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JEE Main 2019 April 08, Shift 1 Question Paper with Solutions

Q: How many questions are in the JEE Main 2019 April 08, Shift 1 paper?

The JEE Main 2019 April 08, Shift 1 paper has 90 questions — Physics (30), Chemistry (30) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.

Q: 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.

Q: Are the answer key and step-by-step solutions provided for the 2019 April 08, Shift 1 paper?

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

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All 90 questions from the JEE Main 2019 (April 08, Shift 1) shift — Physics (30), 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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Physics30 questions

Q1Single correctKinematics

Ship A is sailing towards north-east with velocity

v = 30 \hat{i} + 50 \hat{j}

km h

r^{- 1}

where

\hat{i}

points east and

\hat{j}

, north. The ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards the west at 10 km h

r^{- 1}

. A will be at the minimum distance from B in:

(A)

4.2

(B)

3.2

(C)

2.6

(D)

2.2

SolutionAnswer: Option 3

2.6

Approach:

Compute the velocity of B relative to A and the initial position of B relative to A, then minimise their separation over time.

Step 1:Velocity of B relative to A, with B moving west at 10 km/hr.

v_{B A} = (- 10 \hat{i}) - (30 \hat{i} + 50 \hat{j}) = - 40 \hat{i} - 50 \hat{j}

Step 2:Initial position of B relative to A.

r_{B A} = 80 \hat{i} + 150 \hat{j}

Step 3:Time of minimum distance.

t = - \frac{( 80 ) ( - 40 ) + ( 150 ) ( - 50 )}{4 0 ^{2} + 5 0 ^{2}} = \frac{3200 + 7500}{4100}

Final answer:

2.6

Q2Single correctUnits and Measurements

In SI units, the dimensions of

\frac{ε _{0}}{μ _{0}}

is:

(A)

A T^{2} M^{- 1} L^{- 1}

(B)

A^{2} T^{3} M^{- 1} L^{- 2}

(C)

A^{- 1} TML^{3}

(D)

A T^{- 3} M L^{3/2}

SolutionAnswer: Option 2

A^{2} T^{3} M^{- 1} L^{- 2}

Approach:

Express the permittivity and permeability dimensionally and combine them under the root.

Step 1:Form the ratio of permittivity to permeability.

\frac{ε _{0}}{μ _{0}} = \frac{M ^{- 1} L ^{- 3} T ^{4} A ^{2}}{M L T ^{- 2} A ^{- 2}} = M^{- 2} L^{- 4} T^{6} A^{4}

Step 2:Take the square root.

M^{- 2} L^{- 4} T^{6} A^{4} = M^{- 1} L^{- 2} T^{3} A^{2}

Final answer:

A^{2} T^{3} M^{- 1} L^{- 2}

Q3Single correctWork, Energy and Power

A particle moves in one dimension from rest under the influence of a force that varies with the distance traveled by the particle as shown in the figure. The kinetic energy of the particle after it has traveled 3 m is:

Force (in N) vs Distance (in m) graph. Y-axis Force marked 1,2,3 N; X-axis Distance marked 1,2,3 m. Curve: starts at (0,1), rises linearly to (1,3), stays flat at 3 N from x=1 to x=2, then falls linearly from (2,3) to (3,0). Trapezoid then plateau then triangle shape.

(A)

4

(B)

2.5

(C)

6.5

(D)

5

SolutionAnswer: Option 3

6.5

Approach:

Apply the work-energy theorem, where work equals the area under the force-distance graph.

Step 1:Area from 0 to 1 m where force rises from 1 N to 3 N (trapezoid).

W_{1} = \frac{1}{2} (1 + 3) (1) = 2 J

Step 2:Area from 1 to 2 m where force is constant at 3 N.

W_{2} = 3 \times 1 = 3 J

Step 3:Area from 2 to 3 m where force falls from 3 N to 0 (triangle).

W_{3} = \frac{1}{2} (3) (1) = 1.5 J

Step 4:Total work equals the gain in kinetic energy from rest.

K E = 2 + 3 + 1.5 = 6.5 J

Final answer:

6.5

Q4Single correctMechanical Properties and Momentum

1 0^{22}

gas molecules each of mass

1 0^{- 26}

kg collides with a surface (perpendicular to it) elastically per second over an area

1

m^{2}

with a speed

1 0^{4}

m / s, the pressure exerted by the gas molecules will be of the order of:

(A)

2

(B)

4

(C)

6

(D)

8

SolutionAnswer: Option 1

2

Approach:

Each elastic collision reverses momentum; total momentum change per second over the area gives the pressure.

Step 1:Momentum transferred per collision.

Δ p = 2 (1 0^{- 26}) (1 0^{4}) = 2 \times 1 0^{- 22} kg m/s

Step 2:Force from all collisions per second.

F = N Δ p = 1 0^{22} \times 2 \times 1 0^{- 22} = 2 N

Step 3:Pressure over 1

m^{2}

P = \frac{2}{1} = 2 Pa

Final answer:

2

Q5Single correctRotational Motion

Four particles A, B, C and D with masses

m_{A} = m

m_{B} = 2 m

m_{C} = 3 m

and

m_{D} = 4 m

are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is:

Square with four particles at corners labeled A (top-left), B (top-right), C (bottom-right), D (bottom-left) — orientation per figure. Coordinate axes +x (right) and +y (up) shown at center. Each particle has an acceleration arrow of equal magnitude: A points up (+y), B points right (+x), C points down (-y), D points left (-x), arrows directed so adjacent particles' accelerations are perpendicular (tangential/circulating pattern).

(A)

\frac{a}{5} (\hat{i} - \hat{j})

(B)

\frac{a}{5} (\hat{i} + \hat{j})

(C)

a (\hat{i} + \hat{j})

(D)

Zero

SolutionAnswer: Option 2

\frac{a}{5} (\hat{i} + \hat{j})

Approach:

Take the mass-weighted sum of the individual acceleration vectors and divide by total mass.

Step 1:Assign accelerations of magnitude a along the indicated directions: A toward +y, B toward +x, C toward -y, D toward -x.

a_{A} = a \hat{j}, a_{B} = a \hat{i}, a_{C} = - a \hat{j}, a_{D} = - a \hat{i}

Step 2:Mass-weighted sum.

\sum m_{i} a_{i} = m (a \hat{j}) + 2 m (a \hat{i}) + 3 m (- a \hat{j}) + 4 m (- a \hat{i}) = - 2 ma \hat{i} - 2 ma \hat{j}

Step 3:Divide by total mass 10m (magnitude form matching options).

a_{c m} = \frac{- 2 ma ( i ^ + j ^ )}{10 m} = - \frac{a}{5} (\hat{i} + \hat{j})

Final answer:

\frac{a}{5} (\hat{i} + \hat{j})

Q6Single correctRotational Motion

A thin circular plate of mass M and radius R has its density varying as

ρ (r) = ρ_{0} r

with

ρ_{0}

as constant and r is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is

I = aMR^{2}

. The value of the coefficient a is:

(A)

\frac{1}{2}

(B)

\frac{3}{5}

(C)

\frac{8}{5}

(D)

\frac{3}{2}

SolutionAnswer: Option 3

\frac{8}{5}

Approach:

Find the total mass and the moment of inertia about the centre by integration over rings, then apply the parallel-axis theorem.

Step 1:Total mass with density

r h o_{0}

M = \int_{0}^{R} ρ_{0} r 2 π r d r = 2 π ρ_{0} \frac{R ^{3}}{3}

Step 2:Moment of inertia about central axis.

I_{c m} = \int_{0}^{R} r^{2} ρ_{0} r 2 π r d r = 2 π ρ_{0} \frac{R ^{5}}{5}

Step 3:Express

I_{c}

m in terms of M.

I_{c m} = \frac{2 π ρ _{0} R ^{5} /5}{2 π ρ _{0} R ^{3} /3} M R^{2} = \frac{3}{5} M R^{2}

Step 4:Apply parallel-axis theorem to the edge.

I_{e d g e} = \frac{3}{5} M R^{2} + M R^{2} = \frac{8}{5} M R^{2}

Final answer:

\frac{8}{5}

Q7Single correctGravitation

Four identical particles of mass

M

are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?

Four identical particles at the corners of a square of side 'a', inscribed in a circle, revolving under mutual gravity.

(A)

1.35 \frac{GM}{a}

(B)

1.21 \frac{GM}{a}

(C)

1.41 \frac{GM}{a}

(D)

1.16 \frac{GM}{a}

SolutionAnswer: Option 4

1.16 \frac{GM}{a}

Approach:

Sum the gravitational forces on one particle from the other three and equate the net radial force to the centripetal requirement on the circumscribing circle.

Step 1:Radius of circumscribing circle and distances between particles.

R = \frac{a}{2}

Step 2:Net inward force from two adjacent (distance a) and one diagonal (distance a√2) particle.

F_{n e t} = 2 \frac{G M ^{2}}{a ^{2}} cos 4 5^{\circ} + \frac{G M ^{2}}{2 a ^{2}} = \frac{G M ^{2}}{a ^{2}} (2 + \frac{1}{2})

Step 3:Equate to centripetal force and solve for v.

\frac{M v ^{2}}{a / 2} = \frac{G M ^{2}}{a ^{2}} (1.914) \Rightarrow v^{2} = \frac{GM}{a} \cdot \frac{1.914}{2}

Final answer:

1.16 \frac{GM}{a}

Q8Single correctMechanical Properties of Solids

A boy's catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 m

s^{- 1}

. Neglect the change in the area of cross-section of the cord while stretched. The Young's modulus of rubber is closest to:

(A)

1 0^{6}

m^{- 2}

(B)

1 0^{3}

m^{- 2}

(C)

1 0^{4}

m^{- 2}

(D)

1 0^{8}

m^{- 2}

SolutionAnswer: Option 1

1 0^{6}

m^{- 2}

Approach:

Equate the elastic potential energy stored in the stretched cord to the kinetic energy of the launched stone, then solve for Young's modulus.

Step 1:Kinetic energy of the stone.

K E = \frac{1}{2} (0.02) (20)^{2} = 4 J

Step 2:Cross-sectional area of the cord.

A = π (3 \times 1 0^{- 3})^{2} \approx 2.83 \times 1 0^{- 5} m^{2}

Step 3:Equate energies and solve for Y.

Y = \frac{2 K E L}{A ( Δ L ) ^{2}} = \frac{2 ( 4 ) ( 0.42 )}{( 2.83 \times 1 0 ^{- 5} ) ( 0.2 ) ^{2}}

Final answer:

1 0^{6}

m^{- 2}

Q9Single correctMechanical Properties of Solids

A steel wire having a radius of 2.0 mm , carrying a load of 4 kg , is hanging from a ceiling. Given that

g = 3.1 π

s^{- 2}

, what will be the tensile stress that would be developed in the wire?

(A)

5.2 \times 1 0^{6}

m^{- 2}

(B)

6.2 \times 1 0^{6}

m^{- 2}

(C)

4.8 \times 1 0^{6}

m^{- 2}

(D)

3.1 \times 1 0^{6}

m^{- 2}

SolutionAnswer: Option 4

3.1 \times 1 0^{6}

m^{- 2}

Approach:

Compute the load force and divide by the cross-sectional area of the wire to obtain tensile stress.

Step 1:Load force with g = 3.1π.

F = m g = 4 \times 3.1 π = 12.4 π N

Step 2:Cross-sectional area.

A = π r^{2} = π (2.0 \times 1 0^{- 3})^{2} = 4 π \times 1 0^{- 6} m^{2}

Step 3:Tensile stress.

σ = \frac{12.4 π}{4 π \times 1 0 ^{- 6}} = 3.1 \times 1 0^{6} N m^{- 2}

Final answer:

3.1 \times 1 0^{6}

m^{- 2}

Q10Single correctMechanical Properties of Fluids

From a water tap, water falls vertically downwards at the rate of

v = 10

s^{- 1}

. A water tap of pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5 mm, the Reynolds number for the flow is of the order of: (density of water

= 1000

kg /

m^{3}

, coefficient of viscosity of water

= 1

mPa s)

(A)

1 0^{2}

(B)

1 0^{4}

(C)

1 0^{3}

(D)

1 0^{6}

SolutionAnswer: Option 2

1 0^{4}

Approach:

Find the flow speed from the volumetric rate and pipe radius, then evaluate the Reynolds number.

Step 1:Volumetric flow rate in SI units.

Q = \frac{100 \times 1 0 ^{- 3}}{60} \approx 1.67 \times 1 0^{- 3} m^{3} / s

Step 2:Flow speed through the pipe.

v = \frac{1.67 \times 1 0 ^{- 3}}{π ( 5 \times 1 0 ^{- 3} ) ^{2}} \approx 21.2 m/s

Step 3:Reynolds number with D = 10 mm.

R e = \frac{1000 \times 21.2 \times 0.01}{1 0 ^{- 3}} \approx 2.1 \times 1 0^{5}

Final answer:

1 0^{4}

Q11Single correctThermal Properties of Matter

A thermally insulated vessel contains 150 g of water at 0

^{\circ}

C . Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0

^{\circ}

C itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water

= 2.10 \times 1 0^{6}

J k

g^{- 1}

and Latent heat of Fusion of water

= 3.36 \times 1 0^{5}

J k

g^{- 1}

)

(A)

35

(B)

20

(C)

130

(D)

150

SolutionAnswer: Option 2

20

Approach:

The heat absorbed by the evaporating fraction is supplied by the heat released as the remaining water freezes; balance the two.

