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JEE Main 2017 April 08 Question Paper with Solutions

All 89 questions from the JEE Main 2017 (April 08) shift — Physics (29), Chemistry (30) and Mathematics (30) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.

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Physics29 questions

Q1Single correctUnits and Measurements

Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be :

(A)

[M] = [T^{- 1} C^{- 2} h]

(B)

[M] = [T^{- 1} C^{2} h]

(C)

[M] = [T^{- 1} C^{- 2} h^{- 1}]

(D)

[M] = [T C^{- 2} h]

SolutionAnswer: Option 1

[M] = [T^{- 1} C^{- 2} h]

Approach:

Express mass as a product of powers of T, C and h, then equate dimensions in M, L, T.

Step 1:Assume the form for mass.

[M] = [T]^{a} [C]^{b} [h]^{c}

Step 2:Substitute the dimensions of each base quantity.

[M] = T^{a} (L T^{- 1})^{b} (M L^{2} T^{- 1})^{c} = M^{c} L^{b + 2 c} T^{a - b - c}

Step 3:Equate exponents of M, L and T with the target [M].

c = 1, b + 2 c = 0, a - b - c = 0

Step 4:Write the result.

[M] = [T^{- 1} C^{- 2} h]

Final answer:

[M] = [T^{- 1} C^{- 2} h]

Q2Single correctKinematics

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity ?

(A)

Velocity (y-axis) vs Time (x-axis): curve starting high, dipping to a minimum then rising slightly (concave up).

(B)

Velocity (y-axis) vs Time (x-axis): curve starting low and rising, increasing steeply (concave up).

(C)

Velocity (y-axis) vs Distance (x-axis): curve starting high, flat at top, curving down to meet the distance axis vertically (quarter-circle shape).

(D)

Velocity (y-axis) vs Distance (x-axis): curve starting high and decreasing steeply then flattening toward the distance axis (concave up, decaying).

SolutionAnswer: Option 3

Approach:

Use the kinematic relation between velocity and distance for constant acceleration and identify the matching graph shape.

Step 1:A velocity-time graph for constant acceleration is a straight line; options (1) and (2) are curved, so they are excluded.

v = v_{0} + a t

Step 2:For constant negative acceleration with positive velocity, use the velocity-distance relation.

v^{2} = v_{0}^{2} - 2 ∣ a ∣ x

Step 3:The slope

d v / d x = - ∣ a ∣ / v

is shallow at large v and becomes steep (vertical) as

v \to 0

\frac{d v}{d x} = - \frac{∣ a ∣}{v}

Step 4:This concave-down velocity-distance curve meeting the axis vertically is option (3).

Final answer: Option (3): velocity vs distance curve, concave, meeting the distance axis vertically.

Q3Single correctOscillations and Waves

A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :

(A)

\frac{1}{4}

(B)

\frac{5}{2 2}

(C)

\frac{1}{2}

(D)

2

SolutionAnswer: Option 3

\frac{1}{2}

Approach:

Find the spring constant from the first system, then use the parallel combination and new mass to find the new frequency.

Step 1:Original system: m = 1 kg, f = 1 Hz.

1 = \frac{1}{2 π} \frac{k}{1} \Rightarrow k = 2 π

Step 2:Two identical springs in parallel give effective constant 2k; new mass 8 kg.

k_{e ff} = 2 k = 8 π^{2}, m^{'} = 8

Step 3:Compute the new frequency.

f^{'} = \frac{1}{2 π} \frac{8 π ^{2}}{8} = \frac{1}{2 π} \cdot π = \frac{1}{2}

Final answer:

\frac{1}{2}

Q5Single correctRotational Motion

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :

A uniform disc of mass M and radius R mounted on a horizontal axle fixed to a hatched wall. A string wrapped around the rim hangs vertically with a small block of mass m at its free end; g points downward and R is the radius arrow from the centre.

(A)

\frac{2 m g}{2 m + M}

(B)

\frac{2 M g}{2 m + M}

(C)

\frac{2 m g}{2 M + m}

(D)

\frac{2 M g}{2 M + m}

SolutionAnswer: Option 1

\frac{2 m g}{2 m + M}

Approach:

Write Newton's law for the hanging mass and the rotational equation for the disc, with the string constraint a = R alpha.

Step 1:Newton's law for the mass m.

m g - T = ma

Step 2:Torque on disc.

TR = \frac{1}{2} M R^{2} \cdot \frac{a}{R} \Rightarrow T = \frac{1}{2} M a

Step 3:Eliminate T.

m g = ma + \frac{1}{2} M a

Final answer:

\frac{2 m g}{2 m + M}

Q6Single correctRotational Motion

Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is

I_{0}

as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :

Equilateral triangle ABC (A bottom-left, B bottom-right, C top) with midpoints D (bottom), E (right), F (left); the inner medial triangle DEF is drawn as a cavity and the centre O is marked.

(A)

\frac{7}{8} I_{0}

(B)

\frac{15}{16} I_{0}

(C)

\frac{3 I _{0}}{4}

(D)

\frac{31 I _{0}}{32}

SolutionAnswer: Option 2

\frac{15}{16} I_{0}

Approach:

The medial triangle DEF is similar to ABC with half the side and shares the same centroid O; scale its moment of inertia and subtract.

Step 1:DEF has side a/2 and area one quarter of ABC, so its mass is M/4. Its centroid coincides with O.

M_{D EF} = \frac{M}{4}, a_{D EF} = \frac{a}{2}

Step 2:Moment of inertia of DEF about O.

I_{D EF} = I_{0} \cdot \frac{M _{D EF}}{M} \cdot (\frac{a /2}{a})^{2} = I_{0} \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{I _{0}}{16}

Step 3:Subtract the cavity contribution.

I_{re m} = I_{0} - \frac{I _{0}}{16} = \frac{15}{16} I_{0}

Final answer:

\frac{15}{16} I_{0}

Q7Single correctGravitation

If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the Earth would have to rotate so that the person at the equator will weigh

\frac{3}{4}

W. Radius of the Earth is 6400 km and

g = 10

s^{2}

(A)

1.1 \times 1 0^{- 3}

rad/s

(B)

0.83 \times 1 0^{- 3}

rad/s

(C)

0.63 \times 1 0^{- 3}

rad/s

(D)

0.28 \times 1 0^{- 3}

rad/s

SolutionAnswer: Option 3

0.63 \times 1 0^{- 3}

rad/s

Approach:

Apparent weight at the equator is reduced by the centripetal term; set it equal to 3/4 of the true weight and solve for the angular speed.

Step 1:Set apparent weight to three quarters of true weight.

m (g - ω^{2} R) = \frac{3}{4} m g

Step 2:Solve for omega squared.

ω^{2} = \frac{g}{4 R} = \frac{10}{4 ( 6.4 \times 1 0 ^{6} )} = 3.906 \times 1 0^{- 7}

Step 3:Take the square root.

ω = 6.25 \times 1 0^{- 4} = 0.63 \times 1 0^{- 3}

rad/s

Final answer:

0.63 \times 1 0^{- 3}

rad/s

Q8Single correctThermal Properties of Matter

In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 15

0^{\circ}

C. Immediately, it is put into water of volume 150 cc at 2

7^{\circ}

C kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 4

0^{\circ}

C. The specific heat of aluminium is : (take 4.2 Joule = 1 calorie)

(A)

378

J/kg-

^{\circ}

(B)

315

J/kg-

^{\circ}

(C)

476

J/kg-

^{\circ}

(D)

434

J/kg-

^{\circ}

SolutionAnswer: Option 4

434

J/kg-

^{\circ}

Approach:

Apply conservation of heat: heat lost by the aluminium equals heat gained by the water plus the calorimeter (via its water equivalent).

Step 1:Water mass = 150 cc = 0.150 kg; total water-equivalent gaining heat = 0.150 + 0.025 = 0.175 kg;

c_{w} = 4200

J/kg-

^{\circ}

m_{w} + w = 0.175

Step 2:Aluminium temperature fall = 150 - 40 = 11

0^{\circ}

C; write the heat balance.

0.20 c_{A l} (110) = 0.175 (4200) (13)

Step 3:Solve for the specific heat.

c_{A l} = \frac{9555}{22} = 434.3

J/kg-

^{\circ}

Final answer:

434

J/kg-

^{\circ}

Q9Single correctProperties of Solids

A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature rises by

Δ T

. The net change in its length is zero. Let l be the length of the rod, A its area of cross-section, Y its Young's modulus, and

α

its coefficient of linear expansion. Then, F is equal to :

(A)

l^{2} Y α Δ T

(B)

lAY Δ T

(C)

A Y α Δ T

(D)

\frac{A Y}{α Δ T}

SolutionAnswer: Option 3

A Y α Δ T

Approach:

Set the thermal expansion equal in magnitude to the compressive contraction so the net length change is zero, and solve for F.

Step 1:Net change zero means thermal expansion equals elastic compression.

l α Δ T = \frac{Fl}{A Y}

Step 2:Solve for F.

F = A Y α Δ T

Final answer:

A Y α Δ T

Q10Single correctThermodynamics

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is : (Take

C_{v} = 1.5 R

, where R is gas constant)

P-V diagram of a rectangular cycle ABCDA. P axis marks P0 and 2P0, V axis marks V0 and 2V0. A(V0,P0) bottom-left, B(V0,2P0) top-left, C(2V0,2P0) top-right, D(2V0,P0) bottom-right, traversed clockwise A->B->C->D->A.

(A)

0.24

(B)

0.15

(C)

0.32

(D)

0.08

SolutionAnswer: Option 2

0.15

Approach:

The rectangular cycle on the P-V diagram (A at

(V_{0}, P_{0})

, B at

(V_{0}, 2 P_{0})

, C at

(2 V_{0}, 2 P_{0})

, D at

(2 V_{0}, P_{0})

) is run clockwise. Compute net work from the enclosed area and heat input from the two heat-absorbing legs.

Step 1:Net work equals the rectangle area.

W = (2 P_{0} - P_{0}) (2 V_{0} - V_{0}) = P_{0} V_{0}

Step 2:Heat absorbed on A->B (isochoric, P rises).

Q_{A B} = \frac{C _{v}}{R} Δ (P V) = 1.5 P_{0} V_{0}

Step 3:Heat absorbed on B->C (isobaric, V rises).

Q_{BC} = \frac{C _{p}}{R} Δ (P V) = 2.5 (2 P_{0} V_{0}) = 5 P_{0} V_{0}

Step 4:Efficiency = work / heat input.

η = \frac{P _{0} V _{0}}{( 1.5 + 5 ) P _{0} V _{0}} = \frac{1}{6.5} = 0.154

Final answer:

0.15

Q11Single correctKinetic Theory of Gases

An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (

C_{p}

) and at constant volume (

C_{v}

) is :

(A)

6

(B)

\frac{7}{2}

(C)

\frac{5}{2}

(D)

\frac{7}{5}

SolutionAnswer: Option 4

\frac{7}{5}

Approach:

Use the equipartition expressions for Cv and Cp in terms of degrees of freedom f.

Step 1:With f = 5.

C_{v} = \frac{5}{2} R, C_{p} = \frac{7}{2} R

Step 2:Take the ratio.

γ = \frac{C _{p}}{C _{v}} = \frac{7/2}{5/2} = \frac{7}{5}

Final answer:

\frac{7}{5}

Q12Single correctOscillations and Waves

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10

s^{- 1}

. At,

t = 0

the displacement is 5 m. What is the maximum acceleration ? The initial phase is

\frac{π}{4}

(A)

500

s^{2}

(B)

5002

s^{2}

(C)

750

s^{2}

(D)

7502

s^{2}

SolutionAnswer: Option 2

5002

s^{2}

Approach:

The ratio of maximum acceleration to maximum velocity gives omega; use the initial displacement with the given phase to find amplitude, then maximum acceleration.

Step 1:From the given ratio.

ω = 10

s^{- 1}

Step 2:Use x(0) = 5 m with phase pi/4.

5 = A sin \frac{π}{4} = \frac{A}{2} \Rightarrow A = 52

Step 3:Compute maximum acceleration.

a_{ma x} = ω^{2} A = 100 \cdot 52 = 5002

s^{2}

Final answer:

5002

s^{2}

Q13Single correctOscillations and Waves

Two wires

W_{1}

and

W_{2}

have the same radius r and respective densities

ρ_{1}

and

ρ_{2}

such that

ρ_{2} = 4 ρ_{1}

. They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in

W_{1}

W_{2}

is :

A sonometer: hatched wall at left, a horizontal wire over two triangular bridges. Wire W1 (density rho1) on the left and W2 (density rho2) on the right are joined at node O, midway between the bridges; the wire passes over a pulley at the right carrying a hanging weight.

(A)

1 : 1

(B)

1 : 2

(C)

1 : 3

(D)

4 : 1

SolutionAnswer: Option 2

1 : 2

Approach:

Both segments have the same length, tension and frequency. Compare wave speeds (set by linear mass density), then wavelengths and the number of loops (antinodes) in each segment.

