 4)

y = –x

y=x

(0, k)

r

O

x



40

CAREER POINT

[ CODE – A ]

Circle touches the line By graph radius = 4 – k Perpendicular distance from centre = radius

⇒ 4 − k =

0 − k 2

k 2 ⇒ 16 + k – 8k = 2 2

⇒ k 2 – 16k + 32 = 0 ⇒ k =

16 ± 256 − 4(32) 2

⇒ k =

16 ± 128 2

⇒ k =

16 ± 8 2 2

⇒ k = 8 ± 4 2 ⇒ k = 8 − 4 2 (k should be 0 < k < 4) Radius = 4 – k = 4 − (8 − 4 2 ) = 4 ( 2 − 1)

Q.80

The eccentricity of an ellipse whose centre is at the origin is

1 . If one of its directrices is x = – 4, then the 2

⎛ 3 ⎞ is : ⎟ ⎝ 2 ⎠

equation of the normal to it at ⎜1, (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 : [ JE E Main, Chapter : E llipse, Page 58, Ex. 3, Q. No. 3] [ JE E A dvance, Chapter : E llipse, Page 24, Ex. 3, Q. No. 13] Ans.

[1]

Sol.

x 2 y2 Let the equation of ellipse 2 + 2 = 1 a b given that e =

1 2

⎧ x = −a / e ⎫ ⎪ ⎪ and directrix ⎨ or ⎬ ⎪ x = −4 ⎪ ⎩ ⎭ CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


41

CAREER POINT

⇒

[ CODE – A ]

a = 4 e

⇒ a = 2 Now,

b2 = a2 (1 – e2)

⎛ 1 ⎞ b2 = 4 ⎜1 − ⎟ = 3 ⎝ 4 ⎠ Equation of ellipse x 2 y2 + = 1 4 3 diff. w.r.t. x 2x 2 y dy = 0 + 4 3 dx 1 ⎛ dy ⎞ =− ⎜ ⎟ 2 ⎝ dx ⎠(1, 3 / 2) 3 ⎞ ∴ Equation of normal at ⎛ ⎜1, ⎟ is ⎝ 2 ⎠ y – y1 = −

Q.81

1 ( x − x1 ) ⎛ dy ⎞ ⎜ ⎟ ⎝ dx ⎠

3 = 2(x – 1) 2

⇒

y–

⇒ ⇒

2y – 3 = 4x – 4 4x – 2y = 1

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 )

Ans.

[1]

Sol.

Let the equation of hyperbola

(2) ( 3 , 2 )

(3) (– 2 , – 3 )

(4) ( 3 2 , 2 3 )

x 2 y2 – = 1 a 2 b 2 given that foci (±ae, 0) = (±2, 0)

⇒

ae = 2

Hyperbola passes through P( 2 , 3 )

∴

2 3 – = 1 a 2 b 2

⇒

2 3 – =1 a 2 a 2 (e 2 – 1)



42

CAREER POINT

⇒

[ CODE – A ]

2 3 – =1 2 a 4 – a 2

⇒ 8 – 2a2 – 3a2 = 4a2 – a4 ⇒ a4 – 9a2 + 8 = 0 ⇒ (a2 – 8) (a2 – 1) = 0 ⇒ a2 = 8 b2 = a2(e2 – 1) b2 = 4 – 8

a2 = 1 b2 = a2(e2 – 1) b2 = 4 – 1 = 3

b2 = – 4 (not possible)

∴ Equation of hyperbola

and

x 2 y2 – =1 1 3 tangent at P( 2 , 3 ) T=0 2x –

y = 1 3

By option it passes through ( 2 2 , 3 3 )

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)

10 83

x – 1 y + 2 z – 4 x – 2 y + 1 z + 7 = = and = = , is : 1 – 2 3 2 – 1 – 1

(2)

5 83

(3)

10 74

(4)

20 74

Students may find similar question in CP exercise sheet : [ JE E A dvance, Chapter : 3D, Page 110, E x. 2, Q. No. 17 ] Ans.

[1]

Sol.