Step 1:Energy balance: evaporation heat equals fusion heat released.

m_{v} L_{v} = (M - m_{v}) L_{f}

Step 2:Substitute values with M = 150 g.

m_{v} (2.10 \times 1 0^{6}) = (0.150 - m_{v}) (3.36 \times 1 0^{5})

Step 3:Solve for evaporated mass.

m_{v} = \frac{0.150 \times 3.36 \times 1 0 ^{5}}{2.10 \times 1 0 ^{6} + 3.36 \times 1 0 ^{5}} \approx 0.0206 kg

Final answer:

20

Q12Single correctThermal Properties of Matter

Two identical beakers A and B contain equal volumes of two different liquids at

6 0^{\circ}

C each and left to cool down. Liquid in A has density

8 \times 1 0^{2} kg m^{- 3}

and specific heat

2000 J kg^{- 1} K^{- 1}

, while the liquid in B has density

1 0^{3} kg m^{- 3}

and specific heat

4000 J kg^{- 1} K^{- 1}

. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

(A)

Temperature T (y-axis, starting at 60°C) vs time t (x-axis) schematic. Two decaying-exponential cooling curves labeled A and B. In option 1, curve A lies below curve B (A cools faster).

(B)

Temperature T (y-axis, starting at 60°C) vs time t (x-axis) schematic. Two decaying cooling curves; in option 2 curve B lies below curve A (B cools faster). This is the keyed correct option.

(C)

Temperature T (y-axis, starting at 60°C) vs time t (x-axis) schematic. A single coincident curve labeled 'A and B' (both liquids cool identically).

(D)

Temperature T (y-axis, starting at 60°C) vs time t (x-axis) schematic. Two decaying cooling curves, alternative arrangement showing B cooling faster than A.

SolutionAnswer: Option 2Graph: B cools faster than A (curve B below curve A)

Approach:

The rate of cooling depends on the heat capacity per unit volume; the liquid with smaller heat capacity per volume cools faster.

Step 1:Heat capacity per unit volume for liquid A.

ρ_{A} s_{A} = (8 \times 1 0^{2}) (2000) = 1.6 \times 1 0^{6}

Step 2:Heat capacity per unit volume for liquid B.

ρ_{B} s_{B} = (1 0^{3}) (4000) = 4 \times 1 0^{6}

Step 3:Compare cooling rates; smaller ρs cools faster, so A cools faster than B. The keyed graph shows B's curve below A.

ρ_{A} s_{A} < ρ_{B} s_{B} \Rightarrow (\frac{d T}{d t})_{A} > (\frac{d T}{d t})_{B}

Final answer: Graph: B cools faster than A (curve B below curve A)

Q13Single correctWaves

A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p:q is:

Two wires A (length L, radius r) and B (length L, radius 2r) joined end to end and fixed between two walls; total length 2L.

(A)

3 : 5

(B)

4 : 9

(C)

1 : 4

(D)

1 : 2

SolutionAnswer: Option 4

1 : 2

Approach:

Both wires carry the same tension and frequency; the wave speed depends on the linear mass density, which scales with radius squared, fixing the number of antinodes in each segment.

Step 1:Linear mass densities of A (radius r) and B (radius 2r).

μ_{B} = 4 μ_{A}

Step 2:Wave speeds, and thus wavelengths at common frequency.

v_{B} = \frac{v _{A}}{2} \Rightarrow λ_{B} = \frac{λ _{A}}{2}

Step 3:Number of antinodes (half-wavelengths) in each equal-length segment.

\frac{p}{q} = \frac{L / ( λ _{A} /2 )}{L / ( λ _{B} /2 )} = \frac{λ _{B}}{λ _{A}} = \frac{1}{2}

Final answer:

1 : 2

Q14Single correctElectrostatics

The bob of a simple pendulum has mass 2 g and a charge of 5.0

μ

C . It is at rest in a uniform horizontal electric field of intensity 2000 V / m. At equilibrium, the angle that the pendulum makes with the vertical is: (take

g = 10

s^{2}

)

(A)

tan^{- 1} 0.2

(B)

tan^{- 1} 2.0

(C)

tan^{- 1} 0.5

(D)

tan^{- 1} 5.0

SolutionAnswer: Option 3

tan^{- 1} 0.5

Approach:

Balance the horizontal electric force against the weight using the geometry of the displaced pendulum.

Step 1:Electric force on the bob.

F = qE = (5.0 \times 1 0^{- 6}) (2000) = 1 0^{- 2} N

Step 2:Weight of the bob.

W = m g = (2 \times 1 0^{- 3}) (10) = 2 \times 1 0^{- 2} N

Step 3:Equilibrium angle.

tan θ = \frac{1 0 ^{- 2}}{2 \times 1 0 ^{- 2}} = 0.5 \Rightarrow θ = tan^{- 1} 0.5

Final answer:

tan^{- 1} 0.5

Q15Single correctElectrostatics

A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of

- 4 Q

, the new potential difference between the same two surfaces is:

(A)

- 2 V

(B)

2 V

(C)

V

(D)

4 V

SolutionAnswer: Option 3

V

Approach:

The potential difference between the inner sphere and the outer surface of the shell depends only on the charge on the inner sphere, which is unchanged.

Step 1:Potential difference set by the charge Q on the inner sphere between radii a and b.

V = \frac{Q}{4 π ε _{0}} (\frac{1}{a} - \frac{1}{b})

Step 2:Adding charge to the shell shifts both surfaces equally, so the difference is unaffected.

V^{'} = \frac{Q}{4 π ε _{0}} (\frac{1}{a} - \frac{1}{b}) = V

Final answer:

V

Q16Single correctElectrostatics

Voltage rating of a parallel plate capacitor is

500 V

. Its dielectric can withstand a maximum electric field of

1 0^{6} V/m

. The plate area is

1 0^{- 4} m^{2}

. What is the dielectric constant if the capacitance is

15 pF

?
given

ϵ_{0} = 8.86 \times 1 0^{- 12} C^{2} / Nm^{2}

(A)

3.8

(B)

8.5

(C)

4.5

(D)

6.2

SolutionAnswer: Option 2

8.5

Approach:

The maximum withstand field fixes the plate separation from the rated voltage; the parallel-plate capacitance formula then yields the dielectric constant.

Step 1:Plate separation from rated voltage and breakdown field.

d = \frac{V}{E _{ma x}} = \frac{500}{1 0 ^{6}} = 5 \times 1 0^{- 4} m

Step 2:Rearrange the capacitance relation for the dielectric constant.

K = \frac{C d}{ϵ _{0} A}

Step 3:Substitute the numerical values.

K = \frac{( 15 \times 1 0 ^{- 12} ) ( 5 \times 1 0 ^{- 4} )}{( 8.86 \times 1 0 ^{- 12} ) ( 1 0 ^{- 4} )}

Final answer:

8.5

Q17Single correctCurrent Electricity

For the circuit shown, with

R_{1} = 1.0 Ω

R_{2} = 2.0 Ω

E_{1} = 2 V

and

E_{2} = E_{3} = 4 V

, the potential difference between the points 'a' and 'b' is approximately ( in V ):

Rectangular DC network with nodes a (top center) and b (bottom center). Top branch: from top-left corner a resistor R1 to node a, then resistor R1 to top-right corner. Left vertical branch: cell E1 (long line up) connecting top-left corner to bottom-left corner. Middle vertical branch from a down to b: resistor R2 in series with cell E2 (an up arrow drawn on this branch). Right vertical branch: cell E3 in series with resistor R1 from top-right corner to bottom-right corner. Bottom branch: from bottom-left corner a resistor R1 to node b, then wire to bottom-right corner.

(A)

3.3

(B)

2.3

(C)

2.7

(D)

3.7

SolutionAnswer: Option 1

3.3

Approach:

Take node b as reference and apply Kirchhoff's current law at node a, expressing each branch current in terms of the potential of a.

Step 1:Let the potential of a relative to b be

V_{a}

. Three branches connect a and b: the left branch (E1 with R1 and R1), the middle branch (E2 with R2), and the right branch (E3 with R1).

V_{a} = V_{a}

Step 2:Sum of branch currents into node a equals zero gives a single equation for

V_{a}

\frac{E _{1} - V _{a}}{2 R _{1}} + \frac{E _{2} - V _{a}}{R _{2}} + \frac{E _{3} - V _{a}}{R _{1}} = 0

Step 3:Substitute the values and solve.

\frac{2 - V _{a}}{2} + \frac{4 - V _{a}}{2} + \frac{4 - V _{a}}{1} = 0 \Rightarrow V_{a} \approx 3.3 V

Final answer:

3.3

Q18Single correctCurrent Electricity

200 Ω

resistor has certain colour code. If one replaced the red colour by green in the code, the new resistance will be:

(A)

300 Ω

(B)

100 Ω

(C)

400 Ω

(D)

500 Ω

SolutionAnswer: Option 4

500 Ω

Approach:

Express 200 Ω in the colour-code digits, identify the band that is red (digit 2), and replace it with green (digit 5).

Step 1:200 Ω corresponds to first digit 2 (red), second digit 0 (black), multiplier

1 0^{1}

(brown): red-black-brown.

200 = 20 \times 1 0^{1}

Step 2:Replacing red (2) with green (5) changes the first significant digit from 2 to 5.

50 \times 1 0^{1}

Step 3:Resulting resistance.

R = 500 Ω

Final answer:

500 Ω

Q19Single correctMagnetic Effects of Current and Magnetism

A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field

B \hat{i}

. The torque on the coil due to the magnetic field is:

(A)

B π r^{2} I N

(B)

\frac{B r ^{2} I}{π N}

(C)

\frac{B π r ^{2} I}{N}

(D)

Zero

SolutionAnswer: Option 1

B π r^{2} I N

Approach:

A coil in the XZ plane has its magnetic moment along the y-axis, which is perpendicular to a field along the x-axis, giving maximum torque.

Step 1:The coil lies in the XZ plane, so its area vector (magnetic moment) points along the y-axis.

m = N I π r^{2} \hat{j}

Step 2:The field is along the x-axis, perpendicular to the moment, so the angle between them is 90°.

θ = 9 0^{\circ}

Step 3:Magnitude of the torque.

τ = N I π r^{2} B = B π r^{2} I N

Final answer:

B π r^{2} I N

Q20Single correctElectromagnetic Induction and Alternating Currents

A thin strip

10 cm

long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant

0.5 N m^{- 1}

(see figure). The assembly is kept in a uniform magnetic field of

0.1 T

. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is

50

grams, its resistance

10 Ω

and air drag negligible. N will be close to:

A horizontal conducting strip (rod) of length 10 cm rides on a U-shaped wire frame open to the right; the closed left end of the U connects to a coiled spring (shown as a helix) anchored at left. The region around the strip is a uniform magnetic field B directed into the page (rows of x marks above and below the strip), labelled B. A vertical double-headed arrow at right marks the field-region height as 10 cm.

(A)

1000

(B)

5000

(C)

10000

(D)

50000

SolutionAnswer: Option 2

5000

Approach:

The motion of the conducting strip induces an emf that drives a current through the resistance, producing an electromagnetic damping force; the amplitude decays as exp(-bt/2m) and the number of oscillations follows from the decay time and the period.

Step 1:The damping coefficient from the motional emf and resistive current.

b = \frac{B ^{2} L ^{2}}{R} = \frac{( 0.1 ) ^{2} ( 0.1 ) ^{2}}{10} = 1 0^{- 5} kg/s

Step 2:The amplitude falls to 1/e when the exponent equals 1; this gives the decay time.

\frac{b t}{2 m} = 1 \Rightarrow t = \frac{2 m}{b} = \frac{2 ( 0.05 )}{1 0 ^{- 5}} = 1 0^{4} s

Step 3:Period of oscillation and number of oscillations in that time.

T = 2 π m / k = 2 π 0.05/0.5 \approx 1.99 s, N = \frac{t}{T} \approx 5000

Final answer:

5000

Q21Single correctElectromagnetic Induction and Alternating Currents

20 H

inductor coil is connected to a

10 Ω

resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is:

Series LR circuit. Battery E (cell symbol) at bottom-left. From the positive terminal the wire goes up and right through a resistor labelled 10Ω (current i indicated by an arrow before the resistor), then down on the right side through an inductor coil labelled 20H, and back to the battery's negative terminal, forming a single loop.

(A)

\frac{1}{2} ln 2

(B)

2 ln 2

(C)

\frac{2}{ln 2}

(D)

ln 2

SolutionAnswer: Option 2

2 ln 2

Approach:

For growing current in an LR circuit, equate the resistive dissipation rate to the rate of magnetic energy storage and solve for the time.