Step 1:Same radius, so speed varies with density only.

\frac{v _{1}}{v _{2}} = \frac{ρ _{2}}{ρ _{1}} = 4 = 2

Step 2:Same frequency f throughout, so wavelengths scale with speed.

\frac{λ _{1}}{λ _{2}} = \frac{v _{1}}{v _{2}} = 2

Step 3:Each segment has equal length L with nodes at both ends; number of loops (antinodes) is inversely proportional to wavelength.

\frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{1}{2}

Final answer:

1 : 2

Q14Single correctElectrostatics

There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at the centre of the sphere where the radius vector makes an angle of 6

0^{\circ}

with the direction of the field ?

(A)

589.5

(B)

589.2

(C)

589.4

(D)

589.6

SolutionAnswer: Option 3

589.4

Approach:

In a uniform field the potential varies linearly across the sphere, so the centre potential is the arithmetic mean of the maximum and minimum surface potentials.

Step 1:Surface extremes occur at theta = 0 and theta = pi; their average eliminates the field term and gives the centre value.

V_{P} = \frac{V _{ma x} + V _{min}}{2}

Step 2:Substitute the limits.

V_{P} = \frac{589.8 + 589.0}{2} = 589.4

Final answer:

589.4

Q15Single correctElectrostatics

The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4

μ

C charge, its radius will be : [Take :

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}

N -

m^{2}

C^{2}

]

(A)

20

(B)

32

(C)

28

(D)

16

SolutionAnswer: Option 4

16

Approach:

Use the self-energy of a charged conducting sphere and solve for the radius.

Step 1:Substitute the known values.

4.5 = 9 \times 1 0^{9} \cdot \frac{( 4 \times 1 0 ^{- 6} ) ^{2}}{2 R}

Step 2:Simplify the numerator.

4.5 = \frac{9 \times 1 0 ^{9} \cdot 16 \times 1 0 ^{- 12}}{2 R} = \frac{0.144}{2 R} = \frac{0.072}{R}

Step 3:Solve for R.

R = \frac{0.072}{4.5} = 0.016

= 16

Final answer:

16

Q16Single correctElectronic Devices

What is the conductivity of a semiconductor sample having electron concentration of

5 \times 1 0^{18}

m^{- 3}

, hole concentration of

5 \times 1 0^{19}

m^{- 3}

, electron mobility of

2.0

m^{2}

V^{- 1}

s^{- 1}

and hole mobility of

0.01

m^{2}

V^{- 1}

s^{- 1}

?
(Take charge of electron as

1.6 \times 1 0^{- 19}

(A)

1.68

(Ω -m)^{- 1}

(B)

1.83

(Ω -m)^{- 1}

(C)

0.59

(Ω -m)^{- 1}

(D)

1.20

(Ω -m)^{- 1}

SolutionAnswer: Option 1

1.68

(Ω -m)^{- 1}

Approach:

Apply the conductivity expression for a semiconductor with both electron and hole contributions.

Step 1:Compute electron contribution.

n_{e} μ_{e} = 5 \times 1 0^{18} \times 2.0 = 1.0 \times 1 0^{19}

Step 2:Compute hole contribution.

n_{h} μ_{h} = 5 \times 1 0^{19} \times 0.01 = 5 \times 1 0^{17}

Step 3:Substitute into conductivity formula.

σ = 1.6 \times 1 0^{- 19} \times (1.05 \times 1 0^{19})

Final answer:

1.68

(Ω -m)^{- 1}

Q17Single correctCurrent Electricity

9

V battery with internal resistance of

0.5

Ω

is connected across an infinite network as shown in the figure. All ammeters

A_{1}

A_{2}

A_{3}

and voltmeter V are ideal.
Choose correct statement.

An infinite ladder resistor network. A 9 V battery B with internal resistance 0.5 ohm and an ideal voltmeter V are on the left. Top rail has ammeter A1 then three 1 ohm resistors; vertical 4 ohm shunt branches carry ammeters A2 and A3 (third shunt has no ammeter); bottom rail has three 1 ohm resistors; the network extends to infinity.

(A)

Reading of

A_{1}

2

(B)

Reading of

A_{1}

18

(C)

Reading of V is

9

(D)

Reading of V is

7

SolutionAnswer: Option 1Reading of

A_{1}

2

Approach:

Find the equivalent resistance of the self-similar infinite ladder using the recursive relation, then apply Ohm's law to the full loop.

Step 1:Each repeating cell adds two series

1 Ω

resistors (top and bottom) and a

4 Ω

shunt. Let R be the equivalent resistance of the infinite network.

R = 1 + 1 + \frac{4 R}{4 + R}

Step 2:Clear the fraction and solve the quadratic.

R^{2} - 2 R - 8 = 0 \Rightarrow R = 4 Ω

Step 3:Total resistance includes the internal resistance; compute the main current through

A_{1}

I = \frac{9}{4 + 0.5} = 2 A

Final answer: Reading of

A_{1}

2

Q18Single correctMoving Charges and Magnetism

In a certain region static electric and magnetic fields exist. The magnetic field is given by

B = B_{0} (\hat{i} + 2 \hat{j} - 4 \hat{k})

. If a test charge moving with a velocity

v = v_{0} (3 \hat{i} - \hat{j} + 2 \hat{k})

experiences no force in that region, then the electric field in the region, in SI units, is :

(A)

E = - v_{0} B_{0} (3 \hat{i} - 2 \hat{j} - 4 \hat{k})

(B)

E = - v_{0} B_{0} (\hat{i} + \hat{j} + 7 \hat{k})

(C)

E = v_{0} B_{0} (14 \hat{j} + 7 \hat{k})

(D)

E = - v_{0} B_{0} (14 \hat{j} + 7 \hat{k})

SolutionAnswer: Option 4

E = - v_{0} B_{0} (14 \hat{j} + 7 \hat{k})

Approach:

Zero net Lorentz force requires the electric force to cancel the magnetic force, giving E = -(v x B).

Step 1:Evaluate the cross product of velocity and magnetic field.

v \times B = v_{0} B_{0} [(- 1) (- 4) - (2) (2)] \hat{i} - v_{0} B_{0} [(3) (- 4) - (2) (1)] \hat{j} + v_{0} B_{0} [(3) (2) - (- 1) (1)] \hat{k}

Step 2:Apply the force balance to obtain the electric field.

E = - v \times B = - v_{0} B_{0} (14 \hat{j} + 7 \hat{k})

Final answer:

E = - v_{0} B_{0} (14 \hat{j} + 7 \hat{k})

Q19Single correctMoving Charges and Magnetism

A magnetic dipole in a constant magnetic field has :

(A)

maximum potential energy when the torque is maximum.

(B)

zero potential energy when the torque is minimum.

(C)

zero potential energy when the torque is maximum.

(D)

minimum potential energy when the torque is maximum.

SolutionAnswer: Option 3zero potential energy when the torque is maximum.

Approach:

Compare the angular dependence of the dipole potential energy and torque.

Step 1:Torque is maximum when sine is maximum, i.e. at theta = 90 degrees.

τ_{ma x} at θ = 9 0^{\circ}

Step 2:Evaluate the potential energy at this orientation.

U = - m B cos 9 0^{\circ} = 0

Final answer: zero potential energy when the torque is maximum.

Q20Single correctElectromagnetic Induction

A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current

I = I_{o} cos (ω t)

. The emf induced in the smaller inner loop is nearly :

(A)

\frac{π μ _{o} I _{o}}{2} \cdot \frac{a ^{2}}{b} ω sin (ω t)

(B)

\frac{π μ _{o} I _{o}}{2} \cdot \frac{a ^{2}}{b} ω cos (ω t)

(C)

π μ_{o} I_{o} \frac{a ^{2}}{b} ω sin (ω t)

(D)

\frac{π μ _{o} I _{o} b ^{2}}{a} ω cos (ω t)

SolutionAnswer: Option 1

\frac{π μ _{o} I _{o}}{2} \cdot \frac{a ^{2}}{b} ω sin (ω t)

Approach:

Approximate the field of the large loop as uniform over the small loop, compute the flux linkage, and differentiate to get the induced emf.

Step 1:Flux through the small loop assuming nearly uniform field over its area.

Φ = B π a^{2} = \frac{μ _{o} I}{2 b} π a^{2}

Step 2:Substitute the alternating current and differentiate.

ε = - \frac{π μ _{o} a ^{2}}{2 b} \frac{d}{d t} (I_{o} cos ω t) = \frac{π μ _{o} a ^{2} I _{o} ω}{2 b} sin ω t

Final answer:

\frac{π μ _{o} I _{o}}{2} \cdot \frac{a ^{2}}{b} ω sin (ω t)

Q21Single correctElectromagnetic Waves

Magnetic field in a plane electromagnetic wave is given by

B = B_{0} sin (k x + ω t) \hat{j}

T
Expression for corresponding electric field will be :
Where c is speed of light.

(A)

E = B_{0} c sin (k x + ω t) \hat{k}

V/m

(B)

E = \frac{B _{0}}{c} sin (k x + ω t) \hat{k}

V/m

(C)

E = - B_{0} c sin (k x + ω t) \hat{k}

V/m

(D)

E = B_{0} c sin (k x - ω t) \hat{k}

V/m

SolutionAnswer: Option 1

E = B_{0} c sin (k x + ω t) \hat{k}

V/m

Approach:

Use E0 = cB0 for magnitude and require the Poynting vector E x B to point along the direction of propagation to fix the direction and sign.

Step 1:The phase (kx + omega t) indicates propagation along the negative x direction.

\overset{n}{^} = - \hat{i}

Step 2:E must be perpendicular to B (which is along j). Test E along +k.

\hat{k} \times \hat{j} = - \hat{i}

Step 3:Apply the amplitude relation with the same phase.

E = c B_{0} sin (k x + ω t) \hat{k}

Final answer:

E = B_{0} c sin (k x + ω t) \hat{k}

V/m

Q22Single correctRay Optics

Let the refractive index of a denser medium with respect to a rarer medium be

n_{12}

and its critical angle be

θ_{C}

. At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is

9 0^{\circ}

. Angle A is given by :

(A)

\frac{1}{cos ^{- 1} ( sin θ _{C} )}

(B)

\frac{1}{tan ^{- 1} ( sin θ _{C} )}

(C)

cos^{- 1} (sin θ_{C})

(D)

tan^{- 1} (sin θ_{C})

SolutionAnswer: Option 4

tan^{- 1} (sin θ_{C})

Approach:

Use the 90 degree condition between reflected and refracted rays to relate the refraction angle to A, then apply Snell's law and the critical angle definition.

Step 1:The reflected ray makes angle A and the refracted ray makes angle r; the 90 degree condition gives r = 90 - A.

A + 9 0^{\circ} + r = 18 0^{\circ} \Rightarrow r = 9 0^{\circ} - A

Step 2:Apply Snell's law for travel from denser to rarer medium.

n_{12} sin A = sin (9 0^{\circ} - A) = cos A

Step 3:Substitute the critical angle relation.

tan A = sin θ_{C} \Rightarrow A = tan^{- 1} (sin θ_{C})

Final answer:

tan^{- 1} (sin θ_{C})

Q23Single correctWave Optics

A single slit of width b is illuminated by a coherent monochromatic light of wavelength

λ

. If the second and fourth minima in the diffraction pattern at a distance

1

m from the slit are at

3

cm and

6

cm respectively from the central maximum, what is the width of the central maximum ? (i.e. distance between first minimum on either side of the central maximum)

(A)

1.5

(B)

3.0

(C)

4.5

(D)

6.0

SolutionAnswer: Option 2

3.0

Approach:

Use the single-slit minima condition to find the spacing per order, then double it for the full central maximum width.

Step 1:Use the second minimum position to find the per-order spacing.

y_{2} = \frac{2 λ D}{b} = 3 cm \Rightarrow \frac{λ D}{b} = 1.5 cm

Step 2:Confirm with the fourth minimum.

y_{4} = \frac{4 λ D}{b} = 6 cm

Step 3:Width of central maximum equals distance between first minima on either side.

W = 2 \times \frac{λ D}{b} = 2 \times 1.5 = 3.0 cm

Final answer:

3.0

Q24Single correctDual Nature of Radiation and Matter

The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to

3 n

, the maximum velocity of the ejected photoelectrons will be :

(A)

less than

3 v

(B)

(C)

more than

3 v

(D)

equal to

3 v

SolutionAnswer: Option 3more than

3 v

Approach:

Apply Einstein's photoelectric equation at both frequencies and compare the maximum kinetic energies.

Step 1:Write the kinetic energy at frequency n and at frequency 3n.

\frac{1}{2} m v^{'2} = h (3 n) - ϕ = (hn - ϕ) + 2 hn

Step 2:Since hn = (1/2)

m v^{2}

+ phi > (1/2)

m v^{2}

, the added term exceeds twice the original KE.

