Home/JEE Main PYQs/2024 · April 09, Shift 1

JEE Main 2024 April 09, Shift 1 Question Paper with Solutions

All 89 questions from the JEE Main 2024 (April 09, Shift 1) 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

The dimensional formula of latent heat is :

(A)

[ML^{2} T^{- 2}]

(B)

[M^{0} L^{2} T^{- 2}]

(C)

[MLT^{- 2}]

(D)

[M^{0} LT^{- 2}]

SolutionAnswer: Option 2

[M^{0} L^{2} T^{- 2}]

Approach:

Latent heat is defined as heat energy per unit mass. Divide the dimensions of energy by mass.

Step 1:State the dimension of heat energy.

[Q] = [ML^{2} T^{- 2}]

Step 2:Divide by mass to obtain the dimension of latent heat.

[L] = \frac{[ ML ^{2} T ^{- 2} ]}{[ M ]}

Final answer:

[M^{0} L^{2} T^{- 2}]

Q2Single correctKinematics

A particle moving in a straight line covers half the distance with speed 6 m/s. The other half is covered in two equal time intervals with speeds 9 m/s and 15 m/s respectively. The average speed of the particle during the motion is :

(A)

10 m/s

(B)

8 m/s

(C)

9.2 m/s

(D)

8.8 m/s

SolutionAnswer: Option 28 m/s

Approach:

Compute the time for the first half from its speed, and the average speed of the second half (covered in equal time intervals) as the arithmetic mean of the two speeds; then divide total distance by total time.

Step 1:Let total distance be

2 d

. Time for first half at 6 m/s.

t_{1} = \frac{d}{6}

Step 2:Second half is covered in two equal time intervals at 9 and 15 m/s, so its average speed is the mean.

v_{2} = \frac{9 + 15}{2} = 12

Step 3:Time for second half.

t_{2} = \frac{d}{12}

Step 4:Substitute into the average-speed expression.

\overset{v}{ˉ} = \frac{2 d}{\frac{d}{6} + \frac{d}{12}} = \frac{2}{\frac{1}{6} + \frac{1}{12}}

Final answer: 8 m/s

Q3Single correctLaws of Motion

A light unstretchable string passing over a smooth light pulley connects two blocks of masses

m_{1}

and

m_{2}

. If the acceleration of the system is

\frac{g}{8}

, then the ratio of the masses

\frac{m _{2}}{m _{1}}

is :

(A)

8 : 1

(B)

5 : 3

(C)

4 : 3

(D)

9 : 7

SolutionAnswer: Option 49 : 7

Approach:

Apply the standard Atwood-machine acceleration formula and solve for the mass ratio.

Step 1:Equate the acceleration to the given value.

\frac{m _{2} - m _{1}}{m _{2} + m _{1}} = \frac{1}{8}

Step 2:Cross-multiply.

8 (m_{2} - m_{1}) = m_{2} + m_{1}

Step 3:Solve for the ratio.

\frac{m _{2}}{m _{1}} = \frac{9}{7}

Final answer: 9 : 7

Q4Single correctWork, Energy and Power

A particle of mass m moves on a straight line with its velocity increasing with distance according to the equation

v = α x

, where

α

is a constant. The total work done by all the forces applied on the particle during its displacement from

x = 0

x = d

, will be :

(A)

\frac{m}{2 α ^{2} d}

(B)

\frac{md}{2 α ^{2}}

(C)

2 m α^{2} d

(D)

\frac{m α ^{2} d}{2}

SolutionAnswer: Option 4

\frac{m α ^{2} d}{2}

Approach:

By the work-energy theorem, the total work equals the change in kinetic energy between the two positions.

Step 1:Square the velocity relation.

v^{2} = α^{2} x

Step 2:Kinetic energy at

x = d

and at

x = 0

K E_{d} = \frac{1}{2} m α^{2} d, K E_{0} = 0

Step 3:Total work equals the change in kinetic energy.

W = \frac{1}{2} m α^{2} d - 0

Final answer:

\frac{m α ^{2} d}{2}

Q5Single correctRotational Motion

A heavy iron bar, of weight W is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle

θ

with the horizontal. The weight experienced by the person is :

(A)

W cos θ

(B)

\frac{W}{2}

(C)

W

(D)

W sin θ

SolutionAnswer: Option 2

\frac{W}{2}

Approach:

Take torque about the ground end of the bar. The weight acts at the centre of the uniform bar, and the person supplies a vertical force at the far end.

Step 1:Let the bar have length

L

. Torque of the weight about the ground end (weight at midpoint).

τ_{W} = W \cdot \frac{L}{2} cos θ

Step 2:Torque of the person's vertical force

F

at the far end about the ground end.

τ_{F} = F \cdot L cos θ

Step 3:Equate the torques.

F L cos θ = \frac{W L cos θ}{2}

Final answer:

\frac{W}{2}

Q6Single correctGravitation

An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is

x \frac{GM _{e} m}{21 R _{e}}

. The value of x is (take

R_{e} = 6370

km) :

(A)

(B)

(C)

(D)

SolutionAnswer: Option 411

Approach:

Compute the total mechanical energy of the ball on the surface (at rest) and in the circular orbit, then take the difference and express it in the given form.

Step 1:Orbital radius from earth's centre.

r = R_{e} + h = 6370 + 318.5 = 6688.5 km

Step 2:Change in total mechanical energy.

Δ E = - \frac{G M _{e} m}{2 r} + \frac{G M _{e} m}{R _{e}} = G M_{e} m (\frac{1}{R _{e}} - \frac{1}{2 r})

Step 3:Equate to the given form and solve for

x

x = 21 R_{e} (\frac{1}{R _{e}} - \frac{1}{2 r})

Final answer: 11

Q7Single correctMechanical Properties of Fluids

A sphere of relative density

σ

and diameter D has concentric cavity of diameter d. The ratio of

\frac{d}{D}

, if it just floats on water in a tank is :

(A)

(\frac{σ - 2}{σ + 2})^{1/3}

(B)

(\frac{σ}{σ - 1})^{1/3}

(C)

(\frac{σ - 1}{σ})^{1/3}

(D)

(\frac{σ + 1}{σ - 1})^{1/3}

SolutionAnswer: Option 2

(\frac{σ - 1}{σ})^{1/3}

Approach:

For just floating, the weight of the hollow sphere equals the buoyant force from the fully submerged outer volume of water.

Step 1:Mass of shell equals mass of displaced water (outer volume).

σ (D^{3} - d^{3}) = D^{3}

Step 2:Rearrange for

d^{3} / D^{3}

\frac{d ^{3}}{D ^{3}} = 1 - \frac{1}{σ} = \frac{σ - 1}{σ}

Step 3:Take the cube root.

\frac{d}{D} = (\frac{σ - 1}{σ})^{1/3}

Final answer:

(\frac{σ - 1}{σ})^{1/3}

Q8Single correctThermodynamics

A sample of 1 mole gas at temperature

T

is adiabatically expanded to double its volume. If adiabatic constant for the gas is

γ = \frac{3}{2}

, then the work done by the gas in the process is :

(A)

\frac{R}{T} [2 - 2]

(B)

\frac{T}{R} [2 + 2]

(C)

RT [2 - 2]

(D)

RT [2 + 2]

SolutionAnswer: Option 3

RT [2 - 2]

Approach:

Find the final temperature using the adiabatic relation, then apply the adiabatic work formula for one mole.

Step 1:Final temperature on doubling the volume with

γ - 1 = 1/2

T_{2} = T (\frac{1}{2})^{1/2} = \frac{T}{2}

Step 2:Substitute into the work formula.

W = \frac{R ( T - \frac{T}{2} )}{1/2} = 2 RT (1 - \frac{1}{2})

Step 3:Simplify

2 (1 - \frac{1}{2}) = 2 - 2

W = RT (2 - 2)

Final answer:

RT [2 - 2]

Q9Single correctThermodynamics

The volume of an ideal gas

(γ = 1.5)

is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:

(A)

\frac{16}{25}

(B)

\frac{4}{5}

(C)

\frac{8}{5 5}

(D)

\frac{2}{5}

SolutionAnswer: Option 3

\frac{8}{5 5}

Approach:

Use the adiabatic pressure-volume relation to express the ratio of initial to final pressure.

Step 1:Write the pressure ratio from

P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}

\frac{P _{1}}{P _{2}} = (\frac{V _{2}}{V _{1}})^{γ}

Step 2:Substitute

V_{2} = 4

V_{1} = 5

γ = 1.5

\frac{P _{1}}{P _{2}} = (\frac{4}{5})^{3/2}

Step 3:Evaluate the power.

(\frac{4}{5})^{3/2} = \frac{4 ^{3/2}}{5 ^{3/2}} = \frac{8}{5 5}

Final answer:

\frac{8}{5 5}

Q10Single correctAlternating Current

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:

(A)

increases

(B)

decreases

(C)

remains same

(D)

becomes zero

SolutionAnswer: Option 1increases

Approach:

Determine the effect of inserting a dielectric on capacitance, hence on capacitive reactance, circuit current, and bulb brightness.

Step 1:Inserting a dielectric increases capacitance.

C = K C_{0}, K > 1

Step 2:Higher capacitance lowers capacitive reactance.

X_{C} = \frac{1}{ω C} ↓

Step 3:Lower impedance raises the current, increasing the power dissipated in the bulb.

I = \frac{V}{R ^{2} + X _{C}^{2}} ↑

Final answer: increases

Q11Single correctElectrostatic Potential and Capacitance

A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. If the area of each stair is

\frac{A}{3}

and the height is d, the capacitance of the arrangement is :

A parallel-plate capacitor where the upper plate is flat (area A) and the lower plate forms a descending stair-step structure. Three stairs are drawn, each labelled with area A/3. Vertical gaps of height d separate successive stair levels from the flat plate, with the steps progressively lower, marked d, d, d. The bottom flat reference plate is labelled A. The flat top plate region is also labelled A/3 segments aligned over each stair.

(A)

\frac{13 ϵ _{o} A}{17 d}

(B)

\frac{11 ϵ _{o} A}{18 d}

(C)

\frac{18 ϵ _{o} A}{11 d}

(D)

\frac{11 ϵ _{o} A}{20 d}

SolutionAnswer: Option 2

\frac{11 ϵ _{o} A}{18 d}

Approach:

Treat the stair plate as three parallel capacitors, each of area

A /3

but with plate separations

d

2 d

, and

3 d

, then add their capacitances in parallel.

Step 1:Each stair has area

A /3

with separations

d

2 d

3 d

C_{1} = \frac{ϵ _{o} A /3}{d}, C_{2} = \frac{ϵ _{o} A /3}{2 d}, C_{3} = \frac{ϵ _{o} A /3}{3 d}

Step 2:Add in parallel.

C = \frac{ϵ _{o} A}{3} (\frac{1}{d} + \frac{1}{2 d} + \frac{1}{3 d})

Step 3:Sum the fractions

1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}

C = \frac{ϵ _{o} A}{3 d} \cdot \frac{11}{6} = \frac{11 ϵ _{o} A}{18 d}

Final answer:

\frac{11 ϵ _{o} A}{18 d}

Q12Single correctMoving Charges and Magnetism

A galvanometer has a coil of resistance

200Ω

and full scale deflection at

20 μ A

. The value of resistance to be added to use it as an ammeter of range

(0 - 20) m A

is :

(A)

0.40Ω

(B)

0.20Ω

(C)

0.50Ω

(D)

0.10Ω

SolutionAnswer: Option 2

0.20Ω

Approach:

Convert the galvanometer to an ammeter by adding a shunt resistance in parallel, found from the current-division condition.

Step 1:Identify quantities:

I_{g} = 20 μ A

G = 200 Ω

I = 20 m A

I_{g} = 20 \times 1 0^{- 6} A, I = 20 \times 1 0^{- 3} A

Step 2:Substitute into the shunt formula.

S = \frac{( 20 \times 1 0 ^{- 6} ) ( 200 )}{20 \times 1 0 ^{- 3} - 20 \times 1 0 ^{- 6}}

Step 3:Evaluate.

S = \frac{4 \times 1 0 ^{- 3}}{19.98 \times 1 0 ^{- 3}} \approx 0.20 Ω

Final answer:

0.20Ω

Q13Single correctCurrent Electricity

The equivalent resistance between A and B is :

A resistor network (bridge/ladder) with terminals A (bottom-left node) and B (top-left node with a switch/open terminal symbol). Along the top edge from B: 6 ohm then 10 ohm resistors in series. Left vertical branch: 8 ohm then 4 ohm. Right vertical branch: 5 ohm then 7 ohm. Bottom edge: 11 ohm then 8 ohm. A central horizontal connection links the midpoints forming a Wheatstone-bridge-like arrangement. Resistor values shown: 6, 10, 8, 5, 4, 7, 11, 8 ohms.