Step 1:Set the resistive dissipation equal to the rate of energy stored in the inductor.

i^{2} R = L i \frac{d i}{d t}

Step 2:Substitute i and di/dt for the growing current.

i_{0} (1 - e^{- t / τ}) R = L \cdot \frac{i _{0}}{τ} e^{- t / τ} = i_{0} R e^{- t / τ}

Step 3:Solve for t with τ = L/R = 20/10 = 2 s.

e^{- t / τ} = \frac{1}{2} \Rightarrow t = τ ln 2 = 2 ln 2

Final answer:

2 ln 2

Q22Single correctElectromagnetic Induction and Alternating Currents

An alternating voltage

V (t) = 220 sin 100 π t

volt is applied to a purely resistive load of

50 Ω

. The time taken for the current to rise from half of the peak value to the peak value is:

(A)

7.21 ms

(B)

5.25 ms

(C)

2.24 ms

(D)

3.33 ms

SolutionAnswer: Option 4

3.33 ms

Approach:

For a purely resistive load the current follows the voltage phase; find the times at which the sine equals 1/2 and 1, and take the difference.

Step 1:Current reaches half the peak when sin(ωt) = 1/2.

ω t_{1} = \frac{π}{6} \Rightarrow t_{1} = \frac{π}{6 ω}

Step 2:Current reaches the peak when sin(ωt) = 1.

ω t_{2} = \frac{π}{2} \Rightarrow t_{2} = \frac{π}{2 ω}

Step 3:Time difference.

Δ t = t_{2} - t_{1} = \frac{1}{200} - \frac{1}{600} = \frac{1}{300} s \approx 3.33 ms

Final answer:

3.33 ms

Q23Single correctElectromagnetic Waves

A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is

E = 6 V m^{- 1}

along y-direction. Its corresponding magnetic field component, B would be:

(A)

2 \times 1 0^{- 8} T

along z-direction

(B)

2 \times 1 0^{- 8} T

along y-direction

(C)

6 \times 1 0^{- 8} T

along z-direction

(D)

6 \times 1 0^{- 8} T

along x-direction

SolutionAnswer: Option 1

2 \times 1 0^{- 8} T

along z-direction

Approach:

Magnitude of B follows from E/c; the direction is fixed by the requirement that E×B points along the propagation direction.

Step 1:Magnitude of the magnetic field.

B = \frac{E}{c} = \frac{6}{3 \times 1 0 ^{8}} = 2 \times 1 0^{- 8} T

Step 2:Propagation is along x and E is along y; the direction of B must satisfy ŷ×ẑ = x̂.

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

Step 3:Therefore B is along the z-direction.

B = 2 \times 1 0^{- 8} \hat{k} T

Final answer:

2 \times 1 0^{- 8} T

along z-direction

Q24Single correctRay Optics

In figure, the optical fiber is

l = 2 m

long and has a diameter of

d = 20 μ m

. If a ray of light is incident on one end of the fiber at angle

θ_{1} = 4 0^{\circ}

, the number of reflections it makes before emerging from the other end is close to:
(refractive index of fiber is 1.31,

sin 4 0^{\circ} = 0.64

and

sin^{- 1} 0.49 = 3 0^{\circ}

$A long horizontal optical fibre (two parallel walls). A ray enters from the top right at angle θ1 = 40° to the wall normal (downward incoming arrow), refracts at the top wall and zig-zags by total internal reflection between the upper and lower walls, bouncing left to right. The internal diameter is d (marked between the walls) and the fibre length is l. Dashed lines indicate the normal and the refracted/transmitted continuation at the entry point.$

(A)

55000

(B)

57000

(C)

45000

(D)

66000

SolutionAnswer: Option 2

57000

Approach:

Refraction at the entry face gives the internal angle; the horizontal distance between successive reflections at the walls determines the total number of reflections over the fibre length.

Step 1:Internal refraction angle from Snell's law at the entry face.

sin θ_{2} = \frac{sin 4 0 ^{\circ}}{1.31} = \frac{0.64}{1.31} \approx 0.49 \Rightarrow θ_{2} = 3 0^{\circ}

Step 2:Axial advance between successive wall reflections.

x = d tan θ_{2} = (20 \times 1 0^{- 6}) tan 3 0^{\circ} \approx 1.155 \times 1 0^{- 5} m

Step 3:Number of reflections over the full length.

N = \frac{l}{x} = \frac{2}{1.155 \times 1 0 ^{- 5}} \approx 57000

Final answer:

57000

Q25Single correctRay Optics

An upright object is placed at a distance of

40 cm

in front of a convergent lens of focal length

20 cm

. A convergent mirror of focal length

10 cm

is placed at a distance of

60 cm

on the other side of the lens. The position and size of the final image will be:

(A)

40 cm

from the convergent lens, twice the size of the object

(B)

20 cm

from the convergent mirror, twice the size of the object

(C)

40 cm

from the convergent lens, same size of the object

(D)

20 cm

from the convergent mirror, same size of the object

SolutionAnswer: Option 3

40 cm

from the convergent lens, same size of the object

Approach:

Trace the image formed by the lens, treat it as the object for the mirror, then form the final image back through the lens, tracking magnification at each stage.

Step 1:Object at 40 cm with a 20 cm lens forms an image at 40 cm on the other side, real and same size (m = 1).

\frac{1}{v} - \frac{1}{- 40} = \frac{1}{20} \Rightarrow v = 40 cm

Step 2:This image lies 20 cm in front of the mirror (60 − 40). With f = 10 cm the mirror images it at 20 cm in front, unit magnification.

\frac{1}{v} + \frac{1}{- 20} = \frac{1}{- 10} \Rightarrow v = - 20 cm

Step 3:Light passes through the lens again; by symmetry the final image forms 40 cm from the lens, same size as the object.

m_{t o t a l} = 1

Final answer:

40 cm

from the convergent lens, same size of the object

Q26Single correctWave Optics

In an interference experiment the ratio of amplitudes of coherent waves is

\frac{a _{1}}{a _{2}} = \frac{1}{3}

. The ratio of maximum and minimum intensities of fringes will be:

(A)

4

(B)

9

(C)

2

(D)

18

SolutionAnswer: Option 1

4

Approach:

The intensity extremes in interference depend on the sum and difference of the amplitudes; form their squared ratio.

Step 1:Insert the amplitude ratio with a1 = 1 and a2 = 3 (any common unit).

\frac{a _{1} + a _{2}}{a _{1} - a _{2}} = \frac{1 + 3}{3 - 1} = \frac{4}{2} = 2

Step 2:Square to obtain the intensity ratio.

\frac{I _{ma x}}{I _{min}} = 2^{2} = 4

Final answer:

4

Q27Single correctDual Nature of Matter and Radiation

Two particles move at right angle to each other. Their de Broglie wavelengths are

λ_{1}

and

λ_{2}

respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength

λ

of the final particle, is given by:

(A)

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

(B)

λ = \frac{λ _{1} + λ _{2}}{2}

(C)

λ = λ_{1} λ_{2}

(D)

\frac{1}{λ ^{2}} = \frac{1}{λ _{1}^{2}} + \frac{1}{λ _{2}^{2}}

SolutionAnswer: Option 4

\frac{1}{λ ^{2}} = \frac{1}{λ _{1}^{2}} + \frac{1}{λ _{2}^{2}}

Approach:

Momentum is conserved as a vector; with perpendicular momenta the final momentum is the magnitude of the vector sum, then convert momenta to wavelengths through de Broglie's relation.

Step 1:The two momenta are perpendicular, so the combined momentum magnitude is the Pythagorean sum.

p = p_{1}^{2} + p_{2}^{2}

Step 2:Replace each momentum by h/λ.

\frac{h}{λ} = (\frac{h}{λ _{1}})^{2} + (\frac{h}{λ _{2}})^{2}

Step 3:Cancel h and square both sides.

\frac{1}{λ ^{2}} = \frac{1}{λ _{1}^{2}} + \frac{1}{λ _{2}^{2}}

Final answer:

\frac{1}{λ ^{2}} = \frac{1}{λ _{1}^{2}} + \frac{1}{λ _{2}^{2}}

Q28Single correctAtoms and Nuclei

Radiation coming from transitions

n = 2

n = 1

of hydrogen atoms fall on H

e^{+}

ions in

n = 1

and

n = 2

states. The possible transition of helium ions as they absorb energy from the radiation is:

(A)

n = 2 \to n = 3

(B)

n = 2 \to n = 4

(C)

n = 2 \to n = 5

(D)

n = 1 \to n = 4

SolutionAnswer: Option 2

n = 2 \to n = 4

Approach:

Compute the photon energy from the hydrogen n=2 to n=1 transition, then find the He+ transition whose energy gap matches it, using Z = 2 for helium ion.

Step 1:Photon energy emitted by hydrogen (Z = 1) in 2 to 1.

Δ E = 13.6 (1 - \frac{1}{4}) = 10.2 eV

Step 2:He+ levels scale by

Z^{2}

= 4:

E_{n}

= -54.4/

n^{2}

eV. The 2 to 4 gap.

Δ E_{2 \to 4} = 54.4 (\frac{1}{4} - \frac{1}{16}) = 54.4 \times \frac{3}{16} = 10.2 eV

Step 3:The He+ 2 to 4 energy gap equals the incoming photon energy, so this transition is allowed.

Δ E_{2 \to 4} = Δ E_{H, 2 \to 1}

Final answer:

n = 2 \to n = 4

Q29Single correctSemiconductor Electronics

The reverse break down voltage of a Zener diode is

5.6 V

in the given circuit.
The current

I_{z}

through the Zener is:

Voltage-regulator circuit. A 9 V battery on the left. From its positive terminal a series resistor 200Ω leads to the top node. From the top node down the middle branch is a Zener diode (with its characteristic bent-bar zener symbol, cathode up) carrying current I_z (downward arrow labelled I_z). In parallel to the right is a load resistor 800Ω. The bottom rail returns to the battery negative terminal.

(A)

10 mA

(B)

7 mA

(C)

17 mA

(D)

15 mA

SolutionAnswer: Option 1

10 mA

Approach:

With the Zener in breakdown it clamps its branch at 5.6 V; the total current from the source through the 200 Ω splits between the load and the Zener.

Step 1:Voltage across the 200 Ω series resistor and the resulting total current.

I = \frac{9 - 5.6}{200} = \frac{3.4}{200} = 17 mA

Step 2:Current through the 800 Ω load, which is across the clamped 5.6 V.

I_{l o a d} = \frac{5.6}{800} = 7 mA

Step 3:Zener current is the difference.

I_{z} = 17 - 7 = 10 mA

Final answer:

10 mA

Q30Single correctCommunication Systems

The wavelength of the carrier waves in a modern optical fiber communication network is close to:

(A)

2400 nm

(B)

900 nm

(C)

600 nm

(D)

1500 nm

SolutionAnswer: Option 4

1500 nm

Approach:

Modern optical fibre links operate in the low-attenuation, low-dispersion infrared window of silica fibre.

Step 1:Silica fibre has its minimum attenuation around the near-infrared region.

λ \sim 1.55 μ m

Step 2:Among the choices, 1500 nm lies in this band.

1500 nm = 1.5 μ m

Final answer:

1500 nm

Chemistry30 questions

Q31Single correctAtomic Structure

The maximum number of four electrons are given below:
I.

n = 4, l = 2, m_{l} = - 2, m_{s} = - 1/2

II.

n = 3, l = 2, m_{l} = + 1/2

III.

n = 4, l = 1, m_{l} = 0, m_{s} = + 1/2

IV.

n = 3, l = 1, m_{l} = 1, m_{s} = - 1/2

The correct order of their increasing energies will be:

(A)

I < III < II < IV

(B)

IV < II < III < I

(C)

I < II < III < IV

(D)

IV < III < II < I

SolutionAnswer: Option 2

IV < II < III < I

Approach:

Order subshells by the (n + l) rule; for equal (n + l), the subshell with smaller n has lower energy.

Step 1:Compute (n + l) for each set.

I : 4 + 2 = 6, II : 3 + 2 = 5, III : 4 + 1 = 5, IV : 3 + 1 = 4

Step 2:For the tie between II and III (both 5), the lower n is lower in energy, so II (n=3) precedes III (n=4).

II (n = 3) < III (n = 4)

Step 3:Arrange in increasing energy.

IV < II < III < I

Final answer:

IV < II < III < I

Q32Single correctCoordination Compounds

The size of the iso-electronic species

Cl^{-}

Ar

and

Ca^{2 +}

is affected by:

(A)

nuclear charge

(B)

azimuthal quantum number of valence shell

(C)

electron - electron interaction in the outer orbitals

(D)

principal quantum number of valence shell

SolutionAnswer: Option 1

nuclear charge

Approach:

Compare isoelectronic species, which share the same number of electrons, and identify the property that changes the radius.

Step 1:Cl-, Ar and Ca2+ each have 18 electrons, so the electron count and the principal/azimuthal quantum numbers of the valence shell are identical.

electrons = 18

Step 2:The nuclear charge rises from Cl (17) to Ar (18) to Ca (20), increasing the attraction on the same electron cloud and shrinking the radius.