2 hn > 2 (\frac{1}{2} m v^{2}) \Rightarrow v^{'2} > v^{2} + 2 v^{2} = 3 v^{2}

Final answer: more than

3 v

Q25Single correctAtoms

According to Bohr's theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the

n^{t h}

orbit is proportional to : (n = principal quantum number)

(A)

n^{- 4}

(B)

n^{- 5}

(C)

n^{- 3}

(D)

n^{- 2}

SolutionAnswer: Option 2

n^{- 5}

Approach:

Express the orbital current and the field at the centre in terms of Bohr's velocity and radius dependence on n.

Step 1:Express current in terms of n.

I \propto \frac{v}{r} \propto \frac{1/ n}{n ^{2}} = \frac{1}{n ^{3}}

Step 2:Combine with the field expression.

B \propto \frac{I}{r} \propto \frac{n ^{- 3}}{n ^{2}} = n^{- 5}

Final answer:

n^{- 5}

Q26Single correctNuclei

Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron

= 1.1

MeV and for helium

= 7.0

MeV)

(A)

30.2

MeV

(B)

32.4

MeV

(C)

23.6

MeV

(D)

25.8

MeV

SolutionAnswer: Option 3

23.6

MeV

Approach:

Energy released equals the gain in total binding energy of the products over the reactants.

Step 1:Total binding energy of the helium nucleus (4 nucleons).

B E_{He} = 7.0 \times 4 = 28 MeV

Step 2:Total binding energy of two deuterons (2 nucleons each).

B E_{2 D} = 2 \times (1.1 \times 2) = 4.4 MeV

Step 3:Subtract to find energy released.

Q = 28 - 4.4 = 23.6 MeV

Final answer:

23.6

MeV

Q27Single correctElectronic Devices

The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is :

V-I characteristic of a diode. Vertical axis I (mA), horizontal axis V (Volt). Forward bias rises steeply with dashed guide lines marking I = 10, 15, 20 mA (near V = 7 to 8 V). In reverse bias the current is a small flat -1 microampere, with the point at V = -10 V marked.

(A)

10

(B)

1 0^{- 6}

(C)

1 0^{6}

(D)

100

SolutionAnswer: Option 2

1 0^{- 6}

Approach:

Estimate the forward resistance from the slope in the forward region and the reverse resistance from the reverse leakage point, then take their ratio.

Step 1:Reverse bias: at V = -10 V the reverse current is about 1 microampere.

R_{re v} = \frac{10}{1 \times 1 0 ^{- 6}} = 1 0^{7} Ω

Step 2:Forward bias: in the steep forward region the resistance is of the order of a few ohms to tens of ohms.

R_{f w d} \sim 10 Ω

Step 3:Form the ratio of forward to reverse resistance.

\frac{R _{f w d}}{R _{re v}} \approx \frac{10}{1 0 ^{7}} = 1 0^{- 6}

Final answer:

1 0^{- 6}

Q28Single correctCommunication Systems

A signal of frequency

20

kHz and peak voltage of

5

Volt is used to modulate a carrier wave of frequency

1.2

MHz and peak voltage

25

Volts. Choose the correct statement.

(A)

Modulation index

= 5

, side frequency bands are at

1400

kHz and

1000

kHz

(B)

Modulation index

= 5

, side frequency bands are at

21.2

kHz and

18.8

kHz

(C)

Modulation index

= 0.8

, side frequency bands are at

1180

kHz and

1220

kHz

(D)

Modulation index

= 0.2

, side frequency bands are at

1220

kHz and

1180

kHz

SolutionAnswer: Option 4Modulation index

= 0.2

, side frequency bands are at

1220

kHz and

1180

kHz

Approach:

Compute the modulation index as the ratio of message to carrier amplitude and locate the sidebands at carrier plus or minus message frequency.

Step 1:Modulation index from peak voltages.

m = \frac{5}{25} = 0.2

Step 2:Sidebands at carrier 1200 kHz plus or minus message 20 kHz.

1200 \pm 20 = 1220 kHz, 1180 kHz

Final answer: Modulation index

= 0.2

, side frequency bands are at

1220

kHz and

1180

kHz

Q29Single correctExperimental Skills

In a physical balance working on the principle of moments, when

5

mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct ?

(A)

Left arm is longer than the right arm

(B)

Both the arms are of same length

(C)

Left arm is shorter than the right arm

(D)

Every object that is weighed using this balance appears lighter than its actual weight.

SolutionAnswer: Option 3Left arm is shorter than the right arm

Approach:

Apply the principle of moments about the fulcrum with equal pan masses and an extra weight on the left pan that still leaves the beam horizontal.

Step 1:With equal pan masses but an additional 5 mg on the left, the left side carries more weight while the beam remains horizontal.

(W + 5 mg) L_{l e f t} = W L_{r i g h t}

Step 2:For the moments to balance, the larger left load must act on a shorter arm.

L_{l e f t} = \frac{W}{W + 5 mg} L_{r i g h t} < L_{r i g h t}

Final answer: Left arm is shorter than the right arm

Q30Single correctCurrent Electricity

A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads

1.0

A when two way key

K_{3}

is open. The balance point is at a length

l_{1}

cm from P when two way key

K_{3}

is plugged in between

2

and

1

, while the balance point is at a length

l_{2}

cm from P when key

K_{3}

is plugged in between

3

and

1

. The ratio of two resistances

\frac{R _{1}}{R _{2}}

, is found to be :

A potentiometer circuit. Top branch: ammeter A, driver cell E2, rheostat Rh2, key K2. Middle rail from node P: resistor R1 to node 2, resistor R2 to node 3. Two-way key K3 (terminals 2 and 3 to common 1) with galvanometer G whose jockey lead reaches the potentiometer wire PQ. Bottom driver branch: cell E1, rheostat Rh1, key K1.

(A)

\frac{l _{1}}{l _{1} + l _{2}}

(B)

\frac{l _{2}}{l _{2} - l _{1}}

(C)

\frac{l _{1}}{l _{1} - l _{2}}

(D)

\frac{l _{1}}{l _{2} - l _{1}}

SolutionAnswer: Option 4

\frac{l _{1}}{l _{2} - l _{1}}

Approach:

The balance length is proportional to the potential drop measured across the chosen part of the resistance combination; form the two relations and eliminate the current and potential gradient.

Step 1:With K3 between 2 and 1 the drop across R1 is balanced at length l1.

I R_{1} = k l_{1}

Step 2:With K3 between 3 and 1 the drop across R1 + R2 is balanced at length l2.

I (R_{1} + R_{2}) = k l_{2}

Step 3:Divide and solve for the ratio of resistances.

\frac{R _{1}}{R _{1} + R _{2}} = \frac{l _{1}}{l _{2}} \Rightarrow \frac{R _{1}}{R _{2}} = \frac{l _{1}}{l _{2} - l _{1}}

Final answer:

\frac{l _{1}}{l _{2} - l _{1}}

Chemistry30 questions

Q31Single correctSurface Chemistry

Among the following, correct statement is :

(A)

Brownian movement is more pronounced for smaller particles than for bigger-particles.

(B)

Sols of metal sulphides are lyophilic.

(C)

Hardy Schulze law states that bigger the size of the ions, the greater is its coagulating power.

(D)

One would expect charcoal to adsorb chlorine more than hydrogen sulphide.

SolutionAnswer: Option 1Brownian movement is more pronounced for smaller particles than for bigger-particles.

Approach:

Evaluate each statement against established surface-chemistry facts on Brownian motion, lyophilic/lyophobic sols, the Hardy-Schulze rule and selective adsorption.

Step 1:Brownian motion arises from unbalanced molecular bombardment; smaller, lighter colloidal particles are jostled more, so the zig-zag motion is more pronounced for smaller particles.

Step 2:Metal sulphide sols are inorganic and have little affinity for the dispersion medium, hence lyophobic, not lyophilic.

Step 3:Hardy-Schulze rule: coagulating power depends on the magnitude of the charge of the active (oppositely charged) ion, not on its size.

Step 4:Adsorption increases with ease of liquefaction (higher critical temperature). H2S liquefies more readily than Cl2-equivalent comparison favours H2S, so charcoal adsorbs hydrogen sulphide more, not chlorine.

H_{2} S

Final answer: Brownian movement is more pronounced for smaller particles than for bigger-particles.

Q32Single correctSome Basic Concepts in Chemistry

Excess of NaOH (aq) was added to 100 mL of

FeCl_{3}

(aq) resulting into 2.14 g of

Fe(OH)_{3}

. The molarity of

FeCl_{3}

(aq) is : (Given molar mass of Fe = 56 g mo

l^{- 1}

and molar mass of Cl = 35.5 g mo

l^{- 1}

)

(A)

0.2 M

(B)

0.3 M

(C)

0.6 M

(D)

1.8 M

SolutionAnswer: Option 10.2 M

Approach:

Find moles of Fe(OH)3 precipitated, equate to moles of FeCl3 (1:1), then divide by volume in litres.

Step 1:Molar mass of Fe(OH)3 = 56 + 3(16+1) = 56 + 51 = 107 g

m o l^{- 1}

56 + 3 (17) = 107

Step 2:Moles of Fe(OH)3 = 2.14 / 107 = 0.02 mol.

\frac{2.14}{107} = 0.02

Step 3:Reaction FeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl gives 1 mol FeCl3 per mol Fe(OH)3, so moles FeCl3 = 0.02 mol in 0.1 L.

M = \frac{0.02}{0.1}

Final answer: 0.2 M

Q33Single correctStates of Matter

Among the following, the incorrect statement is :

(A)

At low pressure, real gases show ideal behaviour.

(B)

At very low temperature, real gases show ideal behaviour.

(C)

At very large volume, real gases show ideal behaviour.

(D)

At Boyle's temperature, real gases show ideal behaviour.

SolutionAnswer: Option 2At very low temperature, real gases show ideal behaviour.

Approach:

Recall the conditions under which intermolecular forces and molecular volume become negligible, making real gases approach ideal behaviour.

Step 1:Low pressure and large volume make both correction terms (a

n^{2}

V^{2}

and nb) negligible, so statements (1) and (3) are correct.

Step 2:At Boyle temperature, deviations cancel over a range of pressures and PV = nRT holds, so (4) is correct.

Step 3:At very low temperature, attractive forces dominate (gases approach condensation), causing maximum deviation from ideality; high temperature, not low, favours ideal behaviour.

Final answer: At very low temperature, real gases show ideal behaviour.

Q34Single correctChemical Thermodynamics

For a reaction, A(g)

\to

A(l);

Δ

H =

-

3RT. The correct statement for the reaction is :

(A)

Δ H = Δ U \neq = O

(B)

Δ H = Δ U = O

(C)

∣Δ H ∣ < ∣Δ U ∣

(D)

∣Δ H ∣ > ∣Δ U ∣

SolutionAnswer: Option 4

∣Δ H ∣ > ∣Δ U ∣

Approach:

Use the relation between enthalpy and internal energy with change in moles of gas to compute Delta U and compare magnitudes.

Step 1:Gaseous moles change from 1 (A gas) to 0 (A liquid), so Delta

n_{g}

= 0 - 1 = -1.

Δ n_{g} = - 1

Step 2:Substitute: -3RT = Delta U + (-1)RT, hence Delta U = -3RT + RT = -2RT.

Δ U = Δ H - Δ n_{g} RT = - 3 RT - (- 1) RT = - 2 RT

3 RT > 2 RT

Final answer:

∣Δ H ∣ > ∣Δ U ∣

Q35Single correctElectrochemistry

What is the standard reduction potential (

E^{\circ}

) for

Fe^{3 +} \to

Fe ? Given that :

Fe^{2 +} + 2 e^{-} \to

Fe;

E_{Fe^{2 +} / Fe}^{\circ} = - 0.47

Fe^{3 +} + e^{-} \to Fe^{2 +}

;

E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = + 0.77

(A)

- 0.057

(B)

+ 0.057

(C)

+ 0.30

(D)

- 0.30

SolutionAnswer: Option 1

- 0.057

Approach:

Combine the two half-reactions through Gibbs energies (DeltaG = -nFE), since potentials are not additive but DeltaG values are.

Step 1:Target Fe3+ + 3e- -> Fe is the sum of Fe3+ + e- -> Fe2+ (n=1) and Fe2+ + 2e- -> Fe (n=2).

Step 2:Add Gibbs energies: -3F E = (-1)F(0.77) + (-2)F(-0.47).

- 3 F E^{\circ} = - F (0.77) + (- 2 F) (- 0.47)

Step 3:Solve: -3 E = -0.17, so E = 0.17/3 ... computed as -3E = 0.77 - 0.94 = -0.17 gives E = 0.0567; re-evaluate signs: -3F E = -0.77F + 0.94F = 0.17F, hence E = -0.0567 V.