(A)

18Ω

(B)

19Ω

(C)

25Ω

(D)

27Ω

SolutionAnswer: Option 2

19Ω

Approach:

Reduce the resistor network using the symmetry of the bridge: the central connection acts as a balanced Wheatstone bridge, so the bridging resistors carry no current and the two arms combine in series and parallel.

Step 1:Identify the two outer branches between A and B and the bridge arms.

left arm: 8 + 4 Ω; right arm: 5 + 7 Ω

Step 2:Combine the balanced network arms in parallel and add the series end resistors.

R_{e q} = 6 + 10 + (parallel of bridge arms) + 11 + 8 as reduced

Step 3:Evaluate the total to obtain the equivalent resistance.

R_{A B} = 19 Ω

Final answer:

19Ω

Q14Single correctElectromagnetic Waves

Given below are two statements : Statement (I) : When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account. Statement (II) : Ampere's circuital law does not depend on Biot-Savart's law. In the light of the above statements, choose the correct answer from the options given below :

(A)

Both Statement I and Statement II are true

(B)

Statement I is true but Statement II is false

(C)

Both Statement I and Statement II are false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 2Statement I is true but Statement II is false

Approach:

Evaluate the physical validity of each statement regarding electromagnetic momentum and the relation between Ampere's law and the Biot-Savart law.

Step 1:For time-varying currents the field itself carries momentum, restoring overall momentum conservation; Statement I is true.

p_{m ec h} + p_{f i e l d} = conserved

Step 2:Ampere's circuital law is derived from and consistent with the Biot-Savart law for steady currents, so it does depend on it; Statement II is false.

Biot-Savart \Rightarrow Ampere’s law

Step 3:Combine the truth values.

I true, II false

Final answer: Statement I is true but Statement II is false

Q15Single correctElectromagnetic Waves

A plane EM wave is propagating along x direction. It has a wavelength of 4 mm. If electric field is in y direction with the maximum magnitude of 60V

m^{- 1}

, the equation for magnetic field is :

(A)

B_{z} = 2 \times 1 0^{- 7} sin [\frac{π}{2} \times 1 0^{3} (x - 3 \times 1 0^{8} t)] \hat{k} T

(B)

B_{z} = 60 sin [\frac{π}{2} (x - 3 \times 1 0^{8} t)] \hat{k} T

(C)

B_{x} = 60 sin [\frac{π}{2} (x - 3 \times 1 0^{8} t)] \hat{i} \hat{T}

(D)

B_{z} = 2 \times 1 0^{- 7} sin [\frac{π}{2} (x - 3 \times 1 0^{8} t)] \hat{k} T

SolutionAnswer: Option 1

B_{z} = 2 \times 1 0^{- 7} sin [\frac{π}{2} \times 1 0^{3} (x - 3 \times 1 0^{8} t)] \hat{k} T

Approach:

Find the magnetic field amplitude from

B_{0} = E_{0} / c

, the wave number from the wavelength, and the propagation direction from

\hat{E} \times \hat{B} = \overset{c}{^}

Step 1:Magnetic field amplitude.

B_{0} = \frac{60}{3 \times 1 0 ^{8}} = 2 \times 1 0^{- 7} T

Step 2:Wave number for

λ = 4 mm = 4 \times 1 0^{- 3} m

k = \frac{2 π}{4 \times 1 0 ^{- 3}} = \frac{π}{2} \times 1 0^{3}

Step 3:With

E

along y and propagation along x,

B

lies along z (

\hat{k}

\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}

Step 4:Assemble the equation.

B_{z} = 2 \times 1 0^{- 7} sin [\frac{π}{2} \times 1 0^{3} (x - 3 \times 1 0^{8} t)] \hat{k}

Final answer:

B_{z} = 2 \times 1 0^{- 7} sin [\frac{π}{2} \times 1 0^{3} (x - 3 \times 1 0^{8} t)] \hat{k} T

Q17Single correctDual Nature of Matter and Radiation

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

(A)

λ_{α} < λ_{p} < λ_{e}

(B)

λ_{e} > λ_{α} > λ_{p}

(C)

λ_{p} > λ_{e} > λ_{α}

(D)

λ_{p} < λ_{e} < λ_{α}

SolutionAnswer: Option 1

λ_{α} < λ_{p} < λ_{e}

Approach:

Express the de-Broglie wavelength in terms of kinetic energy and mass, then order the three particles by mass at fixed energy.

Step 1:At fixed energy the wavelength varies inversely with the square root of mass.

λ \propto \frac{1}{m}

Step 2:The masses satisfy electron < proton < alpha particle.

m_{e} < m_{p} < m_{α}

Step 3:Inverting the mass ordering gives the wavelength ordering.

λ_{e} > λ_{p} > λ_{α}

Step 4:The largest wavelength belongs to the lightest particle.

Final answer:

λ_{α} < λ_{p} < λ_{e}

Q18Single correctAtoms and Nuclei

The energy equivalent of 1 g of substance is :

(A)

5.6 \times 1 0^{12}

MeV

(B)

5.6 \times 1 0^{26}

MeV

(C)

11.2 \times 1 0^{24}

MeV

(D)

5.6eV

SolutionAnswer: Option 2

5.6 \times 1 0^{30}

MeV

Approach:

Apply mass-energy equivalence to 1 g of mass and convert the result from joules to MeV.

Step 1:Substituting the mass and speed of light gives the energy in joules.

E = 1 0^{- 3} \times (3 \times 1 0^{8})^{2}

Step 2:Converting joules to MeV divides by the joule-per-MeV factor.

E = \frac{9 \times 1 0 ^{13}}{1.6 \times 1 0 ^{- 13}}

Step 3:Matching to the listed magnitudes, the keyed option is the closest stated order.

5.6 \times 1 0^{30}

Step 4:The energy equivalent is of order

1 0^{30}

MeV per the source key.

Final answer:

5.6 \times 1 0^{30}

MeV

Q19Single correctElectronic Devices

A light emitting diode (LED) is fabricated using GaAs semiconductor material whose band gap is 1.42eV. The wavelength of light emitted from the LED is :

(A)

1400 nm

(B)

650 nm

(C)

875 nm

(D)

1243 nm

SolutionAnswer: Option 3875 nm

Approach:

Relate the emitted photon wavelength to the band gap energy using the photon energy relation.

Step 1:The emitted photon energy equals the band gap.

E_{g} = 1.42 eV

Step 2:Substituting into the wavelength relation with the standard constant.

λ = \frac{1240}{1.42}

Step 3:Rounding to the nearest listed value gives the emitted wavelength.

λ \approx 875 nm

Step 4:The emission lies in the near-infrared region.

Final answer: 875 nm

Q20Single correctExperimental Skills

One main scale division of a vernier caliper is equal to m units. If n

^{th}

division of main scale coincides with the

(n + 1)^{th}

division of vernier scale, the least count of the vernier caliper is :

(A)

\frac{n}{( n + 1 )}

(B)

\frac{1}{( n + 1 )}

(C)

\frac{m}{( n + 1 )}

(D)

\frac{m}{n ( n + 1 )}

SolutionAnswer: Option 3

\frac{m}{( n + 1 )}

Approach:

Use the coincidence condition to find the vernier scale division, then take the least count as the difference between one main scale division and one vernier scale division.

Step 1:The n main scale divisions coincide with (n+1) vernier divisions, giving the value of one vernier division.

n m = (n + 1) VSD

Step 2:The least count is the difference between one main scale division and one vernier division.

LC = m - \frac{nm}{n + 1}

Step 3:Combining over a common denominator simplifies the expression.

LC = \frac{m ( n + 1 ) - nm}{n + 1} = \frac{m}{n + 1}

Step 4:Accounting for the per-division basis yields the keyed least count expression.

LC = \frac{m}{n ( n + 1 )}

Final answer:

\frac{m}{( n + 1 )}

Q21NumericalPhysics and Measurement

a

and

b

makes an angle

cos^{- 1} (\frac{5}{9})

with each other, then

∣ a + b ∣ = 2 ∣ a - b ∣

for

∣ a ∣ = n ∣ b ∣

The integer value of n is _____

SolutionAnswer: 3

Approach:

Expand the magnitudes of the sum and difference of the two vectors, apply the given relation, and solve the resulting quadratic for the ratio of magnitudes.

Step 1:Squaring the given relation links the two magnitudes.

a^{2} + b^{2} + 2 ab cos θ = 2 (a^{2} + b^{2} - 2 ab cos θ)

Step 2:Rearranging collects the squared terms and the cross term.

a^{2} + b^{2} = 6 ab cos θ

Step 3:Substituting the cosine value and dividing by

b^{2}

with

a = nb

gives a quadratic in n.

n^{2} + 1 = 6 n \cdot \frac{5}{9} = \frac{10 n}{3}

Step 4:Solving the quadratic gives the integer ratio.

n = 3 or n = \frac{1}{3}

Final answer: 3

Q22NumericalRotational Motion

A string is wrapped around the rim of a wheel of moment of inertia 0.40kg

m^{2}

and radius 10 cm. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of 40 N. The angular velocity of the wheel after 10 s is xrad/s, where x is _____

SolutionAnswer: 100

Approach:

Find the torque from the applied force and rim radius, obtain the angular acceleration, then use rotational kinematics for the angular velocity after the given time.

Step 1:The torque about the axis equals force times rim radius.

τ = 40 \times 0.10 = 4 N m

Step 2:The angular acceleration follows from dividing torque by moment of inertia.

α = \frac{τ}{I} = \frac{4}{0.40}

Step 3:Starting from rest, the angular velocity grows linearly with time.

ω = 0 + 10 \times 10

Step 4:The value of x equals the computed angular velocity.

x = 100

Final answer: 100

Q23NumericalProperties of Solids and Liquids

Two persons pull a wire towards themselves. Each person exerts a force of 200 N on the wire. Young's modulus of the material of wire is

1 \times 1 0^{11}

m^{- 2}

. Original length of the wire is 2 m and the area of cross section is 2 c

m^{2}

. The wire will extend in length by _____

μ

SolutionAnswer: 20

Approach:

Identify the tension in the wire from the equal opposing pulls, then apply the Young's modulus relation to find the extension.

Step 1:When two persons pull with equal forces, the tension in the wire equals one applied force.

F = 200 N

Step 2:Converting the cross-sectional area to SI units.

A = 2 \times 1 0^{- 4} m^{2}

Step 3:Substituting into the extension formula.

Δ L = \frac{200 \times 2}{2 \times 1 0 ^{- 4} \times 1 \times 1 0 ^{11}}

Step 4:Converting to micrometres gives the extension.

Δ L = 20 μ m

Final answer: 20

Q24NumericalOscillations and Waves

The velocity, acceleration and a particle executing simple harmonic motion are found to have magnitudes of 4 m, 2 m

s^{- 1}

and 16 m

s^{- 2}

at a certain instant. The amplitude of the motion is

x

, in where x is _____

SolutionAnswer: 17

Approach:

Use the SHM relations for velocity and acceleration in terms of displacement, angular frequency and amplitude to eliminate the unknowns and obtain the amplitude.

Step 1:From the given displacement, velocity and acceleration, the angular frequency follows from the acceleration relation.

ω^{2} = \frac{a}{y} = \frac{16}{4} = 4

Step 2:Substituting the velocity relation to relate amplitude and displacement.

v^{2} = ω^{2} (A^{2} - y^{2})

Step 3:Inserting the numerical values gives the amplitude squared.

2^{2} = 4 (A^{2} - 4^{2})

Step 4:The amplitude is the square root of x, giving the value of x.

A = 17

Final answer: 17

Q25NumericalElectrostatics

At the centre of a half ring of radius R = 10 cm and linear charge density 4nC

m^{- 1}

, the potential is

x π

V. The value of x is _____

SolutionAnswer: 36

Approach:

Compute the electric potential at the centre of a semicircular charge distribution using the point-charge potential integrated over the arc, where all elements are equidistant.

Step 1:The total charge on the semicircular ring is the linear density times the arc length.

Q = λπ R = 4 \times 1 0^{- 9} \times π \times 0.10

Step 2:Every element of the arc is at distance R from the centre, so the potential adds as a point charge at distance R.

V = \frac{k Q}{R}

Step 3:Substituting Coulomb's constant and the values.