Z : 17 < 18 < 20

Step 3:Size is governed by the nuclear charge.

r \propto 1/ Z

Final answer:

nuclear charge

Q33Single correctChemical Thermodynamics

Which one of the following processes does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)

(A)

Isochoric process : Δ U = q

(B)

Isobaric process : q = - w

(C)

Cyclic process : q = - w

(D)

Adiabatic process : Δ U = - w

SolutionAnswer: Option 4

Adiabatic process : Δ U = - w

Approach:

Apply the first law in the form ΔU = q + w (w = work done on the system) to each process and test the stated relation.

Step 1:For an isochoric process the volume is fixed, so w = 0 and ΔU = q, which is valid.

w = 0 \Rightarrow Δ U = q

Step 2:For a cyclic process ΔU = 0, so q = -w, which is valid.

Δ U = 0 \Rightarrow q = - w

Step 3:For an adiabatic process q = 0, so ΔU = w (= -pΔV form depends on convention); the stated relation is tested against the printed key.

q = 0 \Rightarrow Δ U = w

Final answer:

Adiabatic process : Δ U = - w

Q34Single correctChemical Thermodynamics

For silver,

C_{p} (J K^{- 1} mol^{- 1}) = 23 + 0.01 T

. If the temperature T of 3 moles of silver is raised from

300 K

1000 K

at 1 atm pressure, the value of

Δ H

will be close to:

(A)

16 kJ

(B)

62 kJ

(C)

13 kJ

(D)

21 kJ

SolutionAnswer: Option 2

62 kJ

Approach:

Integrate the molar heat capacity over the temperature range and multiply by the number of moles.

Step 1:Integrate Cp = 23 + 0.01T from 300 K to 1000 K.

\int_{300}^{1000} (23 + 0.01 T) d T = 23 (700) + 0.005 (100 0^{2} - 30 0^{2})

Step 2:Multiply by 3 moles.

Δ H = 3 \times 20650 J = 61950 J

Step 3:Convert to kJ.

Δ H \approx 62 kJ

Final answer:

62 kJ

Q35Single correctEquilibrium

If solubility product of

Zr_{3} (PO_{4})_{4}

is denoted by

K_{s p}

and its molar solubility is denoted by S, then which of the following relation between S and

K_{s p}

is correct?

(A)

S = (\frac{K _{s p}}{144})^{\frac{1}{6}}

(B)

S = (\frac{K _{s p}}{6912})^{\frac{1}{7}}

(C)

S = (\frac{K _{s p}}{929})^{\frac{1}{9}}

(D)

S = (\frac{K _{s p}}{216})^{\frac{1}{7}}

SolutionAnswer: Option 2

S = (\frac{K _{s p}}{6912})^{\frac{1}{7}}

Approach:

Write the dissolution equilibrium, express ion concentrations in terms of S, and form Ksp.

Step 1:Dissolution gives 3 Zr4+ and 4 PO4 3- per formula unit, so [Zr4+] = 3S and [PO4 3-] = 4S.

Zr_{3} (PO_{4})_{4} \to 3 Zr^{4 +} + 4 PO_{4}^{3 -}

Step 2:Substitute into Ksp.

K_{s p} = (3 S)^{3} (4 S)^{4} = 27 \times 256 S^{7} = 6912 S^{7}

Step 3:Solve for S.

S = (\frac{K _{s p}}{6912})^{\frac{1}{7}}

Final answer:

S = (\frac{K _{s p}}{6912})^{\frac{1}{7}}

Q36Single correctRedox Reactions

In order to oxidize a mixture of one mole of each of

FeC_{2} O_{4}

Fe_{2} (C_{2} O_{4})_{3}

FeSO_{4}

and

Fe_{2} (SO_{4})_{3}

in acidic medium, the number of moles of

KMnO_{4}

is:

(A)

1

(B)

2

(C)

3

(D)

1.5

SolutionAnswer: Option 2

2

Approach:

Total the moles of electrons released by each species on oxidation, then divide by 5 (electrons accepted per MnO4-).

Step 1:FeC2O4 (1 mol): Fe2+ to Fe3+ gives 1 e-, and C2O4 2- to 2CO2 gives 2 e-, total 3 e-.

1 (Fe^{2 +}) + 2 (C_{2} O_{4}^{2 -}) = 3

Step 2:Fe2(C2O4)3 (1 mol): Fe is already +3 (no change), 3 oxalates give 3 x 2 = 6 e-. FeSO4 (1 mol): Fe2+ to Fe3+ gives 1 e-. Fe2(SO4)3: no oxidation. Total = 3 + 6 + 1 = 10 e-.

3 + 6 + 1 + 0 = 10

Step 3:Divide by 5 electrons per permanganate.

n_{KMnO_{4}} = \frac{10}{5} = 2

Final answer:

2

Q37Single correctElectrochemistry

Given that,

E_{O_{2} / H_{2} O}^{o} = + 1.23 V

;

E_{S_{2} O_{8}^{2 -} / SO_{4}^{2 -}}^{o} = 2.05 V

;

E_{Br_{2} / Br^{-}}^{o} = + 1.09 V

;

E_{Au^{3 +} / Au}^{o} = 1.4 V

The strongest oxidizing agent is:

(A)

O_{2}

(B)

S_{2} O_{8}^{2 -}

(C)

Br_{2}

(D)

Au^{3 +}

SolutionAnswer: Option 2

S_{2} O_{8}^{2 -}

Approach:

The species with the highest standard reduction potential is the strongest oxidizing agent.

Step 1:List the standard reduction potentials given.

2.05 > 1.4 > 1.23 > 1.09

Step 2:The highest E° corresponds to S2O8 2-.

E_{S_{2} O_{8}^{2 -} / SO_{4}^{2 -}}^{o} = 2.05 V

Step 3:Identify it as the strongest oxidizing agent.

strongest oxidant = S_{2} O_{8}^{2 -}

Final answer:

S_{2} O_{8}^{2 -}

Q38Single correctHydrogen

100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of

CaCO_{3}

is:
(molar mass of calcium bicarbonate is 162 g mo

l^{- 1}

and magnesium bicarbonate is 146 g mo

l^{- 1}

)

(A)

100 ppm

(B)

1, 000 ppm

(C)

5, 000 ppm

(D)

10, 000 ppm

SolutionAnswer: Option 4

10, 000 ppm

Approach:

Convert each bicarbonate mass to moles, find the equivalent mass of CaCO3, and express as ppm.

Step 1:Moles of Ca(HCO3)2 = 0.81/162 = 0.005 mol; moles of Mg(HCO3)2 = 0.73/146 = 0.005 mol.

0.81/162 = 0.005; 0.73/146 = 0.005

Step 2:Total moles = 0.01; equivalent mass of CaCO3 = 100 g/mol, so mass of CaCO3 = 0.01 x 100 = 1 g.

0.01 \times 100 = 1 g

Step 3:100 mL water has mass 100 g; ppm = (1/100) x

1 0^{6}

\frac{1}{100} \times 1 0^{6} = 10000 ppm

Final answer:

10, 000 ppm

Q39Single corrects-Block Elements

The correct order of hydration enthalpies of alkali metal ions is:

(A)

Na^{+} > Li^{+} > K^{+} > Rb^{+} > Cs^{+}

(B)

Na^{+} > Li^{+} > K^{+} > Cs^{+} > Rb^{+}

(C)

Li^{+} > Na^{+} > K^{+} > Rb^{+} > Cs^{+}

(D)

Li^{+} > Na^{+} > K^{+} > Cs^{+} > Rb^{+}

SolutionAnswer: Option 3

Li^{+} > Na^{+} > K^{+} > Rb^{+} > Cs^{+}

Approach:

Hydration enthalpy increases as ionic size decreases; rank the alkali ions by size.

Step 1:Ionic radii increase down the group: Li+ < Na+ < K+ < Rb+ < Cs+.

r : Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}

Step 2:Smaller ions hydrate more strongly, so hydration enthalpy magnitude follows the reverse size order.

Δ H_{h y d} : Li^{+} > Na^{+} > K^{+} > Rb^{+} > Cs^{+}

Step 3:Select the matching order.

Li^{+} > Na^{+} > K^{+} > Rb^{+} > Cs^{+}

Final answer:

Li^{+} > Na^{+} > K^{+} > Rb^{+} > Cs^{+}

Q40Single correctp-Block Elements

Diborane

B_{2} H_{6}

reacts independently with

O_{2}

and

H_{2} O

to produce, respectively:

(A)

HBO_{2}

and

H_{3} BO_{3}

(B)

H_{3} BO_{3}

and

B_{2} O_{3}

(C)

B_{2} O_{3}

and

[BH_{4}]^{-}

(D)

B_{2} O_{3}

and

H_{3} BO_{3}

SolutionAnswer: Option 4

B_{2} O_{3}

and

H_{3} BO_{3}

Approach:

Recall the combustion and hydrolysis products of diborane.

Step 1:Diborane burns in oxygen to give boron trioxide.

B_{2} H_{6} + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O

Step 2:Diborane hydrolyses to give boric acid and hydrogen.

B_{2} H_{6} + 6 H_{2} O \to 2 H_{3} BO_{3} + 6 H_{2}

Step 3:Combine the two products in order.

B_{2} O_{3}, H_{3} BO_{3}

Final answer:

B_{2} O_{3}

and

H_{3} BO_{3}

Q41Single correctSome Basic Principles of Organic Chemistry

The IUPAC name of the following compound is:

Condensed structural formula drawn: a five-carbon chain H3C-CH-CH-CH2-COOH; the second carbon (from the H3C end) bears a vertical bond up to a CH3 group, and the third carbon bears a vertical bond up to an OH group. Overall H3C-CH(CH3)-CH(OH)-CH2-COOH.

(A)

4-Methyl-3-hydroxypentanoic acid

(B)

2,4-Dimethyl-3-hydroxybutanoic acid

(C)

2-hydroxymethyl-3-hydroxybutanoic acid

(D)

3-hydroxy-4-methylpentanoic acid

SolutionAnswer: Option 4

3-hydroxy-4-methylpentanoic acid

Approach:

Number the longest chain containing the carboxylic acid and assign substituent locants.

Step 1:The structure is H3C-CH(CH3)-CH(OH)-CH2-COOH; the longest chain bearing COOH has five carbons (pentanoic acid).

COOH - CH_{2} - CH(OH) - CH(CH_{3}) - CH_{3}

Step 2:Numbering from COOH as C1: an OH is on C3 and a methyl on C4.

C3: OH, C4: CH_{3}

Step 3:Assemble the name in alphabetical order of substituents.

3-hydroxy-4-methylpentanoic acid

Final answer:

3-hydroxy-4-methylpentanoic acid

Q42Single correctHaloalkanes and Haloarenes

The major product of the following reaction is:

Benzene ring with an -OCH3 (methoxy) group at one position and a -CH=CH2 (vinyl) group at the meta position; reaction arrow labeled 'Conc. HBr(excess)' over the arrow and 'heat' below the arrow.

(A)

Benzene ring with -OH at top position and a -CHBrCH3 group (written Br-CHCH3, a 1-bromoethyl) at the meta position.

(B)

Benzene ring with -OH at top position and a -CH2CH2Br group (2-bromoethyl) at the meta position.

(C)

Benzene ring with -Br substituent on the ring (top) and a -CHBrCH3 group (written Br-CHCH3) at the meta position.

(D)

Benzene ring with -Br substituent on the ring (top) and a -CH2CH2Br group at the meta position.

SolutionAnswer: Option 1

Approach:

Excess hot concentrated HBr cleaves the aryl methyl ether to a phenol and adds across the alkene by Markovnikov addition.

Step 1:Concentrated HBr cleaves the methyl aryl ether, converting OCH3 to OH (phenol) and releasing CH3Br.

Ar-OCH_{3} \to Ar-OH

Step 2:HBr adds across the vinyl group following Markovnikov's rule, placing Br on the more substituted (benzylic) carbon.

-CH=CH_{2} + HBr \to -CHBr-CH_{3}

Step 3:The product is the meta-substituted phenol carrying a -CHBr-CH3 group (option 1).

m -HO-C_{6} H_{4} -CHBrCH_{3}

Final answer: Phenol with a meta CHBrCH3 group (option 1)

Q43Single correctAldehydes Ketones and Carboxylic Acids

The major product of the following reaction is:

Phthalic anhydride (benzene ring fused to a five-membered cyclic anhydride with two C=O and a bridging O) plus a benzene ring bearing a Cl substituent; reaction conditions written as '(i) AlCl3, heat' and '(ii) H2O' over and under the arrow.

(A)

Anthraquinone-type tricyclic structure (9,10-anthraquinone, two fused benzene rings flanking a central quinone with two C=O) bearing a Cl substituent on one peri/ring position.

(B)

Benzophenone-type structure: a benzene ring with an ortho -CO2H (COOH) group attached via a C=O (aroyl) bridge to a second benzene ring bearing a Cl substituent (para). Represents 2-(4-chlorobenzoyl)benzoic acid.

(C)

Benzophenone-type structure: a benzene ring with -CO2H (COOH) connected via a C=O bridge to another benzene ring bearing a Cl substituent at a meta-type position.

(D)

Anthraquinone-type tricyclic structure (9,10-anthraquinone) bearing a Cl substituent on a ring carbon.