E^{\circ} = \frac{0.77 - 2 ( 0.47 )}{3} = \frac{0.77 - 0.94}{3}

then divided by sign

Final answer:

- 0.057

Q36Single correctAtomic Structure

If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of H

e^{+}

is :

(A)

\frac{5 A}{9}

(B)

\frac{9 A}{5}

(C)

\frac{36 A}{5}

(D)

\frac{36 A}{7}

SolutionAnswer: Option 4

\frac{36 A}{7}

Approach:

Use the Rydberg formula for hydrogen-like species to express A from the shortest Lyman line of H, then find the longest Paschen line of He+.

Step 1:Shortest Lyman wavelength of H (Z=1): n1=1, n2=infinity gives 1/A = R, so R = 1/A.

\frac{1}{A} = R (1) (\frac{1}{1} - 0)

Step 2:Longest Paschen line of He+ (Z=2): n1=3, n2=4. 1/lambda = R(4)(1/9 - 1/16) = R(4)(7/144) = 7R/36.

\frac{1}{λ} = R (4) (\frac{1}{9} - \frac{1}{16}) = \frac{7 R}{36}

Step 3:Substitute R = 1/A: lambda = 36/(7R) = 36A/7.

λ = \frac{36}{7 R} = \frac{36 A}{7}

Final answer:

\frac{36 A}{7}

Q37Single correctSolutions

5 g of

Na_{2} SO_{4}

was dissolved in x g of

H_{2} O

. The change in freezing point was found to be 3.8

2^{\circ}

C. If

Na_{2} SO_{4}

is 81.5% ionised, the value of x (

K_{f}

for water = 1.8

6^{\circ}

C kg mo

l^{- 1}

) is approximately : (molar mass of S = 32 g mo

l^{- 1}

and that of Na = 23 g mo

l^{- 1}

)

(A)

15 g

(B)

25 g

(C)

45 g

(D)

65 g

SolutionAnswer: Option 345 g

Approach:

Compute the van't Hoff factor from the degree of ionisation, then apply the depression of freezing point formula to find the mass of water.

Step 1:Molar mass Na2SO4 = 2(23) + 32 + 4(16) = 142 g

m o l^{- 1}

; moles = 5/142 = 0.0352 mol.

\frac{5}{142} = 0.0352

Step 2:Na2SO4 -> 2Na+ +

SO 4^{2}

- gives n=3 ions; i = 1 + 0.815(3-1) = 1 + 1.63 = 2.63.

i = 1 + 0.815 (2) = 2.63

Step 3:Molality m =

De lt a T_{f}

/(i

K_{f}

) = 3.82/(2.63 x 1.86) = 0.781 mol

k g^{- 1}

m = \frac{3.82}{2.63 \times 1.86} = 0.781

Step 4:Mass of water = moles/m = 0.0352/0.781 = 0.0451 kg = 45.1 g.

x = \frac{0.0352}{0.781} = 0.0451 kg

Final answer: 45 g

Q38Single correctEquilibrium

Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is 1

0^{- 5}

, the ratio of salt to acid concentration in the buffer solution will be :

(A)

4 : 5

(B)

1 : 10

(C)

10 : 1

(D)

5 : 4

SolutionAnswer: Option 310 : 1

Approach:

Apply the Henderson-Hasselbalch equation for an acidic buffer using the given pH and ionisation constant.

Step 1:pKa = -log(

1 0^{-} 5

) = 5.

p K_{a} = 5

Step 2:Substitute pH = 6: 6 = 5 + log([salt]/[acid]).

6 = 5 + lo g \frac{[ salt ]}{[ acid ]}

Step 3:log(ratio) = 1 gives [salt]/[acid] = 10, i.e. 10 : 1.

\frac{[ salt ]}{[ acid ]} = 10

Final answer: 10 : 1

Q39Single correctChemical Kinetics

The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.

(A)

9.84 K

(B)

4.92 K

(C)

2.45 K

(D)

19.67 K

SolutionAnswer: Option 24.92 K

Approach:

Use the Arrhenius equation in two-temperature form; both reactions double their rate (same ln2), so equate the two expressions with

E a_{B}

= 2

E a_{A}

Step 1:For A: ln2 = (

E a_{A}

/R)(1/300 - 1/310). For B: ln2 = (

2 E a_{A}

/R)(1/300 - 1/T).

Step 2:Equate right sides: (1/300 - 1/310) = 2(1/300 - 1/T), so (1/300 - 1/T) = (1/2)(1/300 - 1/310).

\frac{1}{300} - \frac{1}{310} = \frac{10}{93000} = 1.075 \times 1 0^{- 4}

Step 3:1/T = 1/300 - 5.376e-5 = 3.2796e-3, so T = 304.92 K; increase = 304.92 - 300 = 4.92 K.

Δ T = 304.92 - 300

Final answer: 4.92 K

Q40Single correctChemical Thermodynamics

The enthalpy change on freezing of 1 mol of water at

5^{\circ}

C to ice at

- 5^{\circ}

C is : (Given

Δ_{f u s}

H = 6 kJ mo

l^{- 1}

0^{\circ}

C_{p} (H_{2} O

, l) = 75.3 J mo

l^{- 1}

K^{- 1}

C_{p} (H_{2} O

, s) = 36.8 J mo

l^{- 1}

K^{- 1}

)

(A)

5.44 kJ mo

l^{- 1}

(B)

5.81 kJ mo

l^{- 1}

(C)

6.56 kJ mo

l^{- 1}

(D)

6.00 kJ mo

l^{- 1}

SolutionAnswer: Option 36.56 kJ mo

l^{- 1}

Approach:

Construct a three-step path (cool water 5 to 0, freeze at 0, cool ice 0 to -5) and sum the enthalpy changes by Hess's law.

Step 1:Cool water from 5 to 0 C: 1 x 75.3 x (-5) = -376.5 J.

75.3 \times (- 5) = - 376.5

Step 2:Freezing at 0 C is the negative of fusion: -6000 J.

- Δ_{f u s} H = - 6000

Step 3:Cool ice from 0 to -5 C: 1 x 36.8 x (-5) = -184 J.

36.8 \times (- 5) = - 184

Step 4:Sum: -376.5 - 6000 - 184 = -6560.5 J = -6.56 kJ

m o l^{- 1}

; magnitude 6.56 kJ

m o l^{- 1}

- 376.5 - 6000 - 184 = - 6560.5 J

Final answer: 6.56 kJ mo

l^{- 1}

Q41Single correctChemical Bonding and Molecular Structure

Which of the following is paramagnetic ?

(A)

NO^{+}

(B)

CO

(C)

O_{2}^{2 -}

(D)

B_{2}

SolutionAnswer: Option 4

B_{2}

Approach:

Use molecular orbital theory to count valence electrons and determine the presence of unpaired electrons for each species.

Step 1:NO+ has 14 electrons; configuration fills bonding/antibonding pairs completely, no unpaired electrons (diamagnetic).

Step 2:CO has 14 electrons, isoelectronic with N2, all paired (diamagnetic).

Step 3:

O 2^{2}

- (peroxide) has 18 electrons; pi* orbitals are completely filled, all paired (diamagnetic).

Step 4:B2 has 10 electrons; configuration places one electron in each of pi(2px) and pi(2py), giving two unpaired electrons (paramagnetic).

π 2 p_{x}^{1} π 2 p_{y}^{1}

Final answer:

B_{2}

Q42Single correctd- and f-Block Elements / Coordination Compounds

The pair of compounds having metals in their highest oxidation state is :

(A)

MnO_{2}

and

CrO_{2} Cl_{2}

(B)

[NiCl_{4}]^{2 -}

and

[CoCl_{4}]^{2 -}

(C)

[Fe(CN)_{6}]^{3 -}

and

[Cu(CN)_{4}]^{2 -}

(D)

[FeCl_{4}]^{-}

and

Co_{2} O_{3}

SolutionAnswer: Option 4

[FeCl_{4}]^{-}

and

Co_{2} O_{3}

Approach:

Assign the oxidation state of the metal in each compound and identify the pair where both metals are at their highest commonly attained oxidation state.

Step 1:Option 1: MnO2 -> Mn(+4) (Mn max is +7, not highest); CrO2Cl2 -> Cr(+6) highest. Not both at highest.

Step 2:Option 2: [NiCl4]2- -> Ni(+2); [CoCl4]2- -> Co(+2). Neither at highest.

Step 3:Option 3: [Fe(CN)6]3- -> Fe(+3); [Cu(CN)4]2- -> Cu(+2). Cu(+2) is not its highest stable state in this context.

Step 4:Option 4: [FeCl4]- -> Fe(+3) (highest common state of iron) and Co2O3 -> Co(+3) (highest common state of cobalt). Both metals are in their highest stable oxidation states.

Fe (+ 3), Co (+ 3)

Final answer:

[FeCl_{4}]^{-}

and

Co_{2} O_{3}

Q43Single correctChemical Bonding and Molecular Structure

p^{3} d^{2}

hybridization is not displayed by :

(A)

BrF_{5}

(B)

SF_{6}

(C)

[CrF_{6}]^{3 -}

(D)

PF_{5}

SolutionAnswer: Option 4

PF_{5}

Approach:

Determine the number of electron domains (steric number) around the central atom; sp3d2 corresponds to six domains (octahedral).

Step 1:BrF5: 5 bond pairs + 1 lone pair = 6 domains -> sp3d2 (square pyramidal).

5 + 1 = 6

Step 2:SF6: 6 bond pairs = 6 domains -> sp3d2 (octahedral).

6

Step 3:[CrF6]3-: octahedral complex, 6 ligands -> sp3d2 (outer orbital).

6

Step 4:PF5: 5 bond pairs = 5 domains -> sp3d (trigonal bipyramidal), not sp3d2.

5

Final answer:

PF_{5}

Q44Single correctEnvironmental Chemistry

Identify the pollutant gases largely responsible for the discoloured and lustreless nature of marble of the Taj Mahal.

(A)

O_{3}

and

CO_{2}

(B)

CO_{2}

and

NO_{2}

(C)

SO_{2}

and

NO_{2}

(D)

SO_{2}

and

O_{3}

SolutionAnswer: Option 3

SO_{2}

and

NO_{2}

Approach:

Recall the environmental chemistry of acid rain and its action on marble (calcium carbonate).

Step 1:SO2 and NO2 from fuel combustion oxidise and dissolve in rain to form sulphuric and nitric acids (acid rain).

SO_{2}, NO_{2}

Step 2:These acids react with the calcium carbonate of marble, corroding and discolouring it, causing the lustreless surface.

CaCO_{3} + acid

Final answer:

SO_{2}

and

NO_{2}

Q45Single correctHydrogen / Redox Reactions

In which of the following reactions, hydrogen peroxide acts as an oxidizing agent ?

(A)

HOCl + H_{2} O_{2} \to H_{3} O^{+} + Cl^{-} + O_{2}

(B)

I_{2} + H_{2} O_{2} + 2 OH^{-} \to 2 I^{-} + 2 H_{2} O + O_{2}

(C)

2 MnO_{4}^{-} + 3 H_{2} O_{2} \to 2 MnO_{2} + 3 O_{2} + 2 H_{2} O + 2 OH^{-}

(D)

PbS + 4 H_{2} O_{2} \to PbSO_{4} + 4 H_{2} O

SolutionAnswer: Option 4

PbS + 4 H_{2} O_{2} \to PbSO_{4} + 4 H_{2} O

Approach:

An oxidising agent is itself reduced; track the oxidation state of oxygen in H2O2 (which is -1). If it falls to -2 (water), H2O2 is the oxidiser; if it rises to 0 (O2), H2O2 is the reducer.

Step 1:In options 1, 2 and 3, H2O2 is converted to O2 (oxygen goes -1 to 0, oxidised), so H2O2 acts as a reducing agent there.

O : - 1 \to 0

Step 2:In option 4, sulphur of PbS goes from -2 to +6 (oxidised), while oxygen of H2O2 goes -1 to -2 in water (reduced).

S : - 2 \to + 6; O : - 1 \to - 2

Step 3:Since H2O2 is reduced and oxidises PbS to PbSO4, it acts as an oxidising agent in reaction (4).

Final answer:

PbS + 4 H_{2} O_{2} \to PbSO_{4} + 4 H_{2} O

Q46Single correctClassification of Elements and Periodicity in Properties

(A)

Both 'A' and 'B' belong to group-1 where 'B' comes below 'A'.

(B)

Both 'A' and 'B' belong to group-1 where 'A' comes below 'B'.

(C)

Both 'A' and 'B' belong to group-2 where 'B' comes below 'A'.

(D)

Both 'A' and 'B' belong to group-2 where 'A' comes below 'B'.

SolutionAnswer: Option 3Both 'A' and 'B' belong to group-2 where 'B' comes below 'A'.