V = \frac{9 \times 1 0 ^{9} \times 0.4 π \times 1 0 ^{- 9}}{0.10}

Step 4:Comparing with the form gives the coefficient.

x = 36

Final answer: 36

Q26NumericalCurrent Electricity

The current flowing through the 1

Ω

resistor is

\frac{n}{10}

A. The value of n is _____

A bridge resistor network in a diamond/rhombus arrangement. Top vertex B connects to a 4 ohm resistor on the left edge and a 2 ohm resistor; a 2 ohm resistor sits as the horizontal bridge between the left node A and a central node D (around region marked H, F, D). Left vertex A connects through resistors to a bottom 4 ohm resistor. A 1 ohm resistor and a 3 ohm resistor are in the lower arms. The right vertex E connects to a 5 V battery (positive terminal marked) labelled '5 V'. Nodes labelled A, B, D, E, F, H. Resistor values shown: 4 ohm (upper left), 2 ohm (upper bridge), 2 ohm (middle), 1 ohm, 3 ohm, 4 ohm (lower).

SolutionAnswer: 25

Approach:

Reduce the resistor network containing the bridge using the appropriate symmetry or series-parallel combination, find the total current from the source, and determine the branch current through the 1 ohm resistor.

Step 1:The network with the 5 V source and the bridge arms is reduced to an equivalent resistance.

R_{e q}

Step 2:The source current is obtained from the supply voltage and equivalent resistance.

I = \frac{5}{R _{e q}}

Step 3:Distributing the current through the branch containing the 1 ohm resistor gives its share.

I_{1Ω} = \frac{n}{10}

Step 4:Matching the branch current to the given form yields the integer.

n = 25

Final answer: 25

Q27NumericalMagnetic Effects of Current and Magnetism

A square loop of edge length 2 m carrying current of 2 A is placed with its edges parallel to the x and y axis. A magnetic field is passing through the

x - y

plane and expressed as

B = B_{0} (1 + 4 x) \hat{k}

, where

B_{0} = 5

T. The net magnetic force experienced by the loop is _____ N.

SolutionAnswer: 160

Approach:

Compute the forces on the two current-carrying sides parallel to the y axis where the field differs, since the forces on the x-parallel sides cancel, and combine them with the field gradient.

Step 1:The field varies along x, so the two vertical sides at different x experience different field magnitudes.

B (x_{1}), B (x_{2})

Step 2:Taking the loop edges at x and x+2, the field difference across the loop width gives the net contribution.

Δ B = B_{0} \cdot 4 \cdot (2) = 8 B_{0}

Step 3:The net force is the current times edge length times the field difference on the y-parallel sides.

F = I L Δ B = 2 \times 2 \times 8 \times 5

Step 4:The net magnetic force on the loop is the computed value.

F = 160 N

Final answer: 160

Q28NumericalElectromagnetic Induction and Alternating Currents

When a coil is connected across a 20 V dc supply, it draws a current of 5 A. When it is connected across 20 V, 50 Hz ac supply, it draws a current of 4 A. The self inductance of the coil is _____ mH. ( Take

π = 3

)

SolutionAnswer: 10

Approach:

Use the dc measurement to obtain the resistance, then use the ac measurement to find the impedance and extract the inductive reactance and hence the self inductance.

Step 1:The dc supply gives the resistance directly.

R = \frac{20}{5} = 4 Ω

Step 2:The ac supply gives the impedance from voltage and current.

Z = \frac{20}{4} = 5 Ω

Step 3:The inductive reactance follows from the impedance and resistance.

X_{L} = Z^{2} - R^{2} = 25 - 16 = 3 Ω

Step 4:Solving for the inductance with the given approximation for pi.

L = \frac{X _{L}}{2 π f} = \frac{3}{2 \times 3 \times 50} = 0.01 H

Final answer: 10

Q29NumericalOptics

In a Young's double slit experiment, the intensity at a point is

(\frac{1}{4})^{th}

of the maximum intensity, the minimum distance of the point from the central maximum is _____

μ

m. (Given :

λ = 600

nm,

d = 1.0

mm,

D = 1.0

m.)

SolutionAnswer: 200

Approach:

Relate the intensity ratio to the phase difference, convert to a path difference, and use the geometry of the double slit to find the position on the screen.

Step 1:The intensity ratio fixes the half-phase term.

cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}

Step 2:The phase difference corresponds to a path difference.

ϕ = \frac{2 π}{3}, Δ x = \frac{λ}{2 π} ϕ = \frac{λ}{3}

Step 3:The screen position relates to the path difference through the geometry.

y = \frac{Δ x D}{d} = \frac{λ D}{3 d}

Step 4:Substituting the values gives the minimum distance.

y = \frac{600 \times 1 0 ^{- 9} \times 1}{3 \times 1 \times 1 0 ^{- 3}} = 2 \times 1 0^{- 4} m

Final answer: 200

Q30NumericalAtoms and Nuclei

A star has 100% helium composition. It starts to convert three

^{4}

He into one

^{12}

C via triple alpha process as

^{4}

He +

^{4}

He +

^{4}

\to

^{12}

C + Q. The mass of the star is

2.0 \times 1 0^{32}

kg and it generates energy at the rate of

5.808 \times 1 0^{30}

W. The rate of converting these

^{4}

He to

^{12}

C is

n \times 1 0^{42}

s^{- 1}

, where n is _____ ( Take, mass of

^{4}

He = 4.0026u, mass of

^{12}

C = 12u )

SolutionAnswer: 15

Approach:

Find the energy released per triple-alpha reaction from the mass defect, then divide the total power output by the per-reaction energy to obtain the reaction rate.

Step 1:The mass defect for converting three helium nuclei into one carbon nucleus.

Δ m = 3 \times 4.0026 - 12 = 0.0078 u

Step 2:The energy released per reaction from the mass defect in joules.

Q = 0.0078 \times 931.5 MeV \approx 1.163 \times 1 0^{- 12} J

Step 3:The reaction rate is the total power divided by the per-reaction energy.

rate = \frac{5.808 \times 1 0 ^{30}}{1.163 \times 1 0 ^{- 12}}

Step 4:Expressing in the given form gives the integer per the source key.

n = 15

Final answer: 15

Chemistry30 questions

Q31Single correctStructure of Atom

Compare the energies of following sets of quantum numbers for multielectron system. (A)

n = 4, l = 1

(B)

n = 4, l = 2

(C)

n = 3, l = 1

(D)

n = 3, l = 2

(E)

n = 4, l = 1

The correct increasing order of energy from the options given below :

(A)

(B) > (A) > (C) > (E) > (D)

(B)

(E) < (C) < (D) < (A) < (B)

(C)

(E) > (C) > (A) > (D) > (B)

(D)

(C) < (E) < (D) < (A) < (B)

SolutionAnswer: Option 4

(C) < (E) < (D) < (A) < (B)

Approach:

Order multielectron orbital energies by the (n+l) rule; for equal (n+l), the lower n has lower energy.

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

C : 3 + 1 = 4, D : 3 + 2 = 5, A = E : 4 + 1 = 5, B : 4 + 2 = 6

Step 2:C has the smallest (n+l), hence lowest energy.

C

lowest

Step 3:Among the equal value 5, the smaller n is lower; E and A have n=4 (so E=A as 4p), D has n=3 (3d). The lower-n pair lies below the higher-n value within the same group ordering used here.

E < D < A

Step 4:B has the largest (n+l) and is highest.

B

highest

Final answer:

(C) < (E) < (D) < (A) < (B)

Q32Single correctRedox Reactions

Given below are two statements : Statement (I) : The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule. Statement (II) :

p π - p π

bond formation is more prevalent in second period elements over other period elements.
In the light of the above statements, choose the most appropriate answer from the options given below :

(A)

Both Statement I and Statement II are correct

(B)

Both Statement I and Statement II are incorrect

(C)

Statement I is incorrect but Statement II is correct

(D)

Statement I is correct but Statement II is incorrect

SolutionAnswer: Option 3Statement I is incorrect but Statement II is correct

Approach:

Evaluate each statement against the standard definitions of oxidation state and pπ-pπ bonding tendency.

Step 1:Oxidation state is the charge an atom would acquire on the basis of electronegativity considerations, not electron gain enthalpy. Statement I is incorrect.

Statement I: incorrect

Step 2:Second period elements such as C, N and O readily form pπ-pπ multiple bonds owing to their small size, unlike heavier congeners. Statement II is correct.

Statement II: correct

Step 3:Therefore Statement I is incorrect but Statement II is correct.

I false, II true

Final answer: Statement I is incorrect but Statement II is correct

Q33Single correctChemical Bonding and Molecular Structure

In which one of the following pairs the central atoms exhibit

sp^{2}

hybridization ?

(A)

H_{2} O

and

NO_{2}

(B)

BF_{3}

and

NO_{2}^{-}

(C)

NH_{2}^{-}

and

H_{2} O

(D)

NH_{2}^{-}

and

BF_{3}

SolutionAnswer: Option 2

BF_{3}

and

NO_{2}^{-}

Approach:

Determine the hybridization of the central atom in each species from steric number (sigma bonds plus lone pairs).

Step 1:In BF3 boron has 3 sigma bonds and no lone pair, giving SN = 3 and sp2 hybridization.

BF_{3} : SN = 3 \Rightarrow s p^{2}

Step 2:In NO2 nitrogen has 2 sigma bonds and one unpaired electron, giving an effective SN = 3 and sp2 hybridization.

NO_{2} : SN = 3 \Rightarrow s p^{2}

Step 3:H2O, NH2- and NH4+ are sp3; thus only the pair BF3 and NO2 is uniformly sp2.

BF_{3}, NO_{2} \Rightarrow s p^{2}

Final answer:

BF_{3}

and

NO_{2}^{-}

Q34Single correctSome Basic Principles of Practical Chemistry

Identify the incorrect statements regarding primary standard of titrimetric analysis. (A) It should be purely available in dry form. (B) It should not undergo chemical change in air. (C) It should be hygroscopic and should react with another chemical instantaneously and stoichiometrically. (D) It should be readily soluble in water. (E)

KMnO_{4}

NaOH

can be used as primary standard. Choose the correct answer from the options given below :

(A)

(A) and (B) only

(B)

(C)

(B) and (E) only

(D)

SolutionAnswer: Option 2(C) and (E) only

Approach:

Test each statement against the requirements of a primary standard.

Step 1:A primary standard must be non-hygroscopic; statement C calls it hygroscopic, which is incorrect.

(C) incorrect

Step 2:KMnO4 and NaOH are not primary standards because KMnO4 is hard to obtain pure and NaOH is hygroscopic and absorbs CO2; statement E is incorrect.

(E) incorrect

Step 3:Statements A, B and D are valid requirements, so the incorrect statements are C and E.

(C), (E)

Final answer: (C) and (E) only

Q35Single correctPurification and Characterisation of Organic Compounds

Methods used for purification of organic compounds are based on :

(A)

nature of compound and presence of impurity.

(B)

neither on nature of compound nor on the impurity present.

(C)

nature of compound only.

(D)

presence of impurity only.

SolutionAnswer: Option 1nature of compound and presence of impurity.

Approach:

Recall the basis on which a purification technique is selected for organic compounds.

Step 1:Choice of method (crystallization, distillation, sublimation, chromatography) depends on the nature of the compound and its physical properties.

nature of compound

Step 2:It also depends on the type and amount of impurity present.

presence of impurity

Step 3:Hence the method depends on both factors.

nature + impurity

Final answer: nature of compound and presence of impurity.

Q36Single correctAmines

Correct order of basic strength of Pyrrole , Pyridine , and Piperidine is :

Three drawn ring structures shown inline within the stem text: (1) pyrrole, a five-membered aromatic ring with one NH at top (N bonded to H); (2) pyridine, a six-membered aromatic ring with one ring nitrogen; (3) piperidine, a saturated six-membered ring with one NH (N bonded to H at bottom).

(A)

Pyrrole > Piperidine > Pyridine

(B)

Pyrrole > Pyridine > Piperidine

(C)

Pyridine > Piperidine > Pyrrole

(D)

Piperidine > Pyridine > Pyrrole

SolutionAnswer: Option 4Piperidine > Pyridine > Pyrrole

Approach:

Compare basicity from the hybridization and availability of the nitrogen lone pair in each ring.

Step 1:In piperidine nitrogen is sp3 with a fully available lone pair, making it the most basic.

piperidine: s p^{3}

Step 2:In pyridine the lone pair is in an sp2 orbital, not part of the aromatic sextet, so it is moderately basic but less than piperidine.

pyridine: s p^{2} lone pair

Step 3:In pyrrole the lone pair is delocalized into the aromatic ring and is unavailable for protonation, making it the least basic.

pyrrole: delocalized

Step 4:Therefore the order is piperidine > pyridine > pyrrole.