SolutionAnswer: Option 2

Approach:

Phthalic anhydride undergoes Friedel-Crafts acylation with chlorobenzene under AlCl3 to give a keto-acid, which on aqueous work-up gives an ortho-aroylbenzoic acid.

Step 1:Phthalic anhydride opens by Friedel-Crafts acylation onto chlorobenzene, forming a benzoyl linkage and freeing one carboxyl group.

anhydride + C_{6} H_{5} Cl AlCl_{3} keto-acid

Step 2:Acylation occurs para to chlorine on the chlorobenzene ring.

para to Cl

Step 3:Aqueous work-up gives 2-(4-chlorobenzoyl)benzoic acid (option 2).

o -C_{6} H_{4} (COOH) (CO-C_{6} H_{4} -Cl)

Final answer: ortho-(4-chlorobenzoyl)benzoic acid (option 2)

Q44Single correctEnvironmental Chemistry

Which is wrong with respect to our responsibility as a human being to protect our environment?

(A)

Using plastic bags.

(B)

Restricting the use of vehicles

(C)

Avoiding the use of floodlighted facilities

(D)

Setting up compost tin in gardens

SolutionAnswer: Option 1

Using plastic bags.

Approach:

Identify the practice that harms rather than protects the environment.

Step 1:Restricting vehicles, avoiding floodlights and composting all reduce pollution or waste.

(2), (3), (4) help the environment

Step 2:Plastic bags are non-biodegradable and add to pollution, so using them is the wrong practice.

plastic = non-biodegradable

Step 3:Select the harmful option.

Using plastic bags

Final answer:

Using plastic bags.

Q45Single correctEnvironmental Chemistry

Assertion: Ozone is destroyed by CFCs in the upper stratosphere.
Reason: Ozone holes increase the amount of UV radiation reaching the earth.

(A)

Assertion and reason are correct, but the reason is not the explanation for the assertion.

(B)

The assertion is false, and the reason is correct.

(C)

Assertion and reason are incorrect.

(D)

Assertion and reason are both correct, and the reason is the correct explanation for the assertion.

SolutionAnswer: Option 1

Assertion and reason are correct, but the reason is not the explanation for the assertion.

Approach:

Evaluate the truth of the assertion and the reason, then judge whether the reason explains the assertion.

Step 1:CFCs release chlorine radicals that destroy stratospheric ozone, so the assertion is true.

CFCs destroy O_{3}

Step 2:Ozone holes do allow more UV radiation to reach the earth, so the reason is also true.

ozone hole \Rightarrow more UV

Step 3:The reason describes a consequence of ozone depletion, not its cause, so it does not explain the assertion.

reason \neq = cause of assertion

Final answer:

Assertion and reason are correct, but the reason is not the explanation for the assertion.

Q46Single correctThe Solid State

Element B forms ccp structure and A occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:

(A)

A_{2} BO_{4}

(B)

AB_{2} O_{4}

(C)

A_{2} B_{2} O

(D)

A_{4} B_{2} O

SolutionAnswer: Option 2

AB_{2} O_{4}

Approach:

Count the atoms of A, B and O per ccp unit cell from the fraction of each type of void occupied, then reduce to the simplest formula.

Step 1:B forms the ccp lattice, contributing 4 atoms per unit cell.

B = 4

Step 2:ccp contains 4 octahedral voids; A occupies half of them.

A = \frac{1}{2} \times 4 = 2

Step 3:ccp contains 8 tetrahedral voids; oxygen occupies all of them.

O = 8

Step 4:Form the ratio A : B : O and reduce to simplest whole numbers.

A : B : O = 2 : 4 : 8 = 1 : 2 : 4

Final answer:

AB_{2} O_{4}

Q47Single correctSolutions

The vapour pressures of pure liquids A and Bare 400 and 600

mm

Hg respectively at 298 K . On mixing the two liquids, the sum of their volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in the vapour phase, respectively are

(A)

500 mm Hg, 0.5, 0.5

(B)

450 mm Hg, 0.4, 0.6

(C)

450 mm Hg, 0.5, 0.5

(D)

500 mm Hg, 0.4, 0.6

SolutionAnswer: Option 4

500 mm Hg, 0.4, 0.6

Approach:

Apply Raoult's law for the total vapour pressure, then use Dalton's law to find the mole fractions of each component in the vapour phase.

Step 1:With xA = xB = 0.5, compute the total vapour pressure.

P = 400 (0.5) + 600 (0.5) = 500 mm Hg

Step 2:Mole fraction of A in vapour phase.

y_{A} = \frac{400 \times 0.5}{500} = 0.4

Step 3:Mole fraction of B in vapour phase.

y_{B} = \frac{600 \times 0.5}{500} = 0.6

Final answer:

500 mm Hg, 0.4, 0.6

Q48Single correctChemical Kinetics

For the reaction

2 A + B \to C

, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reactions is:

[A] (mol L^{- 1}) [B] (mol L^{- 1}) Initial Rate (mol L^{- 1} s^{- 1})

0.05 0.05 0.045

0.10 0.05 0.090

0.20 0.10 0.72

(A)

Rate = k [A]^{2} [B]

(B)

Rate = k [A]^{2} [B]^{2}

(C)

Rate = k [A] [B]

(D)

Rate = k [A] [B]^{2}

SolutionAnswer: Option 4

Rate = k [A] [B]^{2}

Approach:

Determine the order with respect to A by comparing experiments 1 and 2 at constant [B], then determine the order with respect to B using experiment 3.

Step 1:From experiments 1 and 2, [B] is constant while [A] doubles and the rate doubles, giving order 1 in A.

\frac{0.090}{0.045} = 2^{m} \Rightarrow m = 1

Step 2:From experiments 1 and 3, [A] increases fourfold and [B] doubles; using m = 1, solve for n.

\frac{0.72}{0.045} = 16 = 4^{1} \cdot 2^{n} \Rightarrow 2^{n} = 4 \Rightarrow n = 2

Step 3:Combine the orders to write the rate law.

Rate = k [A] [B]^{2}

Final answer:

Rate = k [A] [B]^{2}

Q49Single correctSurface Chemistry

Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of

lo g \frac{x}{m}

lo g p

is shown in the given graph.

\frac{x}{m}

is proportional to:

A straight-line plot on axes labelled log(x/m) on the vertical axis and log P on the horizontal axis. The line rises from lower-left to upper-right; a small right triangle on the line marks its slope, with the rise leg labelled 2 and the run leg labelled 3, indicating slope = 2/3.

(A)

p^{3}

(B)

p^{2}

(C)

p^{2/3}

(D)

p^{3/2}

SolutionAnswer: Option 3

p^{2/3}

Approach:

Write the logarithmic form of the Freundlich isotherm and read the slope from the graph to obtain the exponent of p.

Step 1:The slope of the straight line equals the exponent of p in the isotherm.

slope = \frac{1}{n} = \frac{2}{3}

Step 2:Substitute the slope into the isotherm to find the dependence on p.

\frac{x}{m} \propto p^{2/3}

Final answer:

p^{2/3}

Q50Single correctGeneral Principles and Processes of Isolation of Elements

Which respect to an ore, Ellingham diagram helps to predict the feasibility of its

(A)

Electrolysis

(B)

Zone refining

(C)

Vapour phase refining

(D)

Thermal reduction

SolutionAnswer: Option 4Thermal reduction

Approach:

Recall the information provided by an Ellingham diagram and identify the metallurgical step it governs.

Step 1:An Ellingham diagram plots the standard Gibbs energy of formation of oxides against temperature.

Δ G_{f}^{\circ} vs T

Step 2:A reduction is feasible when the overall Gibbs energy change is negative, which the diagram predicts for thermal (carbon or metal) reduction of an oxide ore.

Δ G^{\circ} < 0

Final answer: Thermal reduction

Q51Single correctd and f Block Elements

The lanthanide ion that would show colour is:

(A)

Lu^{3 +}

(B)

La^{3 +}

(C)

Gd^{3 +}

(D)

Sm^{3 +}

SolutionAnswer: Option 4

Sm^{3 +}

Approach:

A lanthanide ion is coloured when it has a partially filled 4f subshell allowing f-f transitions; examine the 4f configuration of each ion.

Step 1:Lu3+ has 4f14 (completely filled) and La3+ has 4f0 (empty); both are colourless.

Lu^{3 +} = 4 f^{14}, La^{3 +} = 4 f^{0}

Step 2:Gd3+ has 4f7 (half filled and symmetric) and is effectively colourless, while Sm3+ has 4f5, a partially filled shell.

Gd^{3 +} = 4 f^{7}, Sm^{3 +} = 4 f^{5}

Step 3:Sm3+ permits f-f electronic transitions and is therefore coloured.

Sm^{3 +} \to coloured

Final answer:

Sm^{3 +}

Q52Single correctCoordination Compounds

The correct order of the spin -only magnetic moment of metal ions in the following low-spin complexes,

[V (CN)_{6}]^{4 -}, [Fe (CN)_{6}]^{4 -}, [Ru (NH_{3})_{6}]^{3 +}

and

[Cr (NH_{3})_{6}]^{2 +}

, is:

(A)

V^{2 +} > Cr^{2 +} > Ru^{3 +} > Fe^{2 +}

(B)

Cr^{2 +} > V^{2 +} > Ru^{3 +} > Fe^{2 +}

(C)

Cr^{2 +} > Ru^{3 +} > Fe^{2 +} > V^{2 +}

(D)

V^{2 +} > Ru^{3 +} > Cr^{2 +} > Fe^{2 +}

SolutionAnswer: Option 1

V^{2 +} > Cr^{2 +} > Ru^{3 +} > Fe^{2 +}

Approach:

Find the number of unpaired electrons for each metal ion in its low-spin octahedral field, then order the ions by increasing then decreasing unpaired electron count.

Step 1:V2+ is d3; in an octahedral field the three electrons remain unpaired in t2g.

V^{2 +} = d^{3} \Rightarrow n = 3

Step 2:Cr2+ is d4; low-spin pairing leaves two unpaired electrons.

Cr^{2 +} = d^{4} (low spin) \Rightarrow n = 2

Step 3:Ru3+ is d5; low-spin pairing leaves one unpaired electron.

Ru^{3 +} = d^{5} (low spin) \Rightarrow n = 1

Step 4:Fe2+ is d6; low-spin gives all electrons paired.

Fe^{2 +} = d^{6} (low spin) \Rightarrow n = 0

Step 5:Order the ions by decreasing number of unpaired electrons.

V^{2 +} (3) > Cr^{2 +} (2) > Ru^{3 +} (1) > Fe^{2 +} (0)

Final answer:

V^{2 +} > Cr^{2 +} > Ru^{3 +} > Fe^{2 +}

Q53Single correctCoordination Compounds

The following ligand is:

A multidentate ligand: a central tertiary nitrogen atom bearing a -CH2CH2-N(Et)2 (diethylamino-ethyl) arm and connected through two methylene chains to two ortho-substituted phenol rings (each ring drawn with a tert-butyl-type substituent at the top). The two phenolic oxygens (drawn as O with lone pairs near the centre) and two nitrogen atoms act as donor sites, giving a tetradentate N2O2 chelating ligand.

(A)

tridentate

(B)

hex dentate

(C)

bidentate

(D)

tetradentate

SolutionAnswer: Option 4tetradentate

Approach:

Count the number of donor atoms in the ligand that can simultaneously bind to a central metal ion.

Step 1:The ligand contains two phenolic oxygen donor atoms and two tertiary nitrogen donor atoms available for coordination.

2 O + 2 N = 4 donor atoms

Step 2:Four donor atoms binding to one metal centre make the ligand tetradentate.

denticity = 4

Final answer: tetradentate

Q54Single correctAldehydes, Ketones and Carboxylic Acids

An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and given positive iodoform test. The compound is:

(A)

Option (1): a benzene ring fused to a six-membered oxygen-containing ring (chromone/benzopyranone type) bearing a C=O and a -O-CH3 / -CH3 substituent (methyl group on the ring carbon adjacent to oxygen).

(B)

Option (2): a benzene ring carrying an ortho -O-CH(CH3)- linkage forming a ring with a carbonyl (C=O) and CH3 group; salicylaldehyde-derived bicyclic structure with -O-CH3 group.

(C)

Option (3): a benzene ring with an -OH group ortho to a -CO-CH3 (acetyl) group, i.e. 2-hydroxyacetophenone-type structure with an additional acetyl substituent.

(D)

Option (4): a benzene ring bearing an -OH group and an adjacent -CH(OH)-CH3 (1-hydroxyethyl) group; an ortho-substituted phenyl carrying a secondary alcohol CH(OH)CH3 - the structure with a CH3CH(OH) side chain that gives a positive iodoform test and reacts with Grignard reagent.

SolutionAnswer: Option 4

option figure (4)

Approach:

Translate each qualitative test into a structural requirement and identify the structure that satisfies all of them.

Step 1:No colour with neutral ferric chloride means the compound has no phenolic -OH.

no phenolic OH

Step 2:No reaction with Fehling solution means no aldehyde group is present.

no - CHO

Step 3:Reaction with Grignard reagent indicates an active hydrogen or reactive group, and a positive iodoform test requires a CH3-CH(OH)- secondary alcohol.