Approach:

Identify the group from the position of the large jump in successive ionization enthalpies, then order the two elements within the group using the first ionization enthalpy.

Step 1:For both A and B the third ionization enthalpy is far larger than the second (A: 1757 to 14847; B: 1450 to 7731), so removal of the third electron breaks a noble-gas core. Each element therefore has two easily-removable valence electrons.

large jump after 2^{n d} IE

Step 2:Within a group ionization enthalpy decreases on moving down. A has the higher first IE (899 > 737), so A lies higher and B lies below A.

899 > 737

Final answer: Both 'A' and 'B' belong to group-2 where 'B' comes below 'A'.

Q47Single correctGeneral Principles and Processes of Isolation / p-Block Elements

Consider the following standard electrode potentials (V in aqueous solution):

Element Al:

M^{3 +} / M = - 1.66

M^{+} / M = + 0.55

Element Tl:

M^{3 +} / M = + 1.26

M^{+} / M = - 0.34

Based on these data, which of the following statements is correct?

(A)

Tl^{+}

is more stable than

Al^{3 +}

(B)

Al^{3 +}

is more stable than

Tl^{3 +}

(C)

Tl^{+}

is more stable than

Al^{+}

(D)

Tl^{3 +}

is more stable than

Al^{3 +}

SolutionAnswer: Option 3

Tl^{+}

is more stable than

Al^{+}

Approach:

Interpret the sign and magnitude of the standard reduction potentials: a large positive potential means the ion is readily reduced (less stable in that oxidation state); compare the +1 states of Al and Tl.

Step 1:For thallium the inert pair effect makes the +1 state stable; its M+/M potential is negative (-0.34), so Tl+ is not easily reduced and is stable.

E^{\circ} (Tl^{+} / Tl) = - 0.34

Step 2:For aluminium the +1 state is unstable; its M+/M potential is positive (+0.55), so Al+ is readily reduced to Al (and disproportionates), making Al+ unstable.

E^{\circ} (Al^{+} / Al) = + 0.55

Step 3:Comparing the two monovalent ions, Tl+ is more stable than Al+.

- 0.34 < + 0.55

Final answer:

Tl^{+}

is more stable than

Al^{+}

Q48Single corrects-Block Elements

A metal 'M' reacts with nitrogen gas to afford '

M_{3}

N'. '

M_{3}

N' on heating at high temperature gives back 'M' and on reaction with water produces a gas 'B'. Gas 'B' reacts with aqueous solution of

CuSO_{4}

to form a deep blue compound. 'M' and 'B' respectively are :

(A)

Li and

NH_{3}

(B)

Ba and

N_{2}

(C)

Na and

NH_{3}

(D)

Al and

N_{2}

SolutionAnswer: Option 1Li and

NH_{3}

Approach:

Use the M3N stoichiometry to fix the oxidation state of M, identify gas B from the deep-blue CuSO4 test, and select the metal that reacts directly with nitrogen.

Step 1:The nitride M3N requires M to be monovalent (M+), so M is an alkali metal (Li or Na).

3 M^{+} + N^{3 -}

Step 2:Gas B gives a deep blue compound with CuSO4, which is the test for ammonia forming the tetraamminecopper(II) complex. Hence B is NH3.

[Cu(NH_{3})_{4}]^{2 +} (deep blue)

Step 3:Among the alkali metals only lithium reacts directly with nitrogen to give Li3N. Therefore M = Li.

6 Li + N_{2} \to 2 Li_{3} N

Final answer: Li and

NH_{3}

Q49Single correctp-Block Elements (Group 16)

The number of

S=O

and

S - OH

bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are :

(A)

(2 and 2) and (2 and 2)

(B)

(2 and 4) and (2 and 2)

(C)

(4 and 2) and (2 and 4)

(D)

(4 and 2) and (4 and 2)

SolutionAnswer: Option 4(4 and 2) and (4 and 2)

Approach:

Draw the structures of the two oxoacids and count the S=O double bonds and S-OH single bonds in each.

Step 1:Peroxodisulphuric acid (Marshall's acid) is HO-SO2-O-O-SO2-OH. Each sulphur bears two S=O bonds and one S-OH; the two sulphurs are linked by a peroxide bridge.

HO-SO_{2} -O-O-SO_{2} -OH

Step 2:Pyrosulphuric acid (oleum, disulphuric acid) is HO-SO2-O-SO2-OH. Each sulphur again bears two S=O bonds and one S-OH, joined by a single oxygen bridge.

HO-SO_{2} -O-SO_{2} -OH

Final answer: (4 and 2) and (4 and 2)

Q50Single correctEnvironmental / Qualitative Inorganic Analysis

A solution containing a group-IV cation gives a precipitate on passing

H_{2} S

. A solution of this precipitate in dil.HCl produces a non-toxic gas/compound on reaction with potassium ferrocyanide. The cation is :

(A)

Co^{2 +}

(B)

Ni^{2 +}

(C)

Mn^{2 +}

(D)

Zn^{2 +}

SolutionAnswer: Option 4

Zn^{2 +}

Approach:

Use the solubility of the group-IV sulphides in dilute HCl and the confirmatory potassium ferrocyanide test to identify the cation.

Step 1:Among the group-IV cations, the sulphide that dissolves in dilute HCl to give a clean solution is ZnS; CoS and NiS are insoluble in dilute HCl.

ZnS + 2 HCl \to ZnCl_{2} + H_{2} S

Step 2:The dil.HCl solution gives a bluish-white (non-toxic) precipitate of zinc ferrocyanide with potassium ferrocyanide, confirming Zn2+.

Zn_{2} [Fe(CN)_{6}]

Final answer:

Zn^{2 +}

Q51Single correctPurification and Characterisation of Organic Compounds

A mixture containing the following four compounds is extracted with 1M HCl. The compound that goes to aqueous layer is :

Four organic structures labelled (I), (II), (III), (IV), each a cyclohexane ring: (I) ring bearing -S-CH3 (methyl thioether); (II) ring bearing -N(H)-CH3 (secondary amine); (III) ring bearing -O-CH3 (methyl ether); (IV) ring bearing -C(=O)-CH3 (methyl ketone).

(A)

(I)

(B)

(II)

(C)

(III)

(D)

(IV)

SolutionAnswer: Option 2(II)

Approach:

Determine which functional class among the four (thiol I, amine II, ether III, ketone IV) is basic enough to be protonated by dilute HCl and pass into the aqueous layer as a salt.

Step 1:Extraction with 1M HCl removes basic compounds by forming water-soluble ammonium salts.

base + H^{+} \to conjugate acid (water soluble)

Step 2:Of the four compounds, only the amine (II) is basic; the thiol (I), ether (III) and ketone (IV) are not protonated by dilute HCl and remain in the organic layer.

amine (II) \to salt

Final answer: (II)

Q52Single correctChemistry in Everyday Life

The reason for 'drug induced poisoning' is :

(A)

Binding reversibly at the active site of the enzyme

(B)

Bringing conformational change in the binding site of enzyme

(C)

Binding irreversibly to the active site of the enzyme

(D)

Binding at the allosteric sites of the enzyme

SolutionAnswer: Option 3Binding irreversibly to the active site of the enzyme

Approach:

Recall the mechanism by which poisons/drugs permanently inhibit enzyme action.

Step 1:Drug-induced poisoning results when the drug binds strongly (often covalently and irreversibly) to the active site, blocking the substrate permanently and inhibiting enzyme function.

irreversible binding at active site

Final answer: Binding irreversibly to the active site of the enzyme

Q53Single correctHydrocarbons / Haloalkanes and Haloarenes

Which of the following compounds will not undergo Friedel Craft's reaction with benzene ?

(A)

CH2=CH-C(=O)Cl, acryloyl chloride (prop-2-enoyl chloride).

(B)

(C)

(D)

CH2=C(CH3)-COCl, methacryloyl chloride (2-methylprop-2-enoyl chloride).

SolutionAnswer: Option 3

CH_{2} =CH-CH_{2} -CH_{2} Cl

(but-3-enyl chloride)

Approach:

Friedel-Crafts proceeds through an acylium ion (from acid chlorides) or a stabilized carbocation (from suitable alkyl halides). Identify the compound that can give neither.

Step 1:Options (1) acryloyl chloride and (4) methacryloyl chloride are acid chlorides; they form acylium ions and undergo Friedel-Crafts acylation.

RCOCl \to RCO^{+}

Step 2:Option (2) is allyl chloride; ionization gives a resonance-stabilized allyl cation, so it undergoes Friedel-Crafts alkylation.

CH_{2} =CH-CH_{2}^{+}

Step 3:Option (3) is but-3-enyl chloride; the C-Cl is on a primary sp3 carbon not adjacent to the double bond, so ionization would give an unstabilized primary carbocation. It does not undergo Friedel-Crafts.

CH_{2} =CH-CH_{2} -CH_{2}^{+} (1^{\circ})

Final answer:

CH_{2} =CH-CH_{2} -CH_{2} Cl

(but-3-enyl chloride)

Q54Single correctBiomolecules

Among the following, the essential amino acid is :

(A)

Alanine

(B)

Valine

(C)

Aspartic acid

(D)

Serine

SolutionAnswer: Option 2Valine

Approach:

Recall which listed amino acids the human body cannot synthesise and must obtain from diet.

Step 1:Essential amino acids cannot be synthesised by the body. Among the options, valine is essential, whereas alanine, aspartic acid and serine are non-essential.

Valine = essential

Final answer: Valine

Q55Single correctAldehydes, Ketones and Carboxylic Acids / Alcohols

The major product expected from the following reaction is :

Benzene ring with four contiguous substituents -CO2H, -CH2OH, -C(=O)NH2 and -OH (1,2,3,4); reacted with HCl(g)/CCl4.

(A)

Phthalide (isobenzofuran-1(3H)-one) with -CO2H ortho to the CH2 fusion carbon and -OH ortho to the carbonyl fusion carbon.

(B)

Benzene ring with -CO2H, -CH2OH, -C(=O)NH2 and a Cl substituent (aromatic chlorination product; amide and CH2OH intact).

(C)

Phthalide with a -C(=O)NH2 ortho to the CH2 fusion carbon and an -OH adjacent to the amide.

(D)

Phthalide with -CO2H ortho to the CH2 fusion carbon and Cl ortho to the carbonyl fusion carbon.

SolutionAnswer: Option 1Phthalide (5-membered lactone) bearing the -CO2H and -OH groups, formed by cyclization of -CH2OH with the ortho -CONH2

Approach:

Anhydrous HCl(g) in CCl4 promotes intramolecular cyclization of the benzylic -CH2OH with the ortho carbonyl to give a five-membered lactone (phthalide), expelling ammonia from the amide as ammonium chloride.

Step 1:The benzylic primary alcohol is ortho to the amide carbonyl. Under HCl(g)/CCl4 the -CH2OH attacks the protonated amide carbonyl, cyclizing to a five-membered lactone (phthalide ring) and releasing NH3 (as NH4Cl).

-CH_{2} -O-C(=O)- (phthalide)

Step 2:The carboxylic acid (-CO2H) and phenolic -OH on the ring are retained. The product is the phthalide bearing -CO2H and -OH (option 1).

phthalide with -CO_{2} H and -OH

Final answer: Phthalide (5-membered lactone) bearing the -CO2H and -OH groups

Q56Single correctHaloalkanes and Haloarenes / Hydrocarbons

The major product of the following reaction is :

Reaction scheme: CH3-CH(Br)-CH2-CH(Br)-CH2-CH3 (2,4-dibromohexane) heated with KOH in CH3OH.

(A)

CH_{2} =CHCH=CHCH_{2} CH_{3}

(B)

CH_{2} =CHCH_{2} CH=CHCH_{3}

(C)

CH_{3} CH=CH-CH=CHCH_{3}

(D)

CH_{3} CH=CHCH_{2} CH=CH_{2}

SolutionAnswer: Option 3

CH_{3} CH=CH-CH=CHCH_{3}

Approach:

Alcoholic KOH causes double dehydrohalogenation of 2,4-dibromohexane; each elimination follows Saytzeff and the diene that is internal and conjugated is the most stable major product.

Step 1:The substrate is 2,4-dibromohexane, CH3-CHBr-CH2-CHBr-CH2-CH3. Alcoholic KOH eliminates two molecules of HBr.

CH_{3} -CHBr-CH_{2} -CHBr-CH_{2} -CH_{3}

Step 2:Eliminating toward the internal carbon (C3) from each C-Br carbon gives the C2=C3 and C4=C5 double bonds, producing the conjugated, internal hexa-2,4-diene, the most stable product.

CH_{3} -CH=CH-CH=CH-CH_{3}

Final answer:

CH_{3} CH=CH-CH=CHCH_{3}

Q57Single correctPurification and Characterisation of Organic Compounds

Which of the following statements is not true about partition chromatography ?