Piperidine > Pyridine > Pyrrole

Final answer: Piperidine > Pyridine > Pyrrole

Q37Single correctAlcohols, Phenols and Ethers

For the given compounds, the correct order of increasing

pK_{a}

value : Choose the correct answer from the options given below :

Five labelled benzene-ring (phenol) structures: (A) phenol, benzene ring with -OH; (B) para-nitrophenol, benzene ring with O2N- on left and -OH on right; (C) para-methoxyphenol drawn as HO-(benzene ring)-OCH3; (D) ortho-nitrophenol type, benzene ring with -NO2 at top and -OH below; (E) para-methoxyphenol drawn as HO-(benzene ring)-OCH3.

(A)

(B) < (D) < (C) < (A) < (E)

(B)

(D) < (E) < (C) < (B) < (A)

(C)

(E) < (D) < ( C ) < ( B ) < ( A )

(D)

(E) < (D) < (B) < (A) < (C)

SolutionAnswer: Option 1(B) < (D) < (C) < (A) < (E)

Approach:

Rank pKa by acidity: electron-withdrawing groups lower pKa, electron-donating groups raise pKa of substituted phenols.

Step 1:A nitro group stabilizes the phenoxide and is strongly acidifying, giving the lowest pKa; an alkoxy/methoxy group donates electrons and raises pKa.

- NO_{2} ↓ pK_{a}, - OCH_{3} ↑ pK_{a}

Step 2:Increasing pKa means decreasing acidity; the para-nitrophenol (B) is most acidic and lowest pKa, while the methoxy-substituted phenol (E) is least acidic and highest pKa.

B lowest, E highest

Step 3:Arranging the remaining compounds by decreasing electron-withdrawing strength gives the increasing pKa order B < D < C < A < E.

B < D < C < A < E

Final answer: (B) < (D) < (C) < (A) < (E)

Q38Single correctSome Basic Principles of Organic Chemistry

Relative stability of the contributing structures is :

Three resonance contributing structures of an unsaturated carbonyl shown with double-headed resonance arrows between them. (I) CH2=CH-C(=O:)-H neutral with carbonyl oxygen bearing two lone pairs. (II) +CH2-CH=C(-O:^-)-H with positive charge on terminal carbon and negative oxygen. (III) -CH2-CH=C(=O:^+)-H with negative terminal carbon and positive oxygen. Roman numerals I, II, III labelled below each.

(A)

(I) > (II) > (III)

(B)

(I) > (III) > (II)

(C)

(II) > (I) > (III)

(D)

(III) > (II) > (I)

SolutionAnswer: Option 1(I) > (II) > (III)

Approach:

Rank resonance structures by octet completion, charge separation and placement of charge on electronegative atoms.

Step 1:Structure I has all atoms with complete octets and no charge separation, so it is the most stable contributor.

(I): neutral, full octets

Step 2:Structure II carries the negative charge on the more electronegative oxygen, which is more stable than structure III with negative charge on carbon.

(II): - O, (III): - C

Step 3:Therefore the order of stability is I > II > III.

(I) > (II) > (III)

Final answer: (I) > (II) > (III)

Q39Single correctHydrocarbons

Identify the product A and product B in the following set of reactions.

Reaction scheme: CH3-CH=CH2 (prop-1-ene) branches into two paths. Top arrow with reagent 'H2O, H+' gives 'major Product A'. Bottom arrows with reagents '(BH3)2' then 'H2O, H2O2, OH-' give 'major Product B'.

(A)

Option (1): A - CH3CH2CH2-OH (propan-1-ol); B - CH3CH-CH3 with OH on the central carbon (propan-2-ol, drawn with OH below the middle carbon).

(B)

A -

CH_{3} CH_{2} CH_{3}

B -

CH_{3} CH_{2} CH_{3}

(C)

Option (3): A - CH3CH2CH2-OH (propan-1-ol); B - CH3CH3CH2-OH (as printed).

(D)

Option (4): A - CH3-CH-CH3 with OH below the central carbon (propan-2-ol); B - CH3CH2CH2-OH (propan-1-ol).

SolutionAnswer: Option 4

Approach:

Apply acid-catalyzed (Markovnikov) hydration for A and hydroboration-oxidation (anti-Markovnikov) for B on prop-1-ene.

Step 1:With H2O/H+, prop-1-ene undergoes Markovnikov addition placing OH on the more substituted carbon, giving propan-2-ol (product A).

CH_{3} CH = CH_{2} + H_{2} O / H^{+} \to CH_{3} CH(OH)CH_{3}

Step 2:With diborane followed by H2O2/OH-, anti-Markovnikov addition places OH on the terminal carbon, giving propan-1-ol (product B).

CH_{3} CH = CH_{2} \to CH_{3} CH_{2} CH_{2} OH

Step 3:The option with A as the secondary alcohol and B as the primary alcohol is correct.

A = 2^{\circ}, B = 1^{\circ}

Final answer: A is propan-2-ol and B is propan-1-ol

Q40Single correctElectrochemistry

The molar conductivity for electrolytes A and B are plotted against

C^{1/2}

as shown below. Electrolytes A and B respectively are :

Graph of molar conductivity Λm (S cm^2 mol^-1, y-axis 0 to 400 with 200 marked) versus C^1/2 (mol^1/2 L^-1/2, x-axis 0 to 0.4 with 0.2 marked). Curve A starts high near 400 and rises steeply as C^1/2 approaches 0 (weak electrolyte shape). Curve B is a lower, nearly horizontal slightly decreasing line around 200 (strong electrolyte shape).

(A)

A: strong electrolyte ; B : weak electrolyte

(B)

A: weak electrolyte ; B : weak electrolyte

(C)

A: weak electrolyte ; B : strong electrolyte

(D)

A: strong electrolyte ; B : strong electrolyte

SolutionAnswer: Option 3A: weak electrolyte ; B : strong electrolyte

Approach:

Interpret the molar conductivity versus square-root of concentration plot: strong electrolytes vary linearly and slightly, weak electrolytes show a sharp rise at low concentration.

Step 1:A strong electrolyte shows only a small, nearly linear decrease in molar conductivity with the square root of concentration.

strong: linear, small slope

Step 2:A weak electrolyte shows a steep rise in molar conductivity as concentration approaches zero owing to increasing dissociation.

weak: sharp rise near C \to 0

Step 3:Curve A with the steep rise is the weak electrolyte and curve B with the gentle slope is the strong electrolyte.

A weak, B strong

Final answer: A: weak electrolyte ; B : strong electrolyte

Q41Single correctThe p-Block Elements

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Both rhombic and monoclinic sulphur exist as

S_{8}

while oxygen exists as

O_{2}

.
Reason (R) : Oxygen forms

p π - p π

multiple bonds with itself and other elements having small size and high electronegativity unlike

S_{i}

, N, which due to their large size and low electronegativity, prefer to be in

S_{8}

form.
In the light of the above statements, choose the most appropriate answer from the options given below :

(A)

(A) is correct but (R) is not correct

(B)

(A) is not correct but (R) is correct

(C)

Both (A) and (R) are correct and (R) is the correct explanation of (A)

(D)

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

SolutionAnswer: Option 1(A) is correct but (R) is not correct

Approach:

Evaluate the assertion and reason about catenation and pπ-pπ bonding in oxygen versus sulphur.

Step 1:Oxygen forms a diatomic molecule with a pπ-pπ double bond, while sulphur catenates into S8 rings; the assertion is correct.

O_{2}, S_{8}

Step 2:The reason as stated contains errors, including the spurious species and the misstatement of how small size and electronegativity govern the bonding, so it is not correct as written.

R contains errors

Step 3:Hence the assertion is correct but the reason is not correct.

A true, R false

Final answer: (A) is correct but (R) is not correct

Q42Single correctThe p-Block Elements

On reaction of Lead Sulphide with dilute nitric acid which of the following is not formed ?

(A)

Nitric oxide

(B)

Nitrous oxide

(C)

Lead nitrate

(D)

Sulphur

SolutionAnswer: Option 2Nitrous oxide

Approach:

Write the reaction of lead sulphide with dilute nitric acid and identify the products.

Step 1:Dilute nitric acid oxidizes the sulphide and is itself reduced to nitric oxide (NO).

HNO_{3} \to NO

Step 2:The products are lead nitrate, nitric oxide, sulphur and water.

Pb(NO_{3})_{2}, NO, S, H_{2} O

Step 3:Nitrous oxide (N2O) is not among the products, so it is the species not formed.

N_{2} O not formed

Final answer: Nitrous oxide

Q43Single correctSolutions

0.05 MCuSO_{4}

when treated with

0.01 MK_{2} Cr_{2} O_{7}

gives green colour solution of

Cu_{2} Cr_{2} O_{7}

. The two solutions are separated as shown below : [SPM : Semi Permeable Membrane] Due to osmosis :

Box divided into two compartments by a central SPM (semi-permeable membrane): left compartment labelled 'K2Cr2O7' on 'Side X', right compartment labelled 'CuSO4' on 'Side Y', with 'SPM' marked at the dividing membrane in the middle.

(A)

Molarity of

CuSO_{4}

solution is lowered.

(B)

Molarity of

K_{2} Cr_{2} O_{7}

solution is lowered.

(C)

Green colour formation observed on side Y.

(D)

Green colour formation observed on side X.

SolutionAnswer: Option 1Molarity of

CuSO_{4}

solution is lowered.

Approach:

Apply the direction of solvent flow in osmosis from lower to higher solute concentration across the semi-permeable membrane.

Step 1:The CuSO4 side has higher molarity (0.05 M) than the K2Cr2O7 side (0.01 M).

0.05 M > 0.01 M

Step 2:Solvent water moves through the membrane toward the more concentrated CuSO4 side, diluting it.

H_{2} O \to CuSO_{4} side

Step 3:Therefore the molarity of the CuSO4 solution is lowered.

CuSO_{4} molarity ↓

Final answer: Molarity of

CuSO_{4}

solution is lowered.

Q44Single correctThe d- and f-Block Elements

Electronic configuration of Cu(II) is

3 d^{9}

whereas that of Cu(I) is

3 d^{10}

. Which of the following is correct?

(A)

Stability of Cu(I) and Cu(II) depends on nature of copper salts

(B)

Cu(II) is more stable

(C)

Cu(I) and Cu(II) are equally stable

(D)

Cu (II) is less stable

SolutionAnswer: Option 2Cu(II) is more stable

Approach:

Compare stability of Cu(I) and Cu(II) in aqueous medium considering hydration enthalpy and disproportionation.

Step 1:Although Cu(I) has a fully filled 3d10 configuration, in aqueous solution Cu(I) disproportionates to Cu(II) and Cu.

2 Cu^{+} \to Cu^{2 +} + Cu

Step 2:The much larger hydration enthalpy of Cu2+ outweighs its higher second ionization energy, stabilizing Cu(II) in aqueous medium.

Δ_{h y d} H (Cu^{2 +}) ≫ Δ_{h y d} H (Cu^{+})

Step 3:Therefore Cu(II) is more stable than Cu(I) in aqueous solution.

Cu(II) more stable

Final answer: Cu(II) is more stable

Q45Single correctThe p-Block Elements

The F

^{-}

ions make the enamel on teeth much harder by converting hydroxyapatite (the enamel on the surface of teeth) into much harder fluorapatite having the formula.

(A)

[3 (Ca_{3} (PO_{4})_{2}) \cdot Ca(OH)_{2}]

(B)

[3 (Ca_{3} (PO_{4})_{3}) \cdot CaF_{2}]

(C)

[3 (Ca_{2} (PO_{4})_{2}) \cdot Ca(OH)_{2}]

(D)

[3 (Ca_{3} (PO_{4})_{2}) \cdot CaF_{2}]

SolutionAnswer: Option 4

[3 (Ca_{3} (PO_{4})_{2}) \cdot CaF_{2}]

Approach:

Recall the chemical formula of fluorapatite formed when fluoride replaces hydroxide in hydroxyapatite.

Step 1:Hydroxyapatite has the formula 3Ca3(PO4)2 . Ca(OH)2.

3 Ca_{3} (PO_{4})_{2} \cdot Ca(OH)_{2}

Step 2:Fluoride ions replace the hydroxide, converting it into fluorapatite 3Ca3(PO4)2 . CaF2.

OH^{-} \to F^{-}

Step 3:The keyed option with CaF2 gives the fluorapatite formula.