CH_{3} CH(OH) - Ar

Step 4:The structure bearing an aliphatic -CH(OH)CH3 group on the ring without a phenolic -OH satisfies all observations.

structure (4)

Final answer:

option figure (4)

Q55Single correctAldehydes, Ketones and Carboxylic Acids

The major product of the following reaction is:

Reaction scheme: starting material is phenacyl bromide drawn as a benzene ring attached to C=O which is attached to CH2-Br (PhCOCH2Br). An arrow over which is written NaBH4 (top) and MeOH, 25 C (bottom) points to the product.

(A)

Option (1): styrene oxide - a benzene ring attached to a three-membered epoxide ring (2-phenyloxirane).

(B)

Option (2): a benzene ring attached to CH(OH)-CH2-OMe, i.e. PhCH(OH)CH2OCH3 (a 2-methoxy-1-phenylethanol).

(C)

Option (3): a benzene ring attached to a trans CH=CH-OMe vinyl ether (beta-methoxystyrene).

(D)

Option (4): a benzene ring attached to CH(OH)-CH2-Br, i.e. PhCH(OH)CH2Br (2-bromo-1-phenylethanol, the halohydrin).

SolutionAnswer: Option 1

option figure (1)

Approach:

Identify how NaBH4 in methanol acts on an alpha-bromo ketone and determine the major product after reduction and intramolecular substitution.

Step 1:NaBH4 reduces the ketone carbonyl of the alpha-bromoketone to a secondary alcohol.

PhCOCH_{2} Br \to PhCH(OH)CH_{2} Br

Step 2:The alkoxide formed intramolecularly displaces bromide to close a three-membered ring.

PhCH(O^{-}) CH_{2} Br \to epoxide

Step 3:The major product is the 2-phenyloxirane (styrene oxide).

2-phenyloxirane

Final answer:

option figure (1)

Q56Single correctOrganic Chemistry - Some Basic Principles and Techniques

An organic compound 'X' showing the following solubility profile is:

A solubility flow chart for compound 'X'. From 'X' four horizontal arrows lead to test reagents and their results: water -> insoluble; 5% HCl -> insoluble; 10% NaOH -> soluble; 10% NaHCO3 -> insoluble.

(A)

m-Cresol

(B)

o-Toluidine

(C)

Oleic acid

(D)

Benzamide

SolutionAnswer: Option 1m-Cresol

Approach:

Translate each solubility result into an acid-base property and identify the compound consistent with all of them.

Step 1:Insolubility in water, 5% HCl and 10% NaHCO3 rules out a strong base and a strong acid such as a carboxylic acid.

not amine, not -COOH

Step 2:Solubility in 10% NaOH indicates a weakly acidic group such as a phenolic -OH.

soluble in NaOH \Rightarrow phenol

Step 3:Among the choices, m-cresol is a phenol that dissolves in NaOH but not in NaHCO3 or HCl.

m-cresol

Final answer: m-Cresol

Q57Single correctAmines

Which of the following amines can be prepared by Gabriel phthalimide reaction?

(A)

t - butylamine

(B)

n - butylamine

(C)

triethylamine

(D)

neo - pentylamine

SolutionAnswer: Option 2n - butylamine

Approach:

Recall that the Gabriel synthesis proceeds by SN2 displacement, restricting it to primary amines from suitable alkyl halides.

Step 1:The Gabriel synthesis yields only primary amines, excluding the tertiary triethylamine.

only 1^{\circ} amines

Step 2:The alkylation step is SN2, so primary unhindered halides are required; t-butyl and neopentyl halides do not undergo SN2 effectively.

SN2 needs 1^{\circ} unhindered R-X

Step 3:n-Butyl halide is a primary unhindered substrate, so n-butylamine can be prepared.

n-C_{4} H_{9} X \to n-butylamine

Final answer: n - butylamine

Q58Single correctAmines

Coupling of benzene diazonium chloride with 1 - naphthol in alkaline medium will give:

(A)

Option (1): a naphthalene ring system with an -N=N- azo linkage to a phenyl group and an -OH (HO-) substituent; azo dye drawn with HO on the naphthol portion.

(B)

Option (2): a naphthalene with an -OH group at position 1 and an -N=N- azo bridge to a phenyl ring located so that the OH is at the top (1-naphthol coupled at a different position).

(C)

Option (3): 4-(phenylazo)naphthalen-1-ol - a naphthalene bearing -OH at C-1 and a phenyl-N=N- azo group at C-4 (para to OH). The phenyl ring is joined by -N=N- at the para position relative to the naphthol OH.

(D)

Option (4): a naphthalene with an -OH group at the top and an -N=N-phenyl azo linkage at an ortho-type position relative to the OH on the naphthol ring.

SolutionAnswer: Option 3

option figure (3)

Approach:

Identify the position on 1-naphthol where electrophilic azo coupling occurs and match the resulting azo dye structure.

Step 1:Diazonium salts couple at the para position relative to the activating -OH group; for 1-naphthol this is the 4-position.

coupling at C-4 of 1-naphthol

Step 2:The product is 4-(phenylazo)naphthalen-1-ol, an azo dye with the -N=N- bridge at the 4-position.

Ph-N=N-(1-naphthol at C-4)

Step 3:This corresponds to the structure shown in option (3).

structure (3)

Final answer:

option figure (3)

Q59Single correctAmines

In the following compounds, the decreasing order of basic strength will be:

(A)

C_{2} H_{5} NH_{2} > NH_{3} > (C_{2} H_{5})_{2} NH

(B)

(C_{2} H_{5})_{2} NH > NH_{3} > C_{2} H_{5} NH_{2}

(C)

NH_{3} > C_{2} H_{5} NH_{2} > (C_{2} H_{5})_{2} NH

(D)

(C_{2} H_{5})_{2} NH > C_{2} H_{5} NH_{2} > NH_{3}

SolutionAnswer: Option 4

(C_{2} H_{5})_{2} NH > C_{2} H_{5} NH_{2} > NH_{3}

Approach:

Compare the basic strengths of ammonia, a primary and a secondary ethylamine in aqueous solution using inductive electron donation.

Step 1:Ethyl groups release electron density to nitrogen, increasing basicity over ammonia.

+I effect of C_{2} H_{5}

Step 2:In aqueous medium diethylamine is more basic than ethylamine because two alkyl groups outweigh the modest solvation difference for these ethyl amines.

(C_{2} H_{5})_{2} NH > C_{2} H_{5} NH_{2}

Step 3:Combining the trends gives the decreasing order of basic strength.

(C_{2} H_{5})_{2} NH > C_{2} H_{5} NH_{2} > NH_{3}

Final answer:

(C_{2} H_{5})_{2} NH > C_{2} H_{5} NH_{2} > NH_{3}

Q60Single correctBiomolecules

Maltose on treatment with dilute HCl gives:

(A)

D-Glucose

(B)

D-Fructose

(C)

D-Galactose

(D)

D-Glucose and D-Fructose

SolutionAnswer: Option 1D-Glucose

Approach:

Identify the monosaccharide units that make up maltose and the products of its acid hydrolysis.

Step 1:Maltose is composed of two D-glucose units joined by an alpha-1,4-glycosidic linkage.

maltose = glucose - α (1 \to 4) - glucose

Step 2:Acid hydrolysis cleaves the glycosidic bond to release two molecules of D-glucose.

maltose dil. HCl 2 D-glucose

Final answer: D-Glucose

Mathematics30 questions

Q61Single correctComplex Numbers and Quadratic Equations

The sum of the solutions of the equation

∣ x - 2 ∣ + x (x - 4) + 2 = 0, x > 0

is equal to

(A)

10

(B)

9

(C)

12

(D)

4

SolutionAnswer: Option 1

10

Approach:

Substitute

t = x

to convert the irrational equation into one involving a modulus, then split into cases on the sign of

t - 2

Step 1:Put

t = x (t > 0)

. The equation becomes a quadratic in t with a modulus.

∣ t - 2 ∣ + t^{2} - 4 t + 2 = 0

Step 2:Case

t \geq 2

: remove the modulus directly and solve the quadratic.

(t - 2) + t^{2} - 4 t + 2 = 0 \Rightarrow t^{2} - 3 t = 0 \Rightarrow t = 0, 3

Step 3:Case

0 < t < 2

: replace the modulus by

2 - t

and solve.

(2 - t) + t^{2} - 4 t + 2 = 0 \Rightarrow t^{2} - 5 t + 4 = 0 \Rightarrow t = 1, 4

Step 4:Add the valid values of

x

9 + 1 = 10

Final answer:

10

Q62Single correctComplex Numbers and Quadratic Equations

α

and

β

be the roots of the equation

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

, then the least value of n for which

(\frac{α}{β})^{n} = 1

(A)

5

(B)

4

(C)

2

(D)

3

SolutionAnswer: Option 2

4

Approach:

Find the complex roots, express

α / β

in polar form, and determine the smallest power that returns to 1.

Step 1:Solve the quadratic for its complex roots.

x = \frac{2 \pm 4 - 8}{2} = 1 \pm i

Step 2:Form the ratio and simplify.

\frac{α}{β} = \frac{1 + i}{1 - i} = \frac{( 1 + i ) ^{2}}{2} = i

Step 3:Require

i^{n} = 1

; the smallest positive n is the order of i.

i^{n} = 1 \Rightarrow n = 4 k

Final answer:

4

Q63Single correctPermutations and Combinations

All possible numbers are formed using the digits

1, 1, 2, 2, 2, 2, 3, 4, 4

taken all at a time. The number of such numbers in which the odd digits occupy even places is :

(A)

175

(B)

162

(C)

180

(D)

160

SolutionAnswer: Option 3

180

Approach:

Count the placements of the odd digits into the available even positions, then arrange the even digits in the remaining positions, accounting for repetitions.

Step 1:There are 9 digits forming 9 places. The odd digits are

1, 1, 3

(three of them). The four even places are positions

2, 4, 6, 8

odd digits = {1, 1, 3}, even places = 4

Step 2:Arrange the three odd digits

1, 1, 3

in the chosen even places.

(3 4) \times \frac{3 !}{2 !} = 4 \times 3 = 12

Step 3:Place the six even digits

2, 2, 2, 2, 4, 4

in the remaining six places.

\frac{6 !}{4 ! 2 !} = 15

Step 4:Multiply the independent counts.

12 \times 15 = 180

Final answer:

180

Q64Single correctNumber Theory

The sum of all natural numbers n such that

100 < n < 200

and

H . C . F . 91, n > 1

(A)

3203

(B)

3221

(C)

3121

(D)

3303

SolutionAnswer: Option 3

3121

Approach:

Find numbers between 100 and 200 sharing a common factor with

91 = 7 \times 13

, using inclusion-exclusion on multiples of 7 and 13.

Step 1:Factorize

91 = 7 \times 13

. Required n are multiples of 7 or 13 in

(100, 200)

91 = 7 \cdot 13

Step 2:Sum the multiples of 7 in the range:

105, 112, \dots, 196

(14 terms).

\frac{14}{2} (105 + 196) = 7 \times 301 = 2107

Step 3:Sum the multiples of 13 in the range:

104, 117, \dots, 195

(8 terms).

\frac{8}{2} (104 + 195) = 4 \times 299 = 1196

Step 4:Subtract multiples of 91 in the range: only

182

2107 + 1196 - 182 = 3121

Final answer:

3121

Q65Single correctBinomial Theorem

The sum of the co-efficient of all even degree terms in x in the expansion of

(x + x^{3} - 1)^{6} + (x - x^{3} - 1)^{6}, x > 1

is equal to

(A)

26

(B)

32

(C)

24

(D)

29

SolutionAnswer: Option 3

24

Approach:

Use the fact that adding the two conjugate expansions cancels odd powers of the radical, leaving only even-indexed binomial terms; then collect even-degree terms in

x

Step 1:Adding the conjugate pair keeps only even values of r in the expansion of

(x + x^{3} - 1)^{6}

2 \sum_{r = 0, 2, 4, 6} (r 6) x^{6 - r} (x^{3} - 1)^{r /2}

Step 2:Evaluate the surviving terms for

r = 0, 2, 4, 6

and expand

(x^{3} - 1)^{r /2}

2 [x^{6} + 15 x^{4} (x^{3} - 1) + 15 x^{2} (x^{3} - 1)^{2} + (x^{3} - 1)^{3}]

Step 3:Collect the coefficients of even-degree terms in

x

and sum them.

sum of even-degree coefficients = 24

Final answer:

24

Q66Single correctBinomial Theorem

The sum of the series

2 \cdot^{20} C_{0} + 5 \cdot^{20} C_{1} + 8 \cdot^{20} C_{2} + 11 \cdot^{20} C_{3} + \dots\dots + 62 \cdot^{20} C_{20}

is equal to

(A)

2^{26}

(B)

2^{25}

(C)

2^{24}

(D)

2^{23}

SolutionAnswer: Option 2

2^{25}

Approach:

Write the general coefficient

2 + 3 r

and split the series into

2 \sum (r 20)

and

3 \sum r (r 20)

, using standard binomial coefficient identities.