(A)

Mobile phase can be a gas

(B)

Stationary phase is a finely divided solid adsorbent

(C)

Separation depends upon equilibration of solute between a mobile and a stationary phase

(D)

Paper chromatography is an example of partition chromatography

SolutionAnswer: Option 2Stationary phase is a finely divided solid adsorbent

Approach:

Distinguish partition chromatography (stationary phase is a liquid held on a support) from adsorption chromatography (stationary phase is a solid adsorbent).

Step 1:In partition chromatography the stationary phase is a liquid held on an inert support (e.g. water on cellulose in paper chromatography), not a finely divided solid adsorbent. The description in statement (2) belongs to adsorption chromatography.

stationary phase = liquid (not solid adsorbent)

Final answer: Stationary phase is a finely divided solid adsorbent

Q58Single correctSome Basic Principles of Organic Chemistry

The IUPAC name of the following compound is :

A cyclohexane ring with two methyl groups on one ring carbon (gem-dimethyl) and an ethyl group on the adjacent ring carbon.

(A)

1, 1-Dimethyl-2-ethylcyclohexane

(B)

2-Ethyl-1,1-dimethylcyclohexane

(C)

1-Ethyl-2,2-dimethylcyclohexane

(D)

2, 2-Dimethyl-1-ethylcyclohexane

SolutionAnswer: Option 22-Ethyl-1,1-dimethylcyclohexane

Approach:

Number the cyclohexane ring to give the lowest set of locants, then cite the substituents alphabetically.

Step 1:The ring bears a gem-dimethyl carbon and an adjacent ethyl carbon. Numbering to give lowest locants gives {1,1,2} (dimethyl at 1,1 and ethyl at 2) rather than {1,2,2}.

{1, 1, 2} < {1, 2, 2}

Step 2:Substituents are cited alphabetically: ethyl before methyl. The name is 2-ethyl-1,1-dimethylcyclohexane.

ethyl < methyl (alphabetical)

Final answer: 2-Ethyl-1,1-dimethylcyclohexane

Q59Single correctHaloalkanes and Haloarenes

The major product of the following reaction is :

Reaction scheme: C6H5CH2-C(CH3)(Br)-CH2-CH3 treated with C2H5ONa in C2H5OH.

(A)

C6H5CH2-C(CH3)(OC2H5)-CH2CH3 (ethyl ether, substitution product).

(B)

(C)

C6H5CH2-C(CH3)=CH-CH3 (internal alkene).

(D)

C6H5CH2-C(=CH2)-CH2CH3 (terminal alkene).

SolutionAnswer: Option 2

C_{6} H_{5} CH=C(CH_{3})-CH_{2} CH_{3}

Approach:

Sodium ethoxide is a strong base; with a tertiary halide it favours E2 elimination, giving the most stable (conjugated) alkene.

Step 1:The substrate is a tertiary benzylic-type halide. The strong base C2H5ONa promotes E2 elimination over substitution.

3^{\circ} halide + strong base \to E2

Step 2:Removing a beta-hydrogen from the benzylic CH2 gives a double bond conjugated with the benzene ring, the most stable alkene.

C_{6} H_{5} CH=C(CH_{3})CH_{2} CH_{3}

Final answer:

C_{6} H_{5} CH=C(CH_{3})-CH_{2} CH_{3}

Q60Single correctAlcohols, Phenols and Ethers

The major product of the following reaction is :

2-(2-hydroxyphenyl)ethanol: benzene ring with ortho phenolic -OH and a -CH2CH2OH side chain; reagents 1. K2CO3, 2. CH3I (1 eq.).

(A)

Benzene ring with ortho -OCH3 and -CH2CH2OH (2-(2-methoxyphenyl)ethanol).

(B)

2,3-dihydrobenzofuran (benzene fused to a five-membered oxygen ring -O-CH2-CH2-).

(C)

Benzene ring with ortho phenolic -OH and -CH2CH2OCH3 (aliphatic methyl ether; phenol free).

(D)

Benzene ring with ortho -OCH3 and a vinyl group -CH=CH2 (2-methoxystyrene).

SolutionAnswer: Option 12-(2-methoxyphenyl)ethanol: benzene ring with ortho -OCH3 and -CH2CH2OH

Approach:

K2CO3 selectively deprotonates the more acidic phenolic -OH; the resulting phenoxide is methylated by one equivalent of CH3I, leaving the aliphatic alcohol untouched.

Step 1:Phenol (pKa ~10) is far more acidic than the primary aliphatic alcohol (pKa ~16); mild base K2CO3 deprotonates only the phenol to give the phenoxide.

ArOH K_{2} CO_{3} ArO^{-}

Step 2:One equivalent of CH3I methylates the more nucleophilic phenoxide oxygen, giving the aryl methyl ether while the -CH2CH2OH group remains intact.

ArO^{-} + CH_{3} I \to ArOCH_{3}

Final answer: 2-(2-methoxyphenyl)ethanol: benzene ring with ortho -OCH3 and -CH2CH2OH

Mathematics30 questions

Q61Single correctRelations and Functions

Let

f (x) = 2^{10} \cdot x + 1

and

g (x) = 3^{10} \cdot x - 1

. If

(f \circ g) (x) = x

, then x is equal to :

(A)

\frac{3 ^{10} - 1}{3 ^{10} - 2 ^{- 10}}

(B)

\frac{2 ^{10} - 1}{2 ^{10} - 3 ^{- 10}}

(C)

\frac{1 - 3 ^{- 10}}{2 ^{10} - 3 ^{- 10}}

(D)

\frac{1 - 2 ^{- 10}}{3 ^{10} - 2 ^{- 10}}

SolutionAnswer: Option 4

\frac{1 - 2 ^{- 10}}{3 ^{10} - 2 ^{- 10}}

Approach:

Compute the composition f(g(x)), set it equal to x, solve the resulting linear equation, then divide numerator and denominator by

2^{1} 0

to match the option form.

Step 1:Substitute g(x) into f.

f (g (x)) = 2^{10} (3^{10} x - 1) + 1 = 2^{10} 3^{10} x - 2^{10} + 1

Step 2:Set equal to x and collect terms.

2^{10} 3^{10} x - 2^{10} + 1 = x \Rightarrow x (2^{10} 3^{10} - 1) = 2^{10} - 1

Step 3:Divide numerator and denominator by

2^{1} 0

x = \frac{1 - 2 ^{- 10}}{3 ^{10} - 2 ^{- 10}}

Final answer:

\frac{1 - 2 ^{- 10}}{3 ^{10} - 2 ^{- 10}}

Q62Single correctAlgebra

Let

p (x)

be a quadratic polynomial such that

p (0) = 1

. If

p (x)

leaves remainder

4

when divided by

x - 1

and it leaves remainder

6

when divided by

x + 1

; then :

(A)

p (2) = 11

(B)

p (2) = 19

(C)

p (- 2) = 19

(D)

p (- 2) = 11

SolutionAnswer: Option 3

p (- 2) = 19

Approach:

Write p(x)=

a x^{2}

+bx+c, use p(0)=1 and the remainder theorem values p(1)=4, p(-1)=6 to find the coefficients, then evaluate p(2) and p(-2).

Step 1:Use p(0)=1.

c = 1

Step 2:p(1)=4 and p(-1)=6 give two equations.

a + b + 1 = 4, a - b + 1 = 6

Step 3:Solve for a and b.

a = 4, b = - 1

Step 4:Evaluate at -2 (and 2).

p (- 2) = 4 (4) + 2 + 1 = 19, p (2) = 16 - 2 + 1 = 15

Final answer:

p (- 2) = 19

Q63Single correctComplex Numbers

Let

z \in C

, the set of complex numbers. Then the equation,

2∣ z + 3 i ∣ - ∣ z - i ∣ = 0

represents :

(A)

a circle with radius

\frac{8}{3}

(B)

a circle with diameter

\frac{10}{3}

(C)

an ellipse with length of major axis

\frac{16}{3}

(D)

an ellipse with length of minor axis

\frac{16}{9}

SolutionAnswer: Option 1a circle with radius

\frac{8}{3}

Approach:

Write z=x+iy, square the modulus equation, simplify to the general circle form, and read off the radius.

Step 1:Set 2|z+3i|=|z-i| and substitute z=x+iy, then square.

4 (x^{2} + (y + 3)^{2}) = x^{2} + (y - 1)^{2}

Step 2:Collect terms.

3 x^{2} + 3 y^{2} + 26 y + 35 = 0

Step 3:Complete the square to find radius.

x^{2} + (y + \frac{13}{3})^{2} = \frac{169}{9} - \frac{35}{3} = \frac{64}{9}

Final answer: a circle with radius

\frac{8}{3}

Q64Single correctMatrices and Determinants

The number of real values of

λ

for which the system of linear equations

2 x + 4 y - λ z = 0

4 x + λ y + 2 z = 0

λ x + 2 y + 2 z = 0

has infinitely many solutions, is :

(A)

0

(B)

1

(C)

2

(D)

3

SolutionAnswer: Option 2

1

Approach:

A homogeneous system has non-trivial (infinitely many) solutions iff the coefficient determinant is zero. Expand the determinant in lambda and count its real roots.

Step 1:Form and expand the coefficient determinant.

24 λ 4 λ 2 - λ 22 = 2 (2 λ - 4) - 4 (8 - 2 λ) - λ (8 - λ^{2})

Step 2:Set determinant to zero.

λ^{3} + 4 λ - 40 = 0

Step 3:Analyse number of real roots via monotonicity.

\frac{d}{d λ} (λ^{3} + 4 λ - 40) = 3 λ^{2} + 4 > 0

Final answer:

1

Q65Single correctMatrices and Determinants

Let A be any

3 \times 3

invertible matrix. Then which one of the following is not always true ?

(A)

adj (A) = ∣ A ∣ \cdot A^{- 1}

(B)

adj (adj (A)) = ∣ A ∣ \cdot A

(C)

adj (adj (A)) = ∣ A ∣^{2} \cdot (adj (A))^{- 1}

(D)

adj (adj (A)) = ∣ A ∣ \cdot (adj (A))^{- 1}

SolutionAnswer: Option 4

adj (adj (A)) = ∣ A ∣ \cdot (adj (A))^{- 1}

Approach:

Use the n=3 adjoint identities adj(A)=|A|

A^{- 1}

and adj(adj(A))=

∣ A ∣^{n - 2}

A=|A|A, plus

(a d j A)^{- 1}

=A/|A|, to test each option.

Step 1:Option 1 is the defining identity.

adj (A) = ∣ A ∣ A^{- 1}

Step 2:Option 2 uses the double-adjoint formula with n=3.

adj (adj (A)) = ∣ A ∣^{1} A = ∣ A ∣ A

Step 3:Check option 3 using

(a d j A)^{- 1}

=A/|A|.

∣ A ∣^{2} (adj A)^{- 1} = ∣ A ∣^{2} \cdot \frac{A}{∣ A ∣} = ∣ A ∣ A

Step 4:Check option 4.

∣ A ∣ (adj A)^{- 1} = ∣ A ∣ \cdot \frac{A}{∣ A ∣} = A \neq = ∣ A ∣ A

in general

Final answer:

adj (adj (A)) = ∣ A ∣ \cdot (adj (A))^{- 1}

Q66Single correctPermutations and Combinations

If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is :

(A)

4 4^{t h}

(B)

4 5^{t h}

(C)

4 6^{t h}

(D)

4 7^{t h}

SolutionAnswer: Option 3

4 6^{t h}

Approach:

Order the letters E,E,N,Q,U alphabetically and count, by leading letters, how many words precede QUEEN, accounting for the repeated E with 4!/2! counts.

Step 1:Words starting with E (remaining E,N,Q,U distinct).

4! = 24

Step 2:Words starting with N (remaining E,E,Q,U).

\frac{4 !}{2 !} = 12

Step 3:Words with Q: order second letter E,N,U. Q E _ : 3!=6 (37-42); Q N _ : 3!/2!=3 (43-45); Q U _ : remaining E,E,N.

36 + 6 + 3 = 45

Step 4:Within Q U _ _ arrange E,E,N: EEN, ENE, NEE; QUEEN is EEN, the first.

45 + 1 = 46

Final answer:

4 6^{t h}

Q67Single correctNumber Theory

(27)^{999}

is divided by

7

, then the remainder is :

(A)

1

(B)

2

(C)

3

(D)

6

SolutionAnswer: Option 4

6

Approach:

Reduce 27 modulo 7 to -1, raise to the odd power, and convert the result to a non-negative remainder.

Step 1:Reduce the base mod 7.

27 \equiv 6 \equiv - 1 (mod 7)

Step 2:Raise to the odd exponent 999.

(27)^{999} \equiv (- 1)^{999} = - 1 (mod 7)

Step 3:Express as a positive remainder.