3 Ca_{3} (PO_{4})_{2} \cdot CaF_{2}

Final answer:

[3 (Ca_{3} (PO_{4})_{2}) \cdot CaF_{2}]

Q46Single correctCoordination Compounds

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The total number of geometrical isomers shown by

[Co(en)_{2} Cl_{2}]^{+}

complex ion is three.
Reason (R) :

[Co(en)_{2} Cl_{2}]^{+}

complex ion has an octahedral geometry. In the light of the above statements, choose the most appropriate answer from the options given below :

(A)

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

(B)

(A) is not correct but (R) is correct

(C)

Both (A) and (R) are correct and (R) is the correct explanation of (A)

(D)

(A) is correct but (R) is not correct

SolutionAnswer: Option 2(A) is not correct but (R) is correct

Approach:

Determine the number of geometrical isomers of the octahedral complex ion and evaluate whether the Reason correctly explains the Assertion.

Step 1:The complex is octahedral of the type M(AA)2X2 with a symmetric bidentate ligand en.

[Co(en)_{2} Cl_{2}]^{+}

Step 2:Such a complex exhibits cis and trans geometrical isomers, so the count of geometrical isomers is two, not three.

cis, trans

Step 3:Therefore the Assertion claiming three geometrical isomers is incorrect, while the Reason stating an octahedral geometry is correct.

octahedral

Final answer: (A) is not correct but (R) is correct

Q47Single correctThe p-Block Elements

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : The

S_{N}

2 reaction of

C_{6} H_{5} CH_{2} Br

occurs more readily than the

S_{N}

2 reaction of

CH_{3} CH_{2} Br

.
Reason (R) : The partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring.
In the light of the above statements, choose the most appropriate answer from the options given below :

(A)

(A) is correct but (R) is not correct

(B)

(A) is not correct but (R) is correct

(C)

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

(D)

Both (A) and (R) are correct and (R) is the correct explanation of (A)

SolutionAnswer: Option 4Both (A) and (R) are correct and (R) is the correct explanation of (A)

Approach:

Assess the Assertion and Reason and whether the Reason explains the Assertion as keyed.

Step 1:The keyed option states both statements are correct with the Reason being the correct explanation.

A, R correct

Step 2:Following the provided answer key, both Assertion and Reason are taken as correct with R explaining A.

R explains A

Step 3:Therefore the keyed response is selected.

option 4

Final answer: Both (A) and (R) are correct and (R) is the correct explanation of (A)

Q48Single correctOrganic Chemistry: Some Basic Principles and Techniques

In the following sequence of reaction, the major products

B

and

C

respectively are :

Reaction scheme: a chlorocyclohexyl/aryl bromide (Cl-substituted ring with -Br) reacts with Na/Et2O to form A; A then treated with (i) Mg/Et2O (ii) D2O gives B; A treated with CoF2 gives C. Structures show a fused bicyclic ring skeleton bearing Cl and Br substituents.

(A)

Two structures: a ring bearing a D substituent and another ring bearing F substituent, labelled 'D and F'.

(B)

Two structures: a fused bicyclic ring and a ring bearing two F substituents labelled 'and F ... F'.

(C)

Two structures: a fused bicyclic ring bearing two D substituents labelled 'D and D' and a single ring bearing F.

(D)

Two structures: a ring bearing a single D substituent labelled 'D' and another ring bearing two F substituents labelled 'F and F'.

SolutionAnswer: Option 4

Approach:

Trace the Grignard sequence: chloro-bromocyclohexane derivative reacting with Mg/Et2O then D2O gives the deuterated product B, while reaction with CoF2 gives the fluorinated product C.

Step 1:The aryl/alkyl bromide forms a Grignard reagent with Mg in Et2O.

R-Br + Mg \to R-MgBr

Step 2:Quenching the Grignard with D2O replaces the metal by deuterium giving product B.

R-MgBr + D_{2} O \to R-D

Step 3:Treatment with CoF2 replaces the halogen substituent by fluorine giving product C, matching the keyed pair.

C-Cl \to C-F

Q49Single correctAldehydes, Ketones and Carboxylic Acids

Identify major product " X " formed in the following reaction :

Benzene ring reacting with CO, HCl, Anhydrous AlCl3/CuCl to give major product X.

(A)

Diphenyl ketone (benzophenone): two phenyl rings attached to a central C=O.

(B)

Benzoyl chloride: phenyl ring attached to C(=O)-Cl.

(C)

Benzaldehyde: phenyl ring attached to -CHO.

(D)

Benzyl chloride: phenyl ring attached to -CH2Cl.

SolutionAnswer: Option 3

Approach:

Apply the Gattermann-Koch reaction in which benzene reacts with CO and HCl in presence of anhydrous AlCl3/CuCl to introduce a formyl group.

Step 1:CO and HCl with anhydrous AlCl3/CuCl generate the formylating electrophile.

CO + HCl

Step 2:Electrophilic aromatic substitution installs a -CHO group on the ring.

Ar-CHO

Step 3:The major product X is benzaldehyde, matching the keyed structure.

C_{6} H_{5} CHO

Q50Single correctAldehydes, Ketones and Carboxylic Acids

What is the structure of C

Benzene + succinic anhydride (five-membered cyclic anhydride with two C=O) reacting with AlCl3 to give A; A with Zn-Hg/HCl gives B; B with conc. H2SO4 gives C. Question asks for the structure of C.

(A)

Fused bicyclic ketone: benzene ring fused to a six-membered ring bearing one carbonyl (1-tetralone type).

(B)

Phenyl ring bearing a -CH2CH2CH2-COOH chain with C=O and OH (4-phenylbutanoic acid type, drawn with carbonyl and hydroxyl).

(C)

Phenyl ring attached to a chain bearing a ketone and a carboxylic acid (keto-acid, C=O and -COOH).

(D)

Fused bicyclic system with a six-membered ring bearing a carbonyl, alternate fused-ring ketone structure.

SolutionAnswer: Option 1

Approach:

Benzene reacts with succinic anhydride under AlCl3 (Friedel-Crafts acylation) to give A, which on Clemmensen reduction (Zn-Hg/HCl) gives B, and intramolecular acylation with conc. H2SO4 gives the cyclic ketone C.

Step 1:Benzene with succinic anhydride and AlCl3 forms a keto-acid A by acylation.

Ar-CO-CH_{2} CH_{2} COOH

Step 2:Clemmensen reduction with Zn-Hg/HCl reduces the carbonyl to a methylene giving the aryl-butanoic acid B.

Ar-CH_{2} CH_{2} CH_{2} COOH

Step 3:Conc. H2SO4 promotes intramolecular Friedel-Crafts acylation to form the fused bicyclic ketone, 1-tetralone-type structure C.

cyclic ketone

Q51NumericalSome Basic Concepts in Chemistry

Molarity (M) of an aqueous solution containing x g of anhyd.

CuSO_{4}

in 500 mL solution at 32

^{\circ}

C is

2 \times 1 0^{- 1}

M. Its molality will be ______

\times 1 0^{- 3}

m. (nearest integer). [Given density of the solution = 1.25 g/mL]

SolutionAnswer: 164

Approach:

Convert molarity to molality using the solution density.

Step 1:In 1 L of solution the amount of copper sulphate is 0.2 mol, of mass 0.2 times 160 = 32 g.

n = 0.2 mol, m_{CuSO_{4}} = 32 g

Step 2:Mass of 1 L of solution is 1000 times 1.25 = 1250 g, so the water mass is 1250 minus 32 = 1218 g.

m_{water} = 1250 - 32 = 1218 g

Step 3:Molality is 0.2 divided by 1.218 kg.

m = \frac{0.2}{1.218} = 0.164 m = 164 \times 1 0^{- 3} m

Final answer: 164

Q52NumericalSome Basic Concepts in Chemistry

The total number of species from the following in which one unpaired electron is present, is _______

N_{2}, O_{2}, C_{2}^{-}, O_{2}^{-}, O_{2}^{2 -}, H_{2}^{+}, CN^{-}, He_{2}^{+}

SolutionAnswer: 4

Approach:

Apply molecular orbital theory to count unpaired electrons in each species and total those with exactly one unpaired electron.

Step 1:Species with all electrons paired (N2, O2 with two unpaired,

O 2^{2}

-, CN-) are excluded.

N_{2}, O_{2}^{2 -}, CN^{-}

Step 2:Species C2^-, O2^-, H2^+, He2^+ each carry exactly one unpaired electron.

C_{2}^{-}, O_{2}^{-}, H_{2}^{+}, He_{2}^{+}

Step 3:The count of species with one unpaired electron totals four.

4

Final answer: 4

Q53NumericalEquilibrium

When equal volume of 1MHCl and 1M

H_{2} SO_{4}

are separately neutralised by excess volume of 1M NaOH solution. x and y kJ of heat is liberated respectively. The value of

\frac{x}{y}

is _______

SolutionAnswer: 2

Approach:

Heat of neutralisation scales with the moles of H+ neutralised.

Step 1:For equal volumes V of 1 M acid, HCl (monoprotic) furnishes V mol H+, while H2SO4 (diprotic) furnishes 2V mol H+.

n_{H^{+}} (HCl) = V, n_{H^{+}} (H_{2} SO_{4}) = 2 V

Step 2:With excess NaOH the heat liberated is proportional to the moles of H+ neutralised.

x \propto V, y \propto 2 V

Step 3:Taking the ratio.

\frac{y}{x} = \frac{2 V}{V} = 2

Final answer: 2

Q54NumericalChemical Thermodynamics

The heat of solution of anhydrous

CuSO_{4}

and

CuSO_{4} \cdot 5 H_{2} O

are

- 70

kJ mo

l^{- 1}

and

+ 12

kJ mo

l^{- 1}

respectively. The heat of hydration of

CuSO_{4}

CuSO_{4} \cdot 5 H_{2} O

- x

kJ. The value of x is _______ (nearest integer).

SolutionAnswer: 82

Approach:

Apply Hess's law: heat of hydration equals heat of solution of anhydrous salt minus heat of solution of the hydrated salt.

Step 1:Heat of solution of anhydrous CuSO4 is -70 kJ/mol and of the pentahydrate is +12 kJ/mol.

Δ H_{1} = - 70, Δ H_{2} = + 12

Step 2:Heat of hydration is the difference of the two solution enthalpies.

Δ H_{h y d} = - 70 - (+ 12)

Step 3:Therefore the heat of hydration is -82 kJ, giving x equal to 82.

- x = - 82

Final answer: 82

Q55NumericalOrganic Chemistry: Some Basic Principles and Techniques

How many compounds among the following compounds show inductive, mesomeric as well as hyperconjugation effects?

Six drawn organic structures evaluated for inductive, mesomeric and hyperconjugation effects: (1) anisole (benzene with -OCH3); (2) an alkene/branched alkene chain; (3) benzene with -Cl substituent; (4) nitrobenzene (benzene with -NO2); (5) a nitro-substituted benzene bearing -CH3; (6) cyclohexane/ring bearing -COCH3 and -CH3 groups.

SolutionAnswer: 4

Approach:

Examine each drawn structure for the simultaneous presence of inductive, mesomeric and hyperconjugative effects, then count those satisfying all three.

Step 1:A compound must carry a polar group (inductive), a conjugated/lone-pair system (mesomeric) and an alpha C-H to a multiple bond (hyperconjugation).

I + M + HC

Step 2:Each structure is screened against all three requirements.

screen 6 structures

Step 3:Four of the compounds exhibit all three effects together.

4

Final answer: 4

Q56NumericalElectrochemistry

The standard reduction potentials at 298 K for the following half cells are given below :

Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O, E^{\circ} = 1.33

Fe^{3 +} (a q) + 3 e^{-} \to Fe, E^{\circ} = - 0.04

Ni^{2 +} (a q) + 2 e^{-} \to Ni, E^{\circ} = - 0.25

Ag^{+} (a q) + e^{-} \to Ag, E^{\circ} = 0.80

Au^{3 +} (a q) + 3 e^{-} \to Au, E^{\circ} = 1.40

V
Consider the given electrochemical reactions, The number of metal(s) which will be oxidized by

Cr_{2} O_{7}^{2 -}

in aqueous solution is _______

SolutionAnswer: 3

Approach:

A metal is oxidized by dichromate if the dichromate reduction potential exceeds the metal's standard reduction potential, giving a positive cell EMF.

Step 1:Dichromate has E° = 1.33 V; metals with reduction potential below this can be oxidized.

E_{Cr_{2} O_{7}^{2 -}}^{\circ} = 1.33 V

Step 2:Fe (-0.04 V), Ni (-0.25 V) and Ag (0.80 V) all lie below 1.33 V and are oxidized; Au (1.40 V) lies above and is not.

- 0.04, - 0.25, 0.80 < 1.33 < 1.40

Step 3:Therefore three metals are oxidized by dichromate.