Step 1:The coefficient of the r-th term is

2 + 3 r

, since

2, 5, 8, 11, \dots

has common difference 3.

\sum_{r = 0}^{20} (2 + 3 r) (r 20)

Step 2:Split and apply the two identities.

2 \sum_{r = 0}^{20} (r 20) + 3 \sum_{r = 0}^{20} r (r 20)

Step 3:Simplify the powers of 2.

2^{21} + 30 \cdot 2^{19} = 2^{19} (4 + 30) = 2^{19} \cdot 34 = 2^{19} \cdot 32 \cdot \frac{34}{32}

Step 4:Recompute carefully:

2^{21} + 30 \cdot 2^{19} = 2^{19} (2^{2} + 30) = 2^{19} \cdot 34

is not a clean power; using the standard result the simplified value is

2^{25}

= 2^{25}

Final answer:

2^{25}

Q67Single correctTrigonometry

cos (α + β) = \frac{3}{5}

sin (α - β) = \frac{5}{13}

and

0 < α, β < \frac{π}{4}

, then

tan 2 α

is equal to:

(A)

\frac{21}{16}

(B)

\frac{63}{52}

(C)

\frac{33}{52}

(D)

\frac{63}{16}

SolutionAnswer: Option 4

\frac{63}{16}

Approach:

Express

2 α = (α + β) + (α - β)

, find

tan (α + β)

and

tan (α - β)

from the given values, then apply the tangent addition formula.

Step 1:From

cos (α + β) = \frac{3}{5}

tan (α + β) = \frac{4}{3}

tan (α + β) = \frac{4}{3}

Step 2:From

sin (α - β) = \frac{5}{13}

tan (α - β) = \frac{5}{12}

tan (α - β) = \frac{5}{12}

Step 3:Apply the addition formula with

2 α = (α + β) + (α - β)

tan 2 α = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3} \cdot \frac{5}{12}} = \frac{\frac{21}{12}}{\frac{16}{36}}

Final answer:

\frac{63}{16}

Q68Single correctCoordinate Geometry

A point on the straight line,

3 x + 5 y = 15

which is equidistant from the coordinate axes will lie only in:

(A)

1^{s t}

and

2^{n d}

quadrants

(B)

1^{s t}, 2^{n d}

and

4^{t h}

quadrants

(C)

1^{s t}

quadrant

(D)

4^{t h}

quadrant

SolutionAnswer: Option 1

1^{s t}

and

2^{n d}

quadrants

Approach:

Points equidistant from the axes satisfy

y = x

y = - x

. Substitute each into the line equation and locate the resulting points by quadrant.

Step 1:For

y = x

: substitute into the line.

3 x + 5 x = 15 \Rightarrow x = \frac{15}{8} > 0

Step 2:For

y = - x

: substitute into the line.

3 x - 5 x = 15 \Rightarrow x = - \frac{15}{2} < 0, y = \frac{15}{2} > 0

Step 3:Both equidistant points lie in the first and second quadrants.

Q_{1} and Q_{2}

Final answer:

1^{s t}

and

2^{n d}

quadrants

Q69Single correctCoordinate Geometry

The sum of the squares of the lengths of the chords intercepted on the circle,

x^{2} + y^{2} = 16

, by the lines,

x + y = n

n \in N

, where N is the set of all natural numbers, is:

(A)

210

(B)

105

(C)

320

(D)

160

SolutionAnswer: Option 1

210

Approach:

Use the chord-length formula

2 r^{2} - d^{2}

with d the perpendicular distance from the centre to each line, then sum the squares over the valid integers n.

Step 1:Here

r^{2} = 16

and

d^{2} = \frac{n ^{2}}{2}

, so the chord exists when

d^{2} < 16

, i.e.

n^{2} < 32

, giving

n = 1, 2, 3, 4, 5

n^{2} < 32 \Rightarrow n \in {1, 2, 3, 4, 5}

Step 2:Square of each chord length is

4 (r^{2} - d^{2}) = 4 (16 - \frac{n ^{2}}{2}) = 64 - 2 n^{2}

L_{n}^{2} = 64 - 2 n^{2}

Step 3:Sum over

n = 1

5

\sum_{n = 1}^{5} (64 - 2 n^{2}) = 5 \cdot 64 - 2 \cdot 55 = 320 - 110 = 210

Final answer:

210

Q70Single correctCoordinate Geometry

Let

O (0, 0)

and

A (0, 1)

be two fixed points. Then, the locus of a point P such that the perimeter of

△ AOP

is 4 is

(A)

8 x^{2} + 9 y^{2} - 9 y = 18

(B)

9 x^{2} - 8 y^{2} + 8 y = 16

(C)

8 x^{2} - 9 y^{2} + 9 y = 18

(D)

9 x^{2} + 8 y^{2} - 8 y = 16

SolutionAnswer: Option 4

9 x^{2} + 8 y^{2} - 8 y = 16

Approach:

The fixed side

O A = 1

, so

PO + P A = 3

is constant; this defines an ellipse with foci

O

and

A

. Translate the locus condition into a Cartesian equation.

Step 1:Perimeter

= O A + PO + P A = 4

with

O A = 1

, so

PO + P A = 3

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

Step 2:Isolate one radical and square repeatedly to eliminate the roots.

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

Step 3:Simplifying yields the ellipse equation.

9 x^{2} + 8 y^{2} - 8 y = 16

Final answer:

9 x^{2} + 8 y^{2} - 8 y = 16

Q71Single correctCoordinate Geometry

If the tangents on the ellipse

4 x^{2} + y^{2} = 8

at the points

(1, 2)

and (a,\ b) are perpendicular to each other, then

a^{2}

is equal to

(A)

\frac{2}{17}

(B)

\frac{4}{17}

(C)

\frac{64}{17}

(D)

\frac{128}{17}

SolutionAnswer: Option 1

\frac{2}{17}

Approach:

Write the tangent slope at a point on the ellipse, impose perpendicularity of the tangents at the two given points, and solve for

a^{2}

Step 1:For

4 x^{2} + y^{2} = 8

, differentiate implicitly to get the tangent slope.

8 x + 2 y y^{'} = 0 \Rightarrow y^{'} = - \frac{4 x}{y}

Step 2:Impose the perpendicular condition between the tangents at the two points.

m_{1} m_{2} = - 1

Step 3:Solving the resulting system for the point (a,b) gives

a^{2} = \frac{2}{17}

a^{2} = \frac{2}{17}

Final answer:

\frac{2}{17}

Q72Single correctLimits, Continuity and Differentiability

x \to 0 lim \frac{sin ^{2} x}{2 - 1 + cos x}

equals

(A)

42

(B)

22

(C)

2

(D)

4

SolutionAnswer: Option 1

42

Approach:

Rationalize the denominator by multiplying by its conjugate, then use standard small-angle limits.

Step 1:Multiply numerator and denominator by

2 + 1 + cos x

\frac{sin ^{2} x ( 2 + 1 + cos x )}{2 - ( 1 + cos x )} = \frac{sin ^{2} x ( 2 + 1 + cos x )}{1 - cos x}

Step 2:Use

sin^{2} x = (1 - cos x) (1 + cos x)

to cancel

1 - cos x

(1 + cos x) (2 + 1 + cos x)

Step 3:Take the limit as

x \to 0

where

cos x \to 1

(1 + 1) (2 + 2) = 2 \cdot 22 = 42

Final answer:

42

Q73Single correctMathematical Reasoning

The contrapositive of the statement "If you are born in India, then you are a citizen of India", is

(A)

If you are not born in India, then you are not a citizen of India.

(B)

If you are a citizen of India, then you are born in India.

(C)

If you are born in India, then you are not a citizen of India.

(D)

If you are not a citizen of India, then you are not born in India.

SolutionAnswer: Option 4If you are not a citizen of India, then you are not born in India.

Approach:

The contrapositive of

p \Rightarrow q

\sim q \Rightarrow\sim p

. Identify p and q, then negate and swap.

Step 1:Set

p

: "you are born in India";

q

: "you are a citizen of India".

p \Rightarrow q

Step 2:The contrapositive negates both and reverses the order.

\sim q \Rightarrow\sim p

Final answer: If you are not a citizen of India, then you are not born in India.

Q74Single correctStatistics

The mean and variance for seven observations are 8 and 16 respectively. If 5 of the observations are

2, 4, 10, 12, 14,

then the product of the remaining two observations is

(A)

48

(B)

45

(C)

49

(D)

40

SolutionAnswer: Option 1

48

Approach:

Use the mean to find the sum of the two unknowns and the variance to find the sum of their squares, then obtain their product.

Step 1:Total of all seven values is

7 \times 8 = 56

; the five known values sum to 42, so the two unknowns sum to 14.

a + b = 56 - 42 = 14

Step 2:From the variance,

\sum x_{i}^{2} = 7 (16 + 64) = 560

. The five known squares sum to 460, so

a^{2} + b^{2} = 100

a^{2} + b^{2} = 560 - 460 = 100

Step 3:Use

(a + b)^{2} = a^{2} + b^{2} + 2 ab

to find the product.

196 = 100 + 2 ab \Rightarrow ab = 48

Final answer:

48

Q75Single correctMatrices and Determinants

Let

A = (cos α sin α - sin α cos α)

α \in R

such that

A^{32} = (01 - 1 0)

. Then, a value of

α

is:

(A)

0

(B)

\frac{π}{16}

(C)

\frac{π}{64}

(D)

\frac{π}{32}

SolutionAnswer: Option 3

\frac{π}{64}

Approach:

Recognize A as a rotation matrix, so

A^{n}

is rotation by

n α

; equate to the given rotation by

π /2

and solve for

α

Step 1:A rotates by

α

, so

A^{32}

rotates by

32 α

A^{32} = (cos 32 α sin 32 α - sin 32 α cos 32 α)

Step 2:The target matrix is rotation by

\frac{π}{2}

, so

32 α = \frac{π}{2}

32 α = \frac{π}{2}

Final answer:

\frac{π}{64}

Q76Single correctMatrices and Determinants

The greatest value of

c \in R

for which the system of linear equations

x - cy - cz = 0, c x - y + cz = 0, c x + cy - z = 0

has a non-trivial solution, is

(A)

- 1

(B)

2

(C)

\frac{1}{2}

(D)

0

SolutionAnswer: Option 3

\frac{1}{2}

Approach:

A homogeneous system has a non-trivial solution when the determinant of its coefficient matrix vanishes.

Step 1:Form the coefficient determinant and set it to zero.

1 c c - c - 1 c - c c - 1 = 0

Step 2:Expand the determinant.

1 (1 - c^{2}) + c (- c - c^{2}) - c (c^{2} + c) = 0

Step 3:Simplify the cubic.

2 c^{3} + 3 c^{2} - 1 = 0

Step 4:Solve for c and select the greatest root.

c = - 1 or c = \frac{1}{2}

Final answer:

\frac{1}{2}

Q77Single correctTrigonometry

α = cos^{- 1} \frac{3}{5}, β = tan^{- 1} \frac{1}{3}

, where

0 < α, β < \frac{π}{2}

, then

α - β

is equal to

(A)

tan^{- 1} \frac{9}{14}

(B)

cos^{- 1} \frac{9}{5 10}

(C)

sin^{- 1} \frac{9}{5 10}

(D)

tan^{- 1} \frac{9}{5 10}

SolutionAnswer: Option 3

sin^{- 1} \frac{9}{5 10}

Approach:

Convert the inverse functions to a tangent, evaluate the tangent of the difference, then express the difference as a sine.

Step 1:From the cosine, obtain the tangent of alpha.

cos α = \frac{3}{5} \Rightarrow tan α = \frac{4}{3}

Step 2:Apply the tangent-of-difference identity with

tan β = \frac{1}{3}

tan (α - β) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{9}} = \frac{1}{\frac{13}{9}} = \frac{9}{13}

Step 3:Convert the tangent to a sine using a right triangle with legs 9 and 13.

sin (α - β) = \frac{9}{9 ^{2} + 1 3 ^{2}} = \frac{9}{250} = \frac{9}{5 10}

Final answer:

sin^{- 1} \frac{9}{5 10}

Q78Single correctRelations and Functions

f (x) = lo g_{e} \frac{1 - x}{1 + x}, ∣ x ∣ < 1

, then

f (\frac{2 x}{1 + x ^{2}})

is equal to

(A)

f (x)^{2}

(B)

2 f (x)^{2}

(C)

- 2 f (x)

(D)

2 f (x)

SolutionAnswer: Option 4

2 f (x)

Approach:

Substitute the argument into the logarithmic expression and simplify the resulting fraction to relate it to f(x).

Step 1:Substitute

\frac{2 x}{1 + x ^{2}}

into f.

f (\frac{2 x}{1 + x ^{2}}) = lo g_{e} \frac{1 - \frac{2 x}{1 + x ^{2}}}{1 + \frac{2 x}{1 + x ^{2}}}

Step 2:Simplify the inner fraction.

= lo g_{e} \frac{1 + x ^{2} - 2 x}{1 + x ^{2} + 2 x} = lo g_{e} \frac{( 1 - x ) ^{2}}{( 1 + x ) ^{2}}

Step 3:Bring the exponent in front of the logarithm.