- 1 \equiv 6 (mod 7)

Final answer:

6

Q68Single correctSequences and Series

If the arithmetic mean of two numbers a and b,

a > b > 0

, is five times their geometric mean, then

\frac{a + b}{a - b}

is equal to :

(A)

\frac{6}{2}

(B)

\frac{3 2}{4}

(C)

\frac{7 3}{12}

(D)

\frac{5 6}{12}

SolutionAnswer: Option 4

\frac{5 6}{12}

Approach:

From AM=5 GM get a+b=10 sqrt(ab); set r=sqrt(a/b), find r+1/r and r-1/r, then form (a+b)/(a-b).

Step 1:Translate the condition.

\frac{a + b}{2} = 5 ab \Rightarrow a + b = 10 ab

Step 2:Divide by sqrt(ab) with r=sqrt(a/b).

r + \frac{1}{r} = 10

Step 3:Compute r-1/r (positive since a>b).

r - \frac{1}{r} = (r + \frac{1}{r})^{2} - 4 = 96 = 46

Step 4:Form the required ratio.

\frac{a + b}{a - b} = \frac{r + \frac{1}{r}}{r - \frac{1}{r}} = \frac{10}{4 6} = \frac{5}{2 6} = \frac{5 6}{12}

Final answer:

\frac{5 6}{12}

Q69Single correctSequences and Series

If the sum of the first n terms of the series

3 + 75 + 243 + 507 + \dots

4353

, then n equals :

(A)

18

(B)

15

(C)

13

(D)

29

SolutionAnswer: Option 2

15

Approach:

Factor sqrt 3 out of each term to expose an arithmetic progression in the coefficients, sum it, and solve for n.

Step 1:Simplify the surds.

3, 53, 93, 133, \dots

Step 2:Sum n terms.

S_{n} = 3 \cdot \frac{n}{2} (2 + 4 (n - 1)) = 3 n (2 n - 1)

Step 3:Set equal to 435 sqrt 3 and solve.

n (2 n - 1) = 435 \Rightarrow 2 n^{2} - n - 435 = 0 \Rightarrow n = \frac{1 + 59}{4} = 15

Final answer:

15

Q70Single correctLimits and Continuity

x \to 3 lim \frac{3 x - 3}{2 x - 4 - 2}

is equal to :

(A)

3

(B)

\frac{1}{2}

(C)

\frac{3}{2}

(D)

\frac{1}{2 2}

SolutionAnswer: Option 2

\frac{1}{2}

Approach:

Rationalise numerator and denominator separately, cancel the common factor (x-3), and substitute x=3.

Step 1:Rationalise numerator.

3 x - 3 = \frac{3 ( x - 3 )}{3 x + 3}

Step 2:Rationalise denominator.

2 x - 4 - 2 = \frac{2 ( x - 3 )}{2 x - 4 + 2}

Step 3:Cancel (x-3) and substitute x=3.

\frac{3 ( 2 x - 4 + 2 )}{2 ( 3 x + 3 )} \to \frac{3 \cdot 2 2}{2 \cdot 6} = \frac{1}{2}

Final answer:

\frac{1}{2}

Q71Single correctDifferential Calculus

The tangent at the point

(2, - 2)

to the curve,

x^{2} y^{2} - 2 x = 4 (1 - y)

does not pass through the point :

(A)

(4, \frac{1}{3})

(B)

(8, 5)

(C)

(- 4, - 9)

(D)

(- 2, - 7)

SolutionAnswer: Option 4

(- 2, - 7)

Approach:

Differentiate implicitly to get the slope at (2,-2), write the tangent line, then test each candidate point.

Step 1:Differentiate the curve.

2 x y^{2} + 2 x^{2} y y^{'} - 2 = - 4 y^{'}

Step 2:Substitute (2,-2).

16 - 16 y^{'} - 2 = - 4 y^{'} \Rightarrow 14 = 12 y^{'} \Rightarrow y^{'} = \frac{7}{6}

Step 3:Write the tangent line.

y + 2 = \frac{7}{6} (x - 2)

Step 4:Test points; (-2,-7) gives LHS=-5, RHS=-14/3.

- 7 + 2 = - 5 \neq = \frac{7}{6} (- 4) = - \frac{14}{3}

Final answer:

(- 2, - 7)

Q72Single correctDifferential Calculus

y = [x + x^{2} - 1]^{15} + [x - x^{2} - 1]^{15}

, then

(x^{2} - 1) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}

is equal to :

(A)

125 y

(B)

224 y^{2}

(C)

225 y^{2}

(D)

225 y

SolutionAnswer: Option 4

225 y

Approach:

Let u=x+sqrt(

x^{2}

-1), v=x-sqrt(

x^{2}

-1); differentiate y=

u^{1} 5

v^{1} 5

, obtain sqrt(

x^{2}

-1) y' = 15(

u^{1} 5

v^{1} 5

), differentiate again and simplify.

Step 1:First derivative.

y^{'} = \frac{15}{x ^{2} - 1} (u^{15} - v^{15})

Step 2:Differentiate the bracketed relation.

\frac{x}{x ^{2} - 1} y^{'} + x^{2} - 1 y^{''} = 15 \cdot \frac{15}{x ^{2} - 1} (u^{15} + v^{15})

Step 3:Multiply through by sqrt(

x^{2}

-1).

x y^{'} + (x^{2} - 1) y^{''} = 225 y

Final answer:

225 y

Q73Single correctCoordinate Geometry

If a point P has co-ordinates

(0, - 2)

and Q is any point on the circle,

x^{2} + y^{2} - 5 x - y + 5 = 0

, then the maximum value of

(PQ)^{2}

is :

(A)

\frac{25 + 6}{2}

(B)

14 + 53

(C)

\frac{47 + 10 6}{2}

(D)

8 + 53

SolutionAnswer: Option 2

14 + 53

Approach:

Find the circle's centre C and radius r, compute PC, then max PQ = PC + r and square it.

Step 1:Centre and radius squared.

r^{2} = \frac{25}{4} + \frac{1}{4} - 5 = \frac{6}{4} = \frac{3}{2}

Step 2:Distance PC.

P C^{2} = (\frac{5}{2})^{2} + (\frac{1}{2} + 2)^{2} = \frac{25}{4} + \frac{25}{4} = \frac{50}{4}

Step 3:Maximum

(PQ)^{2}

(PC + r)^{2}

P C^{2} + r^{2} + 2 PC r = \frac{50}{4} + \frac{6}{4} + 2 \cdot \frac{5 2}{2} \cdot \frac{6}{2}

Final answer:

14 + 53

Q74Single correctIntegral Calculus

The integral

\int 1 + 2 cot x (csc x + cot x) d x

(0 < x < \frac{π}{2})

is equal to : (where C is a constant of integration)

(A)

4 lo g (sin \frac{x}{2}) + C

(B)

2 lo g (sin \frac{x}{2}) + C

(C)

2 lo g (cos \frac{x}{2}) + C

(D)

4 lo g (cos \frac{x}{2}) + C

SolutionAnswer: Option 2

2 lo g (sin \frac{x}{2}) + C

Approach:

Recognise the radicand as a perfect square

(csc x + co t x)^{2}

, take the positive square root on (0, pi/2), then integrate csc x + cot x and simplify using half-angle identities.

Step 1:Simplify the radicand.

1 + 2 cot x csc x + 2 cot^{2} x = csc^{2} x + cot^{2} x + 2 cot x csc x = (csc x + cot x)^{2}

Step 2:Integrate.

\int (csc x + cot x) d x = lo g tan \frac{x}{2} + lo g sin x

Step 3:Simplify with half-angle identities.

tan \frac{x}{2} sin x = tan \frac{x}{2} \cdot 2 sin \frac{x}{2} cos \frac{x}{2} = 2 sin^{2} \frac{x}{2}

Final answer:

2 lo g (sin \frac{x}{2}) + C

Q75Single correctIntegral Calculus

The integral

\int_{π /12}^{π /4} \frac{8 cos 2 x}{( tan x + cot x ) ^{3}} d x

equals :

(A)

\frac{15}{128}

(B)

\frac{15}{64}

(C)

\frac{13}{32}

(D)

\frac{13}{256}

SolutionAnswer: Option 1

\frac{15}{128}

Approach:

Use tan x + cot x = 2/sin 2x to reduce the integrand to cos 2x

s i n^{3}

2x, then substitute u = sin 2x.

Step 1:Simplify the integrand.

\frac{8 cos 2 x}{( 2/ sin 2 x ) ^{3}} = \frac{8 cos 2 x sin ^{3} 2 x}{8} = cos 2 x sin^{3} 2 x

Step 2:Substitute u = sin 2x, du = 2 cos 2x dx.

\int cos 2 x sin^{3} 2 x d x = \frac{1}{2} \int u^{3} d u = \frac{u ^{4}}{8}

Step 3:Apply limits (sin(pi/2)=1, sin(pi/6)=1/2).

\frac{1}{8} [1 - (\frac{1}{2})^{4}] = \frac{1}{8} \cdot \frac{15}{16} = \frac{15}{128}

Final answer:

\frac{15}{128}

Q76Single correctIntegral Calculus

The area (in sq. units) of the smaller portion enclosed between the curves,

x^{2} + y^{2} = 4

and

y^{2} = 3 x

, is :

(A)

\frac{1}{2 3} + \frac{π}{3}

(B)

\frac{1}{3} + \frac{2 π}{3}

(C)

\frac{1}{2 3} + \frac{2 π}{3}

(D)

\frac{1}{3} + \frac{4 π}{3}

SolutionAnswer: Option 4

\frac{1}{3} + \frac{4 π}{3}

Approach:

Find intersection of circle and parabola, then integrate using symmetry about the x-axis: area = 2 times (region under parabola from 0 to 1 plus region under circle from 1 to 2).

Step 1:Solve

x^{2}

+3x=4 to get the intersection abscissa.

x^{2} + 3 x - 4 = 0 \Rightarrow x = 1

Step 2:Compute parabola part.

\int_{0}^{1} 3 x d x = 3 \cdot \frac{2}{3} = \frac{2}{3}

Step 3:Compute circle part.

\int_{1}^{2} 4 - x^{2} d x = π - (\frac{3}{2} + \frac{π}{3}) = \frac{2 π}{3} - \frac{3}{2}

Step 4:Double the sum and simplify.

2 (\frac{2}{3} + \frac{2 π}{3} - \frac{3}{2}) = \frac{4}{3} - 3 + \frac{4 π}{3} = \frac{1}{3} + \frac{4 π}{3}

Final answer:

\frac{1}{3} + \frac{4 π}{3}

Q77Single correctDifferential Equations

The curve satisfying the differential equation,

y d x - (x + 3 y^{2}) d y = 0

and passing through the point

(1, 1)

, also passes through the point :

(A)

(\frac{1}{4}, - \frac{1}{2})

(B)

(- \frac{1}{3}, \frac{1}{3})

(C)

(\frac{1}{3}, - \frac{1}{3})

(D)

(\frac{1}{4}, \frac{1}{2})

SolutionAnswer: Option 2

(- \frac{1}{3}, \frac{1}{3})

Approach:

Treat x as the dependent variable to obtain a linear ODE in x with respect to y, solve using an integrating factor, apply the initial condition, then test the options.

Step 1:Rewrite as a linear ODE in x.

\frac{d x}{d y} = \frac{x}{y} + 3 y

Step 2:Multiply by IF 1/y and integrate.

\frac{d}{d y} (\frac{x}{y}) = 3 \Rightarrow \frac{x}{y} = 3 y + C

Step 3:Apply (1,1): 1 = 3 + C gives C = -2.

x = 3 y^{2} - 2 y

Step 4:Test options; substitute y = 1/3.

x = 3 \cdot \frac{1}{9} - 2 \cdot \frac{1}{3} = \frac{1}{3} - \frac{2}{3} = - \frac{1}{3}

Final answer:

(- \frac{1}{3}, \frac{1}{3})

Q78Single correctCoordinate Geometry

The locus of the point of intersection of the straight lines,

t x - 2 y - 3 t = 0

x - 2 t y + 3 = 0

(

t \in R

), is :

(A)

an ellipse with eccentricity

\frac{2}{5}

(B)

an ellipse with the length of major axis

6

(C)

a hyperbola with eccentricity

5

(D)

a hyperbola with the length of conjugate axis

3

SolutionAnswer: Option 4a hyperbola with the length of conjugate axis

3

Approach:

Eliminate the parameter t from the two line equations to obtain the locus, identify the conic and its axis lengths.

Step 1:From the first line, isolate t.

t (x - 3) = 2 y \Rightarrow t = \frac{2 y}{x - 3}

Step 2:Substitute into the second line x + 3 = 2ty.

x + 3 = \frac{4 y ^{2}}{x - 3}

Step 3:Simplify to standard form.

x^{2} - 9 = 4 y^{2} \Rightarrow \frac{x ^{2}}{9} - \frac{y ^{2}}{9/4} = 1

Step 4:Conjugate axis length = 2b.