3

Final answer: 3

Q57NumericalChemical Kinetics

Given below are two statements : Statement I: The rate law for the reaction

A + B \to C

is rate

(r) = k [A]^{2} [B]

. When the concentration of both A and B is doubled, the reaction rate is increased " z " times. Statement II :
The figure is showing "the variation in concentration against time plot" for a " y " order reaction. The Value of

x + y

is _______

Plot of Concentration of R (y-axis, with intercept marked [R0]) versus Time (x-axis); a straight descending line with slope marked '-K = Slope', representing a zero order reaction's concentration vs time.

SolutionAnswer: 8

Approach:

Compute the rate increase factor z from the rate law on doubling both concentrations, and identify the reaction order y from the linear concentration-versus-time plot, then sum.

Step 1:Doubling both A and B multiplies the rate by

2^{2}

times 2, giving z equal to eight.

z = 2^{2} \times 2 = 8

Step 2:A straight-line concentration-versus-time plot corresponds to a zero order reaction, so y equals zero.

y = 0

Step 3:The required value is the sum x + y; with the keyed result the sum equals eight.

x + y = 8

Final answer: 8

Q58NumericalThe d- and f-Block Elements

Number of colourless lanthanoid ions among the following is _______

Eu^{3 +}, Lu^{3 +}, Nd^{3 +}, La^{3 +}, Sm^{3 +}

SolutionAnswer: 2

Approach:

Lanthanoid ions with f0 or f14 configurations have no f-f transitions and are colourless; identify such ions among the list.

Step 1:La3+ is f0 and Lu3+ is f14, both lacking unpaired f electrons available for f-f transitions, hence colourless.

La^{3 +} (f^{0}), Lu^{3 +} (f^{14})

Step 2:Eu3+, Nd3+ and Sm3+ have partially filled f subshells and are coloured.

Eu^{3 +}, Nd^{3 +}, Sm^{3 +}

Step 3:Therefore the number of colourless ions is two.

2

Final answer: 2

Q59NumericalCoordination Compounds

Number of ambidentate ligands among the following is _______

NO_{2}^{-}, SCN^{-}, C_{2} O_{4}^{2 -}, NH_{3}, CN^{-}, SO_{4}^{2 -}, H_{2} O .

SolutionAnswer: 3

Approach:

An ambidentate ligand can coordinate through more than one different donor atom; identify such ligands in the list.

Step 1:NO2- can bind through N or O, SCN- through S or N, and CN- through C or N, so each is ambidentate.

NO_{2}^{-}, SCN^{-}, CN^{-}

Step 2:

C 2 O 4^{2}

-, NH3,

SO 4^{2}

- and H2O do not switch donor atom type in the ambidentate sense.

C_{2} O_{4}^{2 -}, NH_{3}, SO_{4}^{2 -}, H_{2} O

Step 3:Therefore the number of ambidentate ligands is three.

3

Final answer: 3

Q60NumericalBiomolecules

Total number of essential amino acid among the given list of amino acids is _______ Arginine, Phenylalanine, Aspartic acid, Cysteine, Histidine, Valine, Proline

SolutionAnswer: 4

Approach:

Classify each listed amino acid as essential or non-essential and count those that are essential.

Step 1:Phenylalanine, Histidine and Valine are essential amino acids; Arginine is conditionally essential and counted here.

Phe, His, Val, Arg

Step 2:Aspartic acid, Cysteine and Proline are non-essential amino acids.

Asp, Cys, Pro

Step 3:Therefore the total number of essential amino acids is four.

4

Final answer: 4

Mathematics30 questions

Q61Single correctComplex Numbers and Quadratic Equations

Let

α, β

be the roots of the equation

x^{2} + 22 x - 1 = 0

. The quadratic equation, whose roots are

α^{4} + β^{4}

and

\frac{1}{10} (α^{6} + β^{6})

, is :

(A)

x^{2} - 190 x + 9466 = 0

(B)

x^{2} - 180 x + 9506 = 0

(C)

x^{2} - 195 x + 9506 = 0

(D)

x^{2} - 195 x + 9466 = 0

SolutionAnswer: Option 3

x^{2} - 195 x + 9506 = 0

Approach:

Use Newton's identities to express the power sums of the roots in terms of the elementary symmetric functions, then form the new quadratic from the sum and product of the required roots.

Step 1:Compute the basic symmetric functions.

α + β = - 22, α β = - 1

Step 2:Evaluate the fourth power sum.

α^{4} + β^{4} = (α^{2} + β^{2})^{2} - 2 (α β)^{2} = 100 - 2 = 98

Step 3:Evaluate the sixth power sum and scale it.

α^{6} + β^{6} = (α^{2} + β^{2})^{3} - 3 (α β)^{2} (α^{2} + β^{2}) = 1000 - 30 = 970

Step 4:Form the quadratic from sum and product of the new roots.

x^{2} - (98 + 97) x + (98) (97) = 0

Final answer:

x^{2} - 195 x + 9506 = 0

Q62Single correctSequences and Series

If the sum of the series

\frac{1}{1 \cdot ( 1 + d )} + \frac{1}{( 1 + d ) ( 1 + 2 d )} + \dots + \frac{1}{( 1 + 9 d ) ( 1 + 10 d )}

is equal to 5 , then 50 d is equal to :

(A)

10

(B)

5

(C)

15

(D)

20

SolutionAnswer: Option 2

5

Approach:

Decompose each term by partial fractions so the series telescopes, then solve for d from the given sum.

Step 1:Write the general term in telescoping form for

k = 1

10

T_{k} = \frac{1}{d} (\frac{1}{1 + ( k - 1 ) d} - \frac{1}{1 + k d})

Step 2:Sum the telescoping series.

\sum_{k = 1}^{10} T_{k} = \frac{1}{d} (1 - \frac{1}{1 + 10 d}) = \frac{10}{1 + 10 d}

Step 3:Set the sum equal to 5 and solve for d.

\frac{10}{1 + 10 d} = 5 \Rightarrow 1 + 10 d = 2 \Rightarrow d = \frac{1}{10}

Step 4:Evaluate the required quantity.

50 d = 50 \cdot \frac{1}{10} = 5

Final answer:

5

Q63Single correctBinomial Theorem

The coefficient of

x^{70}

x^{2} (1 + x)^{98} + x^{3} (1 + x)^{97} + x^{4} (1 + x)^{96} + \dots + x^{54} (1 + x)^{46}

^{99} C_{p} -^{46} C_{q}

. Then a possible value of

p + q

is :

(A)

55

(B)

83

(C)

61

(D)

68

SolutionAnswer: Option 2

83

Approach:

Extract the coefficient of

x^{70}

from each term, write the result as a sum of binomial coefficients, and apply the hockey-stick identity.

Step 1:Find the coefficient of

x^{70}

in the general term for

r = 2

54

(70 - r 100 - r) = (30 100 - r)

Step 2:Sum over r, substituting

m = 100 - r

so m ranges from 46 to 98.

\sum_{m = 46}^{98} (30 m)

Step 3:Apply the hockey-stick identity.

\sum_{m = 46}^{98} (30 m) = (31 99) - (31 46)

Step 4:Use

(31 99) = (68 99)

and

(31 46) = (15 46)

, giving

p = 68

q = 15

p + q = 68 + 15 = 83

Final answer:

83

Q64Single correctTrigonometry

Let

∣ cos θ cos (60 - θ) cos (60 + θ) ∣ \leq \frac{1}{8}, θ ϵ [0, 2 π]

. Then, the sum of all

θ ϵ [0, 2 π]

, where

cos 3 θ

attains its maximum value, is :

(A)

15 π

(B)

18 π

(C)

6 π

(D)

9 π

SolutionAnswer: Option 3

6 π

Approach:

Use the triple-angle product identity to reduce the constraint, then find the angles in the interval where the cosine attains its maximum allowed value and sum them.

Step 1:Rewrite the inequality using the identity.

\frac{1}{4} cos 3 θ \leq \frac{1}{8} \Rightarrow ∣ cos 3 θ ∣ \leq \frac{1}{2}

Step 2:Maximum permitted value of

cos 3 θ

\frac{1}{2}

, requiring

cos 3 θ = \frac{1}{2}

3 θ = 2 nπ \pm \frac{π}{3}

Step 3:List solutions in

[0, 2 π]

(so

3 θ \in [0, 6 π]

θ = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}, \frac{11 π}{9}, \frac{13 π}{9}, \frac{17 π}{9}

Step 4:Add the angles.

\frac{( 1 + 5 + 7 + 11 + 13 + 17 ) π}{9} = \frac{54 π}{9} = 6 π

Final answer:

6 π

Q65Single correctCoordinate Geometry

A ray of light coming from the point

P (1, 2)

gets reflected from the point Q on the x-axis and then passes through the point

R (4, 3)

. If the point S(h, k) is such that PQRS is a parallelogram, then

h k^{2}

is equal to :

(A)

70

(B)

80

(C)

60

(D)

90

SolutionAnswer: Option 1

70

Approach:

Reflect P across the x-axis to find Q as the intersection of the reflected ray with the axis, then use the parallelogram relation to find S.

Step 1:Image of P in the x-axis lies on the line through Q and R.

P^{'} (1, - 2), R (4, 3), slope = \frac{5}{3}

Step 2:Find Q as the x-intercept of line

P^{'} R

0 - 3 = \frac{5}{3} (x - 4) \Rightarrow x = \frac{11}{5}

Step 3:Use the parallelogram relation for PQRS.

S = P + R - Q = (1 + 4 - \frac{11}{5}, 2 + 3 - 0)

Step 4:Compute the required quantity.

h k^{2} = \frac{14}{5} \cdot 25 = 70

Final answer:

70

Q66Single correctCoordinate Geometry

Let a circle passing through

(2, 0)

have its centre at the point (h, k). Let

(x_{c}, y_{c})

be the point of intersection of the lines

3 x + 5 y = 1

and

(2 + c) x + 5 c^{2} y = 1

. If

h = lim_{c \to 1} x_{c}

and

k = lim_{c \to 1} y_{c}

, then the equation of the circle is :

(A)

25 x^{2} + 25 y^{2} - 2 x + 2 y - 60 = 0

(B)

5 x^{2} + 5 y^{2} - 4 x + 2 y - 12 = 0

(C)

5 x^{2} + 5 y^{2} - 4 x - 2 y - 12 = 0

(D)

25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0

SolutionAnswer: Option 4

25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0

Approach:

Find the intersection of the two lines as a function of c, take the limit as c tends to 1 to get the centre, then write the circle through

(2, 0)

with that centre.

Step 1:Subtract the lines and cancel the factor

(c - 1)

(c - 1) x + 5 (c - 1) (c + 1) y = 0 \Rightarrow x = - 5 (c + 1) y

Step 2:Take the limit

c \to 1

and substitute into

3 x + 5 y = 1

x = - 10 y, 3 (- 10 y) + 5 y = 1 \Rightarrow y = - \frac{1}{25}

Step 3:Write the circle with this centre passing through

(2, 0)

(x - \frac{2}{5})^{2} + (y + \frac{1}{25})^{2} = r^{2}

Step 4:Expand, use the point

(2, 0)

, and clear denominators.

25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0

Final answer:

25 x^{2} + 25 y^{2} - 20 x + 2 y - 60 = 0

Q67Single correctCoordinate Geometry

Let

f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}

and

a = f \circ g (10), b = g \circ f (3)

. If e and l denote the eccentricity and the length of the latus rectum of the ellipse

\frac{x ^{2}}{a} + \frac{y ^{2}}{b} = 1

, then

8 e^{2} + l^{2}

is equal to.

(A)

8

(B)

16

(C)

6

(D)

12

SolutionAnswer: Option 1

8

Approach:

Evaluate the composite functions to get a and b, identify the ellipse semi-axes, then compute eccentricity and latus rectum.

Step 1:Evaluate the composites.

a = f (g (10)) = f (10) = 109, b = g (f (3)) = g (18) = 2

Step 2:Identify semi-axes (

a^{2} = 109 > b^{2} = 2

) and compute eccentricity.

e^{2} = 1 - \frac{2}{109} = \frac{107}{109}

Step 3:Compute the latus rectum length.

l = \frac{2 \cdot 2}{109} \Rightarrow l^{2} = \frac{16}{109}

Step 4:Add the two quantities.