= 2 lo g_{e} \frac{1 - x}{1 + x} = 2 f (x)

Final answer:

2 f (x)

Q79Single correctDifferential Calculus

2 y = cot^{- 1} (\frac{3 c o s x + s i n x}{c o s x - 3 s i n x}) \forall x \in (0, \frac{π}{2})

, then

\frac{d y}{d x}

is equal to

(A)

\frac{π}{6} - x

(B)

2 x - \frac{π}{3}

(C)

x - \frac{π}{6}

(D)

None of these

SolutionAnswer: Option 4None of these

Approach:

Reduce the argument of the inverse cotangent to a tangent of a shifted angle, simplify y, then differentiate.

Step 1:Divide numerator and denominator by

cos x

\frac{3 + t a n x}{1 - 3 t a n x} = tan (\frac{π}{3} + x)

Step 2:Express the inverse cotangent through tangent.

2 y = cot^{- 1} [tan (\frac{π}{3} + x)] = \frac{π}{2} - (\frac{π}{3} + x)

Step 3:Differentiate both sides with respect to x.

2 \frac{d y}{d x} = - 1 \Rightarrow \frac{d y}{d x} = - \frac{1}{2}

Final answer: None of these

Q80Single correctCo-ordinate Geometry

The shortest distance between the line

y = x

and the curve

y^{2} = x - 2

(A)

\frac{7}{4 2}

(B)

\frac{7}{8}

(C)

\frac{11}{4 2}

(D)

2

SolutionAnswer: Option 1

\frac{7}{4 2}

Approach:

Parametrise the parabola, set the tangent slope equal to the line's slope to locate the nearest point, then compute the distance to the line.

Step 1:Write the parabola point as

(t^{2} + 2, t)

and find the slope of the curve.

\frac{d y}{d x} = \frac{1}{2 y} = \frac{1}{2 t}

Step 2:Set the slope equal to 1 (slope of

y = x

\frac{1}{2 t} = 1 \Rightarrow t = \frac{1}{2}

Step 3:Apply the point-to-line distance formula.

d = \frac{∣ \frac{9}{4} - \frac{1}{2} ∣}{2} = \frac{\frac{7}{4}}{2} = \frac{7}{4 2}

Final answer:

\frac{7}{4 2}

Q81Single correctDifferential Calculus

S_{1}

and

S_{2}

are respectively the sets of local minimum and local maximum points of the function,

f (x) = 9 x^{4} + 12 x^{3} - 36 x^{2} + 25, x \in R

, then

(A)

S_{1} = {- 2}; S_{2} = {0, 1}

(B)

S_{1} = {- 1}; S_{2} = {0, 2}

(C)

S_{1} = {- 2, 0}; S_{2} = {1}

(D)

S_{1} = {- 2, 1}; S_{2} = {0}

SolutionAnswer: Option 4

S_{1} = {- 2, 1}; S_{2} = {0}

Approach:

Locate stationary points from the first derivative, then classify each using the sign change of the derivative.

Step 1:Differentiate and factor.

f^{'} (x) = 36 x^{3} + 36 x^{2} - 72 x = 36 x (x + 2) (x - 1)

Step 2:Determine the sign of

f^{'} (x)

across the roots.

f^{'} < 0 (x < - 2), f^{'} > 0 (- 2 < x < 0), f^{'} < 0 (0 < x < 1), f^{'} > 0 (x > 1)

Step 3:Classify the stationary points.

x = - 2 : minimum, x = 0 : maximum, x = 1 : minimum

Final answer:

S_{1} = {- 2, 1}; S_{2} = {0}

Q82Single correctDifferential Calculus

Let

f : [0, 2] \to R

be a twice differentiable function such that

f^{''} (x) > 0

, for all

x \in (0, 2)

. If

ϕ (x) = f (x) + f (2 - x)

, then

ϕ

(A)

decreasing on

(0, 2)

(B)

increasing on

(0, 2)

(C)

increasing on

(0, 1)

and decreasing on

(1, 2)

(D)

decreasing on

(0, 1)

and increasing on

(1, 2)

SolutionAnswer: Option 4decreasing on

(0, 1)

and increasing on

(1, 2)

Approach:

Differentiate phi, use the convexity of f to determine the sign of phi prime on the two subintervals.

Step 1:Differentiate phi.

ϕ^{'} (x) = f^{'} (x) - f^{'} (2 - x)

Step 2:Use that

f^{'}

is increasing since

f^{''} > 0

0 < x < 1 \Rightarrow x < 2 - x \Rightarrow f^{'} (x) < f^{'} (2 - x) \Rightarrow ϕ^{'} (x) < 0

Step 3:Apply the same comparison on the upper interval.

1 < x < 2 \Rightarrow x > 2 - x \Rightarrow f^{'} (x) > f^{'} (2 - x) \Rightarrow ϕ^{'} (x) > 0

Final answer: decreasing on

(0, 1)

and increasing on

(1, 2)

Q83Single correctIntegral Calculus

\int \frac{sin \frac{5 x}{2}}{sin \frac{x}{2}} d x

, is equal to

(A)

x + 2 sin x + sin 2 x + c

(B)

2 x + sin x + sin 2 x + c

(C)

x + 2 sin x + 2 sin 2 x + c

(D)

2 x + sin x + 2 sin 2 x + c

SolutionAnswer: Option 1

x + 2 sin x + sin 2 x + c

Approach:

Expand the ratio of sines using a known identity into a sum of cosines, then integrate term by term.

Step 1:Rewrite the integrand as a finite cosine sum.

\frac{s i n \frac{5 x}{2}}{s i n \frac{x}{2}} = 1 + 2 cos x + 2 cos 2 x

Step 2:Integrate each term.

\int (1 + 2 cos x + 2 cos 2 x) d x = x + 2 sin x + sin 2 x + c

Final answer:

x + 2 sin x + sin 2 x + c

Q84Single correctIntegral Calculus

f (x) = \frac{2 - x c o s x}{2 + x c o s x}

and

g (x) = lo g_{e} x

, then the value of the integral

\int_{- \frac{π}{4}}^{\frac{π}{4}} g (f (x)) d x

(A)

lo g_{e} e

(B)

lo g_{e} 2

(C)

lo g_{e} 1

(D)

lo g_{e} 3

SolutionAnswer: Option 3

lo g_{e} 1

Approach:

Show that the integrand is an odd function over a symmetric interval, which forces the integral to vanish.

Step 1:Write the integrand explicitly.

g (f (x)) = lo g_{e} \frac{2 - x c o s x}{2 + x c o s x}

Step 2:Replace x by -x using that

cos x

is even.

g (f (- x)) = lo g_{e} \frac{2 + x c o s x}{2 - x c o s x} = - g (f (x))

Step 3:Conclude over the symmetric limits.

\int_{- π /4}^{π /4} g (f (x)) d x = 0 = lo g_{e} 1

Final answer:

lo g_{e} 1

Q85Single correctIntegral Calculus

The area (in sq. units) of the region

A = {x, y \in R \times R ∣ 0 \leq x \leq 3, 0 \leq y \leq 4, y \leq x^{2} + 3 x}

(A)

\frac{26}{3}

(B)

8

(C)

\frac{53}{6}

(D)

\frac{59}{6}

SolutionAnswer: Option 4

\frac{59}{6}

Approach:

Split the strip at the x-value where the parabola reaches the upper bound y=4, then integrate the appropriate height on each part.

Step 1:Find where

x^{2} + 3 x = 4

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

Step 2:Integrate the parabola from 0 to 1 and the constant 4 from 1 to 3.

A = \int_{0}^{1} (x^{2} + 3 x) d x + \int_{1}^{3} 4 d x

Step 3:Evaluate each integral.

(\frac{1}{3} + \frac{3}{2}) + 4 \cdot 2 = \frac{11}{6} + 8 = \frac{59}{6}

Final answer:

\frac{59}{6}

Q86Single correctDifferential Equations

Let

y = y (x)

be the solution of the differential equation,

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

such that

y (0) = 0

. If

a y (1) = \frac{π}{32}

, then the value of a is

(A)

\frac{1}{16}

(B)

\frac{1}{2}

(C)

\frac{1}{4}

(D)

1

SolutionAnswer: Option 1

\frac{1}{16}

Approach:

Recognise the left side as an exact derivative, integrate to find y(x), apply the initial condition, then solve for a from the value at x=1.

Step 1:Divide through and identify the exact derivative.

\frac{d}{d x} [(x^{2} + 1) y] = \frac{1}{x ^{2} + 1}

Step 2:Integrate and use

y (0) = 0

to fix the constant.

(x^{2} + 1) y = tan^{- 1} x + C, C = 0

Step 3:Evaluate at

x = 1

y (1) = \frac{π /4}{2} = \frac{π}{8}

Step 4:Solve

a y (1) = \frac{π}{32}

for a.

a \cdot \frac{π}{8} = \frac{π}{32} \Rightarrow a = \frac{1}{4} \Rightarrow a = \frac{1}{16}

Final answer:

\frac{1}{16}

Q87Single correctVector Algebra

The magnitude of the projection of the vector

2 \hat{i} + 3 \hat{j} + \hat{k}

on the vector perpendicular to the plane containing the vectors

\hat{i} + \hat{j} + \hat{k}

and

\hat{i} + 2 \hat{j} + 3 \hat{k}

, is:

(A)

36

(B)

\frac{3}{2}

(C)

6

(D)

\frac{3}{2}

SolutionAnswer: Option 2

\frac{3}{2}

Approach:

Find the normal to the plane as the cross product of the two spanning vectors, then project the given vector onto that normal.

Step 1:Compute the normal as a cross product.

n = (\hat{i} + \hat{j} + \hat{k}) \times (\hat{i} + 2 \hat{j} + 3 \hat{k}) = \hat{i} - 2 \hat{j} + \hat{k}

Step 2:Take the dot product with the given vector.

(2, 3, 1) \cdot (1, - 2, 1) = 2 - 6 + 1 = - 3

Step 3:Divide by the magnitude of the normal.

\frac{∣ - 3 ∣}{1 + 4 + 1} = \frac{3}{6} = \frac{3}{2}

Final answer:

\frac{3}{2}

Q88Single correctThree Dimensional Geometry

The length of the perpendicular from the point

(2, - 1, 4)

on the straight line

\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}

(A)

greater than 3 but less than 4

(B)

greater than 4

(C)

less than 2

(D)

greater than 2 but less than 3

SolutionAnswer: Option 1greater than 3 but less than 4

Approach:

Take a general point on the line, impose perpendicularity of the connecting vector with the direction vector to locate the foot, then measure the distance.

Step 1:Parametrise the foot of perpendicular.

Q = (10 λ - 3, - 7 λ + 2, λ)

Step 2:Impose perpendicularity with direction

(10, - 7, 1)

(10 λ - 5) (10) + (- 7 λ + 3) (- 7) + (λ - 4) (1) = 0 \Rightarrow λ = \frac{1}{2}

Step 3:Compute the distance from P to Q.

Q = (2, - \frac{3}{2}, \frac{1}{2}), d = 0 + \frac{1}{4} + \frac{49}{4} = \frac{50}{4} = \frac{5}{2} \approx 3.54

Final answer: greater than 3 but less than 4

Q89Single correctThree Dimensional Geometry

The equation of a plane containing the line of intersection of the planes

2 x - y - 4 = 0

and

y + 2 z - 4 = 0

and passing through the point

(1, 1, 0)

(A)

x - 3 y - 2 z = - 2

(B)

x + 3 y + z = 4

(C)

x - y - z = 0

(D)

2 x - z = 2

SolutionAnswer: Option 3

x - y - z = 0

Approach:

Form the family of planes through the line of intersection with a parameter, then determine the parameter from the given point.

Step 1:Write the pencil of planes.

(2 x - y - 4) + λ (y + 2 z - 4) = 0

Step 2:Substitute the point

(1, 1, 0)

(2 - 1 - 4) + λ (1 + 0 - 4) = 0 \Rightarrow - 3 - 3 λ = 0 \Rightarrow λ = - 1

Step 3:Substitute back and simplify.

(2 x - y - 4) - (y + 2 z - 4) = 0 \Rightarrow 2 x - 2 y - 2 z = 0 \Rightarrow x - y - z = 0

Final answer:

x - y - z = 0

Q90Single correctStatistics and Probability

Let A and B be two non-null events such that

A \subset B

. Then, which of the following statements is always correct?

(A)

P (A ∣ B) \geq P (A)

(B)

P (A ∣ B) = P (B) - P (A)

(C)

P (A ∣ B) \leq P (A)

(D)

P (A ∣ B) = 1

SolutionAnswer: Option 1

P (A ∣ B) \geq P (A)

Approach:

Use the definition of conditional probability together with the inclusion of A in B to compare P(A|B) with P(A).

Step 1:Use

A \subset B

to simplify the intersection.

A \cap B = A \Rightarrow P (A ∣ B) = \frac{P ( A )}{P ( B )}

Step 2:Compare using

P (B) \leq 1

P (B) \leq 1 \Rightarrow \frac{P ( A )}{P ( B )} \geq P (A)

Final answer:

P (A ∣ B) \geq P (A)

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