î jˆ ˆ k ˆ nˆ = 1 – 2 3 = 5î – jˆ(– 7) + 3k 2 – 1 – 1 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 =

| 5 + 21 – 21 + 5 | 10 = 25 + 49 + 9 83



43

CAREER POINT Q.83

[ CODE – A ]

If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line, x y z = = is Q, then PQ is equal to : 1 4 5 (1) 2 42

Ans.

(2)

42

(3) 6 5

(4) 3 5

[1]

Sol.

P(1, –2, 3)

R

given line

(2, 2, 8)

Q

The equation of line PR x – 1 y + 2 z – 3 = = =k 1 4 5 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 = Q.84

4 + 64 + 100 = 168 = 2 42

→

→

→

→

→

→

→

→

ˆ and b = î + ˆj . Let c be a vector such that | c – a | = 3, |( a × b ) × c | = 3 and the angle Let a = 2î + jˆ – 2k →

→

→

→ →

between c and a × b be 30º. Then a · c is equal to : (1) 2

(2) 5

(3)

1 8

(4)

25 8

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

[1]

Sol.

î | a × b | = 2 1 →

→

ˆ jˆ k 1 – 2 1 0



44

CAREER POINT →

[ CODE – A ]

→

ˆ | a × b | = 2î – 2 jˆ + k →

→

| a × b | = 3 →

→

→

given | ( a × b ) × c | = 3 →

→

→

| a × b | | c | sin 30º = 3 →

1 = 3 2

⇒

3| c | ·

⇒

|c |=2

Now

|c – a |=3

→

→

→

→

→

→ →

| c |2 + | a |2 – 2 a . c = 9

⇒ ⇒

Q.85

→ →

4+9–2a .c =9 → →

a .c =2

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

Ans.

[4]

Sol.

Total no. of balls = 25

(3)

6 25

(4)

12 5

(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 = So,

15 3 10 2 = ,q= = 25 5 25 5

σ2 = 10 × σ2 =

3 2 × 5 5

60 12 = 25 5



45

CAREER POINT

Q.86

[ CODE – A ]

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) = 1 . 16 Then the probability that at least one of the events occurs, is : 7 7 3 (1) (2) (3) 16 64 16

1 4

and P(All the three events occur simultaneously) =

(4)

7 32

Students may find similar question in CP exercise sheet : [ JE E Main, Chapter : Pr obability, 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) =

1 4 1 P(Exactly one of B or C occurs) = P(B) + P(C) – 2P(B ∩ C) = 4 1 P(Exactly one of C or A occurs) = P(C) + P(A) – 2P(C ∩ A) = 4 Adding (1), (2) and (3) 3 2[P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A)] = 4 3 P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) = 8

... (1) ... (2) ... (3)

1 16 3 1 7 P(Atleast one of the events occurs) = P(A ∪ B ∪ C) = + = 8 16 16 P(All the three events occurs simultaneously) = P(A ∩ B ∩ C) =

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 : 12 14 7 6 (1) (2) (3) (4) 55 45 55 55

Ans.

[4]

Sol.

n(S) = 11C2 = 55 (0, 4) (0, 8) Favorable events (2, 6), (2, 10) (4, 8), (6, 10)

So, required probability =

Fav. Events 6 = Total Events 55



46

CAREER POINT Q.88

[ CODE – A ]

If 5(tan2x – cos2x) = 2cos2x + 9, then the value of cos4x is : 1 2 7 (1) (2) (3) – 3 9 9

Ans.

[3]

Sol.

5(tan2x – cos2x) = 2cos2x + 9

3 (4) – 5

⎡ sin 2 x ⎤ ⇒ 5 ⎢ 2 – cos 2 x ⎥ = 2(2cos2x – 1) + 9 ⎣ cos x ⎦

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

⇒ cos2x =

⇒ cos2x = 2cos2x – 1 2 – 1 3 3 cos4x = 2cos22x – 1

⇒ cos2x = – 1 = Now

⎛ 1 ⎞ cos4x = 2 ⎜ ⎟ – 1 ⎝ 9 ⎠ cos4x =

– 7 9

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 2 4 6 (1) (2) (3) (4) 4 9 9 7

Ans.

[2]

Sol.

B x C x

α A

4x

β γ P

β = α – γ 1 1 – tan α – tan γ 2 4= 2 tanβ = = 1 9 1 + tan α . tan γ 1+ 8 CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200


47

Solution Jee Main 2017

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