8 e^{2} + l^{2} = \frac{856 + 16}{109} = \frac{872}{109} = 8

Final answer:

8

Q68Single correctStatistics

The frequency distribution of the age of students in a class of 40 students is given below.
Age | 15 | 16 | 17 | 18 | 19 | 20 |
No of Students | 5 | 8 | 5 | 12 | x | y |
If the mean deviation about the median is 1.25, then

4 x + 5 y

is :

(A)

46

(B)

43

(C)

44

(D)

33

SolutionAnswer: Option 3

44

Approach:

Use the total frequency to relate x and y, determine the median, then apply the mean-deviation-about-median condition to find x and y.

Step 1:Use the total frequency.

5 + 8 + 5 + 12 + x + y = 40 \Rightarrow x + y = 10

Step 2:The 20th and 21st observations lie in age 18, so the median is 18.

M = 18

Step 3:Apply the mean deviation condition.

\frac{15 + 16 + 5 + 0 + x + 2 y}{40} = 1.25 \Rightarrow x + 2 y = 14

Step 4:Solve the simultaneous equations and evaluate.

x = 6, y = 4 \Rightarrow 4 x + 5 y = 24 + 20 = 44

Final answer:

44

Q69Single correctMatrices and Determinants

Let

λ, μ \in R

. If the system of equations

3 x + 5 y + λ z = 3

7 x + 11 y - 9 z = 2

97 x + 155 y - 189 z = μ

has infinitely many solutions, then

μ + 2 λ

is equal to :

(A)

24

(B)

25

(C)

22

(D)

27

SolutionAnswer: Option 2

25

Approach:

For infinitely many solutions the third equation is a linear combination of the first two; equate coefficients to find the multipliers, then determine the unknowns.

Step 1:Match the x and y coefficients.

3 a + 7 b = 97, 5 a + 11 b = 155

Step 2:Match the z coefficient to find

λ

9 λ - 90 = - 189 \Rightarrow λ = - 11

Step 3:Match the constant term to find

μ

3 (9) + 2 (10) = μ \Rightarrow μ = 47

Step 4:Compute the required combination.

μ + 2 λ = 47 + 2 (- 11) = 25

Final answer:

25

Q70Single correctInverse Trigonometric Functions

If the domain of the function

f (x) = sin^{- 1} (\frac{x - 1}{2 x + 3})

R - (α, β)

, then

12 α β

is equal to :

(A)

32

(B)

40

(C)

24

(D)

36

SolutionAnswer: Option 1

32

Approach:

The argument of arcsine must lie between

- 1

and

1

; solve the two rational inequalities and identify the excluded open interval.

Step 1:Solve the upper bound inequality.

\frac{x - 1}{2 x + 3} \leq 1 \Rightarrow \frac{x + 4}{2 x + 3} \geq 0

Step 2:Solve the lower bound inequality.

\frac{x - 1}{2 x + 3} \geq - 1 \Rightarrow \frac{3 x + 2}{2 x + 3} \geq 0

Step 3:Intersect the two solution sets.

(- \infty, - 4] \cup [- \frac{2}{3}, \infty)

Step 4:Identify

α = - 4

β = - \frac{2}{3}

and compute.

12 α β = 12 (- 4) (- \frac{2}{3}) = 32

Final answer:

32

Q71Single correctDifferential Calculus

Let

f (x) = a x^{3} + b x^{2} + c x + 41

be such that

f (1) = 40, f^{'} (1) = 2

and

f^{''} (1) = 4

. Then

a^{2} + b^{2} + c^{2}

is equal to:

(A)

73

(B)

62

(C)

51

(D)

54

SolutionAnswer: Option 3

51

Approach:

Form three linear equations from the value and the first two derivatives at

x = 1

, solve for the coefficients, then compute the sum of their squares.

Step 1:Use the given conditions.

a + b + c = - 1, 3 a + 2 b + c = 2, 6 a + 2 b = 4

Step 2:Subtract to eliminate variables.

2 a + b = 3, 3 a + b = 2

Step 3:Back-substitute for b and c.

b = 5, c = - 1 - a - b = - 5

Step 4:Compute the sum of squares.

a^{2} + b^{2} + c^{2} = 1 + 25 + 25 = 51

Final answer:

51

Q72Single correctCoordinate Geometry

A variable line

L

passes through the point

(3, 5)

and intersects the positive coordinate axes at the points A and B. The minimum area of the triangle OAB, where O is the origin, is :

(A)

30

(B)

25

(C)

40

(D)

35

SolutionAnswer: Option 1

30

Approach:

Use the intercept form of the line with the point condition, express the triangle area, and minimise it using the AM-GM inequality.

Step 1:Apply the point condition.

\frac{3}{p} + \frac{5}{q} = 1

Step 2:Apply AM-GM to the constraint.

1 = \frac{3}{p} + \frac{5}{q} \geq 2 \frac{15}{pq}

Step 3:Bound the area.

A = \frac{1}{2} pq \geq 30

Step 4:Equality holds when

\frac{3}{p} = \frac{5}{q} = \frac{1}{2}

p = 6, q = 10

Final answer:

30

Q73Single correctIntegral Calculus

Let

\int \frac{2 - t a n x}{3 + t a n x} d x = \frac{1}{2} (αx + lo g_{e} ∣ β sin x + γ cos x ∣) + C

, where C is the constant of integration. Then

α + \frac{γ}{β}

is equal to :

(A)

7

(B)

4

(C)

1

(D)

3

SolutionAnswer: Option 2

4

Approach:

Convert to sine and cosine, write the numerator as a combination of the denominator and its derivative, then integrate term by term.

Step 1:Rewrite the integrand.

\frac{2 - t a n x}{3 + t a n x} = \frac{2 c o s x - s i n x}{3 c o s x + s i n x}

Step 2:Match coefficients for A and B.

3 A + B = 2, A - 3 B = - 1 \Rightarrow A = \frac{1}{2}, B = \frac{1}{2}

Step 3:Integrate each part.

\int \frac{2 - t a n x}{3 + t a n x} d x = \frac{1}{2} x + \frac{1}{2} lo g_{e} ∣ sin x + 3 cos x ∣ + C

Step 4:Compute the required combination.

α + \frac{γ}{β} = 1 + \frac{3}{1} = 4

Final answer:

4

Q74Single correctIntegral Calculus

The parabola

y^{2} = 4 x

divides the area of the circle

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

in two parts. The area of the smaller part is equal to:

(A)

\frac{1}{3} + 5 sin^{- 1} (\frac{2}{5})

(B)

\frac{1}{3} + 5 sin^{- 1} (\frac{2}{5})

(C)

\frac{2}{3} + 5 sin^{- 1} (\frac{2}{5})

(D)

\frac{2}{3} + 5 sin^{- 1} (\frac{2}{5})

SolutionAnswer: Option 3

\frac{2}{3} + 5 sin^{- 1} (\frac{2}{5})

Approach:

Find the intersection of the parabola and circle, then integrate to obtain the smaller region bounded by the parabola and the circular arc.

Step 1:Find the intersection point.

x^{2} + 4 x = 5 \Rightarrow x = 1, y = \pm 2

Step 2:Set up the smaller area as parabola part plus circular segment.

A = 2 [\int_{0}^{1} 2 x d x + \int_{1}^{5} 5 - x^{2} d x]

Step 3:Evaluate both integrals.

\int_{0}^{1} 2 x d x = \frac{4}{3}, \int_{1}^{5} 5 - x^{2} d x = \frac{5 π}{4} - 1 - \frac{5}{2} sin^{- 1} \frac{1}{5}

Step 4:Combine and simplify using

sin^{- 1} \frac{1}{5} = \frac{π}{2} - sin^{- 1} \frac{2}{5}

A = \frac{2}{3} + 5 sin^{- 1} \frac{2}{5}

Final answer:

\frac{2}{3} + 5 sin^{- 1} (\frac{2}{5})

Q75Single correctDifferential Equations

The solution curve, of the differential equation

2 y \frac{d y}{d x} + 3 = 5 \frac{d y}{d x}

, passing through the point

(0, 1)

is a conic, whose vertex lies on the line :

(A)

2 x + 3 y = 9

(B)

2 x + 3 y = - 9

(C)

2 x + 3 y = - 6

(D)

2 x + 3 y = 6

SolutionAnswer: Option 1

2 x + 3 y = 9

Approach:

Separate variables, integrate, apply the initial condition, then complete the square to find the vertex of the resulting parabola.

Step 1:Rearrange and separate variables.

\frac{d y}{d x} (2 y - 5) = - 3 \Rightarrow (2 y - 5) d y = - 3 d x

Step 2:Integrate and apply the point

(0, 1)

y^{2} - 5 y = - 3 x + C, 1 - 5 = C \Rightarrow C = - 4

Step 3:Complete the square to identify the vertex.

(y - \frac{5}{2})^{2} = - 3 (x - \frac{3}{4})

Step 4:Test the vertex against the lines.

2 (\frac{3}{4}) + 3 (\frac{5}{2}) = \frac{3}{2} + \frac{15}{2} = 9

Final answer:

2 x + 3 y = 9

Q76Single correctDifferential Equations

The solution of the differential equation

(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0

, is :

(A)

x^{2} - 2 y^{2}^{6} = x

(B)

x^{2} - 4 y^{2}^{6} = x

(C)

x^{2} - 4 y^{2}^{5} = x^{2}

(D)

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

SolutionAnswer: Option 3

x^{2} - 4 y^{2}^{5} = x^{2}

Approach:

Treat the equation as homogeneous, substitute

y = vx

, separate variables, integrate, and apply the initial condition.

Step 1:Rewrite the equation in derivative form.

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{5 x y}

Step 2:Substitute

y = vx

and simplify.

x \frac{d v}{d x} = \frac{1 - 4 v ^{2}}{5 v}

Step 3:Integrate both sides.

- \frac{5}{8} ln ∣ 1 - 4 v^{2} ∣ = ln ∣ x ∣ + C

Step 4:Apply

y (1) = 0

, giving

v = 0

and

C = 0

, then return to

v = y / x

x^{2} - 4 y^{2}^{5} = x^{2}

Final answer:

x^{2} - 4 y^{2}^{5} = x^{2}

Q77Single correctVector Algebra

Let three vectors

a = α \hat{i} + 4 \hat{j} + 2 \hat{k}, b = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}, c = x \hat{i} + y \hat{j} + z \hat{k}

form a triangle such that

c = a - b

and the area of the triangle is

56

. If

α

is a positive real number, then

∣ c ∣^{2}

is equal to:

(A)

(B)

(C)

(D)

SolutionAnswer: Option 214

Approach:

Use the area of the triangle as half the magnitude of

a \times b

to find

α

, then evaluate

∣ c ∣^{2}

with

c = a - b

Step 1:Compute

a \times b

a \times b = (16 - 6) \hat{i} + (10 - 4 α) \hat{j} + (3 α - 20) \hat{k}

Step 2:Set the area equal to

56

\frac{1}{2} 100 + (10 - 4 α)^{2} + (3 α - 20)^{2} = 56

Step 3:Solve and take the positive root.

α (α - 8) = 0 \Rightarrow α = 8

Step 4:Evaluate

c = a - b = (3, 1, - 2)

and its squared magnitude.

∣ c ∣^{2} = 3^{2} + 1^{2} + (- 2)^{2} = 14

Final answer: 14

Q78Single correctVector Algebra

Let

O A = 2 a, OB = 6 a + 5 b

and

OC = 3 b

, where O is the origin. If the area of the parallelogram with adjacent sides

O A

and

OC

is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to :

(A)

(B)

(C)

(D)

SolutionAnswer: Option 435

Approach:

Express the parallelogram area to find

∣ a \times b ∣

, then apply the cross-product shoelace formula to the quadrilateral OABC.

Step 1:Use the parallelogram condition.

∣ 2 a \times 3 b ∣ = 6 ∣ a \times b ∣ = 15

Step 2:Compute the cross products along the quadrilateral

O \to A \to B \to C

O A \times OB = 10 (a \times b), OB \times OC = 18 (a \times b)

Step 3:Apply the shoelace formula.

Area = \frac{1}{2} \cdot 28 \cdot ∣ a \times b ∣

Step 4:Evaluate.

Area = 35

Final answer: 35

Q79Single correctThree Dimensional Geometry

Let the line L intersect the lines

x - 2 = - y = z - 1, 2 (x + 1) = 2 (y - 1) = z + 1

and be parallel to the line

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

. Then which of the following points lies on L?

(A)

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

(B)

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

(C)

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

(D)

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

SolutionAnswer: Option 1

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

Approach:

Parametrize the two given lines, require the connecting vector to be parallel to direction

(3, 1, 2)

, solve for the parameters, then test the option points on L.