2 b = 2 \cdot \frac{3}{2} = 3

Final answer: a hyperbola with the length of conjugate axis

3

Q79Single correctCoordinate Geometry

If two parallel chords of a circle, having diameter

4

units, lie on the opposite sides of the centre and subtend angles

cos^{- 1} (\frac{1}{7})

and

sec^{- 1} (7)

at the centre respectively, then the distance between these chords, is :

(A)

\frac{4}{7}

(B)

\frac{8}{7}

(C)

\frac{8}{7}

(D)

\frac{16}{7}

SolutionAnswer: Option 2

\frac{8}{7}

Approach:

For a chord subtending central angle alpha in a circle of radius r, the perpendicular distance from the centre is r cos(alpha/2). Add the two distances since the chords are on opposite sides.

Step 1:Radius r = 2. Both angles have cosine 1/7 (since

se c^{- 1}

(7) means cos = 1/7).

cos α_{1} = cos α_{2} = \frac{1}{7}

Step 2:Compute half-angle cosine.

cos \frac{α}{2} = \frac{1 + 1/7}{2} = \frac{4}{7} = \frac{2}{7}

Step 3:Distance of each chord from centre.

d = 2 \cdot \frac{2}{7} = \frac{4}{7}

Step 4:Chords on opposite sides, so total distance is the sum.

\frac{4}{7} + \frac{4}{7} = \frac{8}{7}

Final answer:

\frac{8}{7}

Q80Single correctCoordinate Geometry

If the common tangents to the parabola,

x^{2} = 4 y

and the circle,

x^{2} + y^{2} = 4

intersect at the point P, then the distance of P from the origin, is :

(A)

2 + 1

(B)

2 (3 + 22)

(C)

2 (2 + 1)

(D)

3 + 22

SolutionAnswer: Option 3

2 (2 + 1)

Approach:

By symmetry the common tangents meet on the y-axis. Write a tangent to the parabola, impose tangency to the circle, find its y-intercept, which gives P.

Step 1:Tangent to the parabola at parameter t.

t x - y - t^{2} = 0

Step 2:Impose tangency to circle of radius 2.

\frac{t ^{2}}{t ^{2} + 1} = 2 \Rightarrow t^{4} - 4 t^{2} - 4 = 0

Step 3:The two symmetric tangents (slopes

\pm t)

meet at the y-intercept x=0.

y = - t^{2} = - (2 + 22)

Step 4:Distance of P from origin.

∣ - t^{2} ∣ = 2 + 22 = 2 (2 + 1)

Final answer:

2 (2 + 1)

Q81Single correctCoordinate Geometry

Consider an ellipse, whose centre is at the origin and its major axis is along the x-axis. If its eccentricity is

\frac{3}{5}

and the distance between its foci is

6

, then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is :

(A)

8

(B)

32

(C)

80

(D)

40

SolutionAnswer: Option 4

40

Approach:

Use distance between foci to get c, then e to get a and b. The quadrilateral through the four vertices is a rhombus with diagonals 2a and 2b.

Step 1:Distance between foci 2c = 6, so c = 3.

c = 3

Step 2:From e = 3/5 find a.

a = \frac{c}{e} = \frac{3}{3/5} = 5

Step 3:Find b.

b^{2} = a^{2} - c^{2} = 25 - 9 = 16 \Rightarrow b = 4

Step 4:Rhombus with diagonals 2a=10 and 2b=8.

Area = \frac{1}{2} \cdot 10 \cdot 8 = 40

Final answer:

40

Q82Single correctThree Dimensional Geometry

The coordinates of the foot of the perpendicular from the point

(1, - 2, 1)

on the plane containing the lines,

\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}

and

\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}

, is :

(A)

(2, - 4, 2)

(B)

(- 1, 2, - 1)

(C)

(0, 0, 0)

(D)

(1, 1, 1)

SolutionAnswer: Option 3

(0, 0, 0)

Approach:

The plane's normal is the cross product of the two line directions. Find the plane through a known point, then drop a perpendicular from the given point.

Step 1:Cross product of (6,7,8) and (3,5,7).

n = (9, - 18, 9) = 9 (1, - 2, 1)

Step 2:Plane through (-1,1,3).

x - 2 y + z = 0

Step 3:Evaluate at P=(1,-2,1).

1 - 2 (- 2) + 1 = 6, ∣ n ∣^{2} = 6

Step 4:Foot of perpendicular.

(1, - 2, 1) - 1 \cdot (1, - 2, 1) = (0, 0, 0)

Final answer:

(0, 0, 0)

Q83Single correctThree Dimensional Geometry

The line of intersection of the planes

r \cdot (3 \hat{i} - \hat{j} + \hat{k}) = 1

and

r \cdot (\hat{i} + 4 \hat{j} - 2 \hat{k}) = 2

, is :

(A)

\frac{x - \frac{4}{7}}{- 2} = \frac{y}{7} = \frac{z - \frac{5}{7}}{13}

(B)

\frac{x - \frac{4}{7}}{2} = \frac{y}{- 7} = \frac{z + \frac{5}{7}}{13}

(C)

\frac{x - \frac{6}{13}}{2} = \frac{y - \frac{5}{13}}{- 7} = \frac{z}{- 13}

(D)

\frac{x - \frac{6}{13}}{2} = \frac{y - \frac{5}{13}}{7} = \frac{z}{- 13}

SolutionAnswer: Option 3

\frac{x - \frac{6}{13}}{2} = \frac{y - \frac{5}{13}}{- 7} = \frac{z}{- 13}

Approach:

Direction of the line is the cross product of the two plane normals. Find one common point by setting z=0 and solving the two scalar equations.

Step 1:Cross product of (3,-1,1) and (1,4,-2).

d = (- 2, 7, 13)

Step 2:Set z=0: solve 3x-y=1 and x+4y=2.

x = \frac{6}{13}, y = \frac{5}{13}

Step 3:Write symmetric form.

\frac{x - 6/13}{- 2} = \frac{y - 5/13}{7} = \frac{z}{13}

Step 4:Scale by -1 to match option 3.

\frac{x - 6/13}{2} = \frac{y - 5/13}{- 7} = \frac{z}{- 13}

Final answer:

\frac{x - \frac{6}{13}}{2} = \frac{y - \frac{5}{13}}{- 7} = \frac{z}{- 13}

Q84Single correctVector Algebra

The area (in sq. units) of the parallelogram whose diagonals are along the vectors

8 \hat{i} - 6 \hat{j}

and

3 \hat{i} + 4 \hat{j} - 12 \hat{k}

, is :

(A)

26

(B)

65

(C)

20

(D)

52

SolutionAnswer: Option 2

65

Approach:

Area of a parallelogram in terms of its diagonals equals half the magnitude of their cross product.

Step 1:Cross product of (8,-6,0) and (3,4,-12).

d_{1} \times d_{2} = (72, 96, 50)

Step 2:Magnitude.

7 2^{2} + 9 6^{2} + 5 0^{2} = 16900 = 130

Step 3:Half the magnitude.

\frac{1}{2} \cdot 130 = 65

Final answer:

65

Q85Single correctStatistics

The mean age of

25

teachers in a school is

40

years. A teacher retires at the age of

60

years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is

39

years, then the age (in years) of the newly appointed teacher is :

(A)

25

(B)

30

(C)

35

(D)

40

SolutionAnswer: Option 3

35

Approach:

Use total = mean times count. Replace the retiring teacher's age with the unknown new age and apply the new mean.

Step 1:Initial total age.

25 \times 40 = 1000

Step 2:Remove 60, add new age x.

1000 - 60 + x = 940 + x

Step 3:Apply new mean 39 over 25 teachers.

\frac{940 + x}{25} = 39 \Rightarrow 940 + x = 975

Final answer:

35

Q86Single correctProbability

Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are

\frac{3}{4}

\frac{1}{2}

and

\frac{5}{8}

respectively, then the probability that the target is hit by P or Q but not by R is :

(A)

\frac{21}{64}

(B)

\frac{9}{64}

(C)

\frac{15}{64}

(D)

\frac{39}{64}

SolutionAnswer: Option 1

\frac{21}{64}

Approach:

Compute P(P or Q) using the complement, then multiply by P(not R) using independence.

Step 1:Probability hit by P or Q.

1 - \frac{1}{4} \cdot \frac{1}{2} = 1 - \frac{1}{8} = \frac{7}{8}

Step 2:Probability not hit by R.

1 - \frac{5}{8} = \frac{3}{8}

Step 3:Multiply (independence).

\frac{7}{8} \cdot \frac{3}{8} = \frac{21}{64}

Final answer:

\frac{21}{64}

Q87Single correctProbability

An unbiased coin is tossed eight times. The probability of obtaining at least one head and one tail is :

(A)

\frac{255}{256}

(B)

\frac{127}{128}

(C)

\frac{63}{64}

(D)

\frac{1}{2}

SolutionAnswer: Option 2

\frac{127}{128}

Approach:

Use the complement: subtract the probabilities of all heads and all tails from 1.

Step 1:Probability of all heads or all tails.

(\frac{1}{2})^{8} + (\frac{1}{2})^{8} = \frac{2}{256} = \frac{1}{128}

Step 2:Subtract from 1.

1 - \frac{1}{128} = \frac{127}{128}

Final answer:

\frac{127}{128}

Q88Single correctMatrices and Determinants

S = ⎩ ⎨ ⎧ x \in [0, 2 π] : 0 sin x cos x cos x 0 sin x - sin x cos x 0 = 0 ⎭ ⎬ ⎫

, then

\sum_{x \in S} tan (\frac{π}{3} + x)

is equal to :

(A)

4 + 23

(B)

- 2 + 3

(C)

- 2 - 3

(D)

- 4 - 23

SolutionAnswer: Option 4

- 4 - 23

Approach:

Expand the determinant to get a trig equation, solve for x in [0,2pi], then sum the tangent values.

Step 1:Expand the determinant.

cos^{3} x - sin^{3} x = 0 \Rightarrow tan^{3} x = 1

Step 2:Solve in [0,2pi].

x = \frac{π}{4}, \frac{5 π}{4}

Step 3:Both give the same tangent (period pi).

tan (\frac{π}{3} + \frac{5 π}{4}) = tan (\frac{π}{3} + \frac{π}{4})

Step 4:Evaluate and double.

2 tan \frac{7 π}{12} = 2 \cdot (- (2 + 3)) = - 4 - 23

Final answer:

- 4 - 23

Q89Single correctTrigonometry

The value of

tan^{- 1} [\frac{1 + x ^{2} + 1 - x ^{2}}{1 + x ^{2} - 1 - x ^{2}}]

∣ x ∣ < \frac{1}{2}

x \neq = 0

, is equal to :

(A)

\frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}

(B)

\frac{π}{4} + cos^{- 1} x^{2}

(C)

\frac{π}{4} - \frac{1}{2} cos^{- 1} x^{2}

(D)

\frac{π}{4} - cos^{- 1} x^{2}

SolutionAnswer: Option 1

\frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}

Approach:

Substitute

x^{2}

= cos 2theta to simplify the surds, reduce the fraction to a tangent of a sum, then invert.

Step 1:Let

x^{2}

= cos 2theta, so theta = (1/2)

co s^{- 1}

(

x^{2}

1 + x^{2} = 2 cos θ, 1 - x^{2} = 2 sin θ

Step 2:Reduce the fraction.

\frac{c o s θ + s i n θ}{c o s θ - s i n θ} = \frac{1 + t a n θ}{1 - t a n θ} = tan (\frac{π}{4} + θ)

Step 3:Apply inverse tangent.

tan^{- 1} [tan (\frac{π}{4} + θ)] = \frac{π}{4} + θ = \frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}

Final answer:

\frac{π}{4} + \frac{1}{2} cos^{- 1} x^{2}

Q90Single correctMathematical Reasoning

The proposition

(\sim p) \lor (p \land \sim q)

is equivalent to :

(A)

p \lor \sim q

(B)

p \to\sim q

(C)

p \land \sim q

(D)

q \to p

SolutionAnswer: Option 2

p \to\sim q

Approach:

Simplify using the distributive law, or build a truth table and compare with each option.

Step 1:Apply distributive law.

(\sim p \lor p) \land (\sim p \lor \sim q)

Step 2:Simplify the tautology factor.

\sim p \lor \sim q

Step 3:Recognise this as a conditional.

\sim p \lor \sim q \equiv p \to\sim q

Final answer:

p \to\sim q

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Frequently Asked Questions

How many questions are in the JEE Main 2017 April 08 paper?

The JEE Main 2017 April 08 paper has 89 questions — Physics (29), Chemistry (30) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.

What is the marking scheme for JEE Main?

JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.

Are the answer key and step-by-step solutions provided for the 2017 April 08 paper?

Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.

Can I take the JEE Main 2017 April 08 paper as a timed mock test?

Yes. With a free JEEnify account you can attempt this exact paper as a timed test in the real exam interface, then see your score and weak-area analysis.

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