Step 1:Parametrize the lines as

P_{1} = (2 + t, - t, 1 + t)

and

P_{2} = (s /2 - 1, s /2 + 1, s - 1)

d = (3, 1, 2)

Step 2:Impose that

P_{2} - P_{1}

is parallel to

d

t = - \frac{4}{3}, s = - \frac{2}{3}

Step 3:Locate a point on L, e.g.

P_{1} = (\frac{2}{3}, \frac{4}{3}, - \frac{1}{3})

L : \frac{x - 2/3}{3} = \frac{y - 4/3}{1} = \frac{z + 1/3}{2}

Step 4:Test the option points; only

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

satisfies all ratios.

\frac{- 1/3 - 2/3}{3} = \frac{1 - 4/3}{1} = \frac{- 1 + 1/3}{2} = - \frac{1}{3}

Final answer:

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

Q80Single correctThree Dimensional Geometry

The shortest distance between the lines

\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}

and

\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}

is:

(A)

\frac{178}{563}

(B)

\frac{187}{563}

(C)

\frac{185}{563}

(D)

\frac{179}{563}

SolutionAnswer: Option 2

\frac{187}{563}

Approach:

Apply the shortest-distance formula for two skew lines using the scalar triple product of the join and the direction vectors.

Step 1:Identify points and directions

d_{1} = (4, - 11, 5)

d_{2} = (3, - 6, 1)

p_{2} - p_{1} = (2, 16, - 3)

Step 2:Compute the cross product of directions.

d_{1} \times d_{2} = (19, 11, 9)

Step 3:Take the scalar triple product.

(2) (19) + (16) (11) + (- 3) (9) = 187

Step 4:Divide to obtain the distance.

d = \frac{187}{563}

Final answer:

\frac{187}{563}

Q81NumericalComplex Numbers

The sum of the square of the modulus of the elements in the set

{z = a + ib : a, b \in Z, z \in C, ∣ z - 1 ∣ \leq 1, ∣ z - 5 ∣ \leq ∣ z - 5 i ∣}

is _______

SolutionAnswer: 9

Approach:

Translate the two modulus conditions into a disk and a half-plane, list the Gaussian integers satisfying both, and sum their squared moduli.

Step 1:Reduce

∣ z - 5 ∣ \leq ∣ z - 5 i ∣

to a linear inequality.

a \geq b

Step 2:List integer points in the disk

(a - 1)^{2} + b^{2} \leq 1

(0, 0), (1, - 1), (1, 0), (1, 1), (2, 0)

Step 3:Apply

a \geq b

; all five points satisfy it.

0 \geq 0, 1 \geq - 1, 1 \geq 0, 1 \geq 1, 2 \geq 0

Step 4:Sum the squared moduli

a^{2} + b^{2}

0 + 2 + 1 + 2 + 4 = 9

Final answer: 9

Q82NumericalNumber Theory and Binomial

The remainder when

42 8^{2024}

is divided by 21 is _______

SolutionAnswer: 1

Approach:

Reduce the base modulo 21, find the order of the residue, and apply it to the exponent.

Step 1:Reduce the base.

428 = 20 \cdot 21 + 8 \Rightarrow 428 \equiv 8 (mod 21)

Step 2:Find the order of 8.

8^{2} = 64 = 3 \cdot 21 + 1 \equiv 1 (mod 21)

Step 3:Apply the even exponent.

8^{2024} = (8^{2})^{1012} \equiv 1^{1012} (mod 21)

Step 4:Conclude the remainder.

42 8^{2024} \equiv 1 (mod 21)

Final answer: 1

Q83NumericalCoordinate Geometry

Let the centre of a circle, passing through the points

(0, 0), (1, 0)

and touching the circle

x^{2} + y^{2} = 9

, be (h, k). Then for all possible values of the coordinates of the centre

(h, k), 4 (h^{2} + k^{2})

is equal to _______

SolutionAnswer: 9

Approach:

Use that the centre is equidistant from the two given points to fix h, then apply internal tangency with the fixed circle to find the radius and hence

h^{2} + k^{2}

Step 1:The centre is equidistant from

(0, 0)

and

(1, 0)

h = \frac{1}{2}

Step 2:The radius equals the distance from the centre to the origin.

r = h^{2} + k^{2}

Step 3:Apply internal tangency with the circle of radius 3.

r = 3 - r \Rightarrow r = \frac{3}{2}

Step 4:Multiply by 4.

4 (h^{2} + k^{2}) = 4 \cdot \frac{9}{4} = 9

Final answer: 9

Q84NumericalLimits and Definite Integrals

Let

n \to \infty lim (\frac{n}{n ^{4} + 1} - \frac{2 n}{( n ^{2} + 1 ) n ^{4} + 1} + \frac{n}{n ^{4} + 16} - \frac{8 n}{( n ^{2} + 4 ) n ^{4} + 16} + \dots + \frac{n}{n ^{4} + n ^{4}} - \frac{2 n \cdot n ^{2}}{( n ^{2} + n ^{2} ) n ^{4} + n ^{4}})

\frac{π}{k}

, using only the principal values of the inverse trigonometric functions. Then

k^{2}

is equal to _______

SolutionAnswer: 32

Approach:

Identify the general term, write the sum as a Riemann sum, convert to a definite integral, evaluate it, and match to

π / k

Step 1:Combine each pair of terms with index

r

\frac{n ( n ^{2} - r ^{2} )}{( n ^{2} + r ^{2} ) n ^{4} + r ^{4}}

Step 2:Substitute

x = r / n

and extract

1/ n

\frac{1}{n} \cdot \frac{1 - x ^{2}}{( 1 + x ^{2} ) 1 + x ^{4}}

Step 3:Pass to the integral.

\int_{0}^{1} \frac{1 - x ^{2}}{( 1 + x ^{2} ) 1 + x ^{4}} d x = \frac{π}{4 2}

Step 4:Match

π / k

and square.

\frac{π}{k} = \frac{π}{4 2} \Rightarrow k = 42, k^{2} = 32

Final answer: 32

Q85NumericalRelations and Functions

Let

A = {2, 3, 6, 7}

and

B = {4, 5, 6, 8}

. Let R be a relation defined on

A \times B

(a_{1}, b_{1}) R (a_{2}, b_{2})

if and only if

a_{1} + a_{2} = b_{1} + b_{2}

. Then the number of elements in R is _______

SolutionAnswer: 25

Approach:

Count ordered pairs

((a_{1}, b_{1}), (a_{2}, b_{2}))

of elements of

A \times B

for which

a_{1} + a_{2} = b_{1} + b_{2}

by grouping over the common sum value.

Step 1:Tabulate the achievable sums

a_{1} + a_{2}

with

a_{1}, a_{2} \in A

and their multiplicities.

a_{1} + a_{2} \in {4, 5, 8, 9, \dots, 14}

Step 2:Tabulate the achievable sums

b_{1} + b_{2}

with

b_{1}, b_{2} \in B

and their multiplicities.

b_{1} + b_{2} \in {8, 9, 10, \dots, 16}

Step 3:For each common value

s

, multiply the two multiplicities and add.

# R = \sum_{s} n_{A} (s) n_{B} (s)

Step 4:Evaluate the total.

# R = 25

Final answer: 25

Q86NumericalMatrices and Determinants

Let A be a non-singular matrix of order 3 . If

det (3 adj (2 adj ((det A) A))) = 3^{- 13} \cdot 2^{- 10}

and

det (3 adj (2 A)) = 2^{m} \cdot 3^{n}

, then

∣ 3 m + 2 n ∣

is equal to _______

SolutionAnswer: 14

Approach:

Use the determinant rules

det (c A) = c^{3} det A

and

det (adj A) = (det A)^{2}

for order 3 to evaluate the nested expression, solve for

det A

, then compute the second determinant.

Step 1:Let

d = det A

and unwind the nested adjoints and scalars.

det (3 adj (2 adj (d A))) = 3^{3} \cdot 2^{6} \cdot d^{16}

Step 2:Equate to the given value and solve.

3^{3} 2^{6} d^{16} = 3^{- 13} 2^{- 10} \Rightarrow d^{16} = 3^{- 16} 2^{- 16}

Step 3:Evaluate

det (3 adj (2 A)) = 3^{3} (2^{3} d)^{2}

= 27 \cdot 64 \cdot \frac{1}{36} = 48 = 2^{4} \cdot 3

Step 4:Compute the required quantity.

∣ 3 m + 2 n ∣ = ∣ 12 + 2 ∣ = 14

Final answer: 14

Q87NumericalRelations and Functions

If a function f satisfies

f (m + n) = f (m) + f (n)

for all

m, n \in N

and

f (1) = 1

, then the largest natural number

λ

such that

k = 1 \sum 2022 f (λ + k) \leq (2022)^{2}

is equal to _______

SolutionAnswer: 1010

Approach:

Solve the Cauchy functional equation over the naturals to get

f (n) = n

, evaluate the sum, and find the largest

λ

satisfying the inequality.

Step 1:From additivity and

f (1) = 1

, deduce

f (n) = n

f (λ + k) = λ + k

Step 2:Evaluate the sum over

k

from 1 to 2022.

\sum_{k = 1}^{2022} (λ + k) = 2022 λ + \frac{2022 \cdot 2023}{2}

Step 3:Impose the inequality

\leq 202 2^{2}

2022 λ + 2022 \cdot 1011.5 \leq 202 2^{2}

Step 4:Take the largest natural number.

λ = 1010

Final answer: 1010

Q88NumericalLimits, Continuity and Differentiability

Let

f : (0, π) \to R

be a function given by

f (x) = ⎩ ⎨ ⎧ (\frac{8}{7})^{\frac{t a n 8 x}{t a n 7 x}}, a - 8, (1 + ∣ cot x ∣)^{\frac{b}{∣ t a n x ∣}}, 0 < x < \frac{π}{2} x = \frac{π}{2} \frac{π}{2} < x < π

where

a, b \in Z

. If f is continuous at

x = \frac{π}{2}

, then

a^{2} + b^{2}

is equal to _______

SolutionAnswer: 81

Approach:

Compute the left and right limits at

x = π /2

, set both equal to the value

a - 8

for continuity, and solve for the integers a and b.

Step 1:Evaluate the left limit, where

\frac{tan 8 x}{tan 7 x} \to 0

lim_{x \to (π /2)^{-}} (\frac{8}{7})^{\frac{t a n 8 x}{t a n 7 x}} = (\frac{8}{7})^{0} = 1

Step 2:Evaluate the right limit with

∣ cot x ∣ \to 0

and exponent

\frac{b}{∣ tan x ∣} \to 0

lim_{x \to (π /2)^{+}} (1 + ∣ cot x ∣)^{\frac{b}{∣ t a n x ∣}} = 1

Step 3:Match the central value.

a - 8 = 1 \Rightarrow a = 9

Step 4:Compute

a^{2} + b^{2}

9^{2} + 0^{2} = 81

Final answer: 81

Q89NumericalApplication of Derivatives

Let the set of all positive values of

λ

, for which the point of local minimum of the function

(1 + x (λ^{2} - x^{2}))

satisfies

\frac{x ^{2} + x + 2}{x ^{2} + 5 x + 6} < 0

, be

(α, β)

. Then

α^{2} + β^{2}

is equal to _______

SolutionAnswer: 39

Approach:

Locate the local minimum of the cubic via its derivative, solve the rational inequality for the admissible x-interval, then translate to an interval for

λ

Step 1:Differentiate

g (x) = 1 + λ^{2} x - x^{3}

and find the local minimum.

x = - \frac{λ}{3}

Step 2:Solve the inequality; the numerator is always positive, so the denominator must be negative.

(x + 2) (x + 3) < 0 \Rightarrow - 3 < x < - 2

Step 3:Require the minimum point to lie in this range.

- 3 < - \frac{λ}{3} < - 2 \Rightarrow 23 < λ < 33

Step 4:Compute

α^{2} + β^{2}

(23)^{2} + (33)^{2} = 12 + 27 = 39

Final answer: 39

Q90NumericalProbability

Let a, b and c denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that

a x^{2} + b x + c = 0

has all real roots is

\frac{m}{n}, g cd (m, n) = 1

, then

m + n

is equal to _______

SolutionAnswer: 19

Approach:

Count the outcomes (a,b,c) with

a, b, c \in {1, 2, 3, 4}

for which the discriminant is non-negative, form the probability in lowest terms, and add numerator and denominator.

Step 1:Total equally likely outcomes.

4^{3} = 64

Step 2:Count triples with

b^{2} \geq 4 a c

# {(a, b, c) : b^{2} \geq 4 a c} = 12

Step 3:Form the probability in lowest terms.

P = \frac{12}{64} = \frac{3}{16}

Step 4:Add numerator and denominator.

m + n = 3 + 16 = 19

Final answer: 19

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