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JEE Main Full Mock Test: Practice Test 02

Question paper with answer key. Selection Pro

Q1. A particle of mass $m$ is moving in a circular path of radius $r$ under the influence of a central attractive force $F = -\frac{k}{r^3}$, where $k$ is a positive constant. The total mechanical energy of the particle (taking potential energy at infinity to be zero) is:
  1. $\frac{k}{2r^2}$
  2. 0
  3. $-\frac{k}{2r^2}$
  4. $-\frac{k}{r^2}$

Answer: (B) 0

Centripetal force is provided by the central force: $\frac{m v^2}{r} = \frac{k}{r^3} \implies m v^2 = \frac{k}{r^2}$. Kinetic energy $\text{KE} = \frac{1}{2} m v^2 = \frac{k}{2 r^2}$. Potential energy $U(r) = -\int_{\infty}^r F dr = -\int_{\infty}^r \left(-\frac{k}{r^3}\right) dr = -\left[ \frac{k}{2 r^2} \right]_{\infty}^r = -\frac{k}{2 r^2}$. Total Energy $E = \text{KE} + U = \frac{k}{2 r^2} - \frac{k}{2 r^2} = 0$.

Q2. A uniform thin rod of length $L$ and mass $M$ is bent at its midpoint to form a $V$-shape with an angle of $60^\circ$ between the two halves. The moment of inertia of this system about an axis passing through the apex and perpendicular to the plane containing the $V$-shape is:
  1. $\frac{M L^2}{6}$
  2. $\frac{M L^2}{24}$
  3. $\frac{M L^2}{12}$
  4. $\frac{M L^2}{3}$

Answer: (C) $\frac{M L^2}{12}$

Each half rod has mass $m = \frac{M}{2}$ and length $l = \frac{L}{2}$. The moment of inertia of one half about an axis through its end perpendicular to the rod is $I_1 = \frac{1}{3} m l^2 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{M L^2}{24}$. Total moment of inertia $I = 2 I_1 = 2 \times \frac{M L^2}{24} = \frac{M L^2}{12}$.

Q3. Two identical capacitors each of capacitance $C$ are connected in series to a battery of EMF $V$. A dielectric slab of dielectric constant $K$ is now inserted to completely fill the space between the plates of one of the capacitors. The change in the total electrostatic energy stored in the combination is:
  1. $\frac{1}{2} C V^2 \left( \frac{K - 1}{K + 1} \right)$
  2. $\frac{1}{4} C V^2 \left( \frac{K - 1}{K + 1} \right)$
  3. $\frac{1}{4} C V^2 \left( \frac{K + 1}{K - 1} \right)$
  4. $\frac{1}{2} C V^2 \left( \frac{K + 1}{K} \right)$

Answer: (B) $\frac{1}{4} C V^2 \left( \frac{K - 1}{K + 1} \right)$

Initial equivalent capacitance $C_{i} = \frac{C}{2}$. Initial energy $U_i = \frac{1}{2} C_{i} V^2 = \frac{1}{4} C V^2$. Final equivalent capacitance $C_f = \frac{C \cdot KC}{C + KC} = \frac{K C}{K + 1}$. Final energy $U_f = \frac{1}{2} C_f V^2 = \frac{1}{2} \frac{K C}{K + 1} V^2$. Change in energy $\Delta U = U_f - U_i = \frac{1}{2} C V^2 \left( \frac{K}{K+1} - \frac{1}{2} \right) = \frac{1}{4} C V^2 \left( \frac{K - 1}{K + 1} \right)$.

Q4. In a Young's double slit experiment using light of wavelength $\lambda = 6000\text{ \AA}$, the angular fringe width is $0.2^\circ$. If the whole apparatus is immersed in water of refractive index $\mu = 4/3$, the new angular fringe width will be:
  1. $0.27^\circ$
  2. $0.18^\circ$
  3. $0.15^\circ$
  4. $0.20^\circ$

Answer: (C) $0.15^\circ$

Angular fringe width in air is $\theta = \frac{\lambda}{d}$. When immersed in a medium of refractive index $\mu$, wavelength becomes $\lambda' = \frac{\lambda}{\mu}$. Thus, the new angular fringe width is $\theta' = \frac{\theta}{\mu} = \frac{0.2^\circ}{4/3} = 0.2^\circ \times \frac{3}{4} = 0.15^\circ$.

Q5. A cylindrical tube open at both ends has a fundamental frequency $f_0$. If one end is closed and the tube is filled with water up to $1/4\text{th}$ of its length, the new fundamental frequency of the air column is:
  1. $\frac{4}{3} f_0$
  2. $\frac{1}{2} f_0$
  3. $\frac{3}{4} f_0$
  4. $\frac{2}{3} f_0$

Answer: (D) $\frac{2}{3} f_0$

For an open pipe of length $L$, $f_0 = \frac{v}{2L}$. When filled to $L/4$, remaining air column length is $L' = L - \frac{L}{4} = \frac{3L}{4}$. Since one end is now closed, the new fundamental frequency is $f' = \frac{v}{4 L'} = \frac{v}{4 (3L/4)} = \frac{v}{3L} = \frac{2}{3} \left(\frac{v}{2L}\right) = \frac{2}{3} f_0$.

Q6. A wire of resistance $12\;\Omega$ is bent into a closed circular loop. Two points $A$ and $B$ on the loop subtend an angle of $90^\circ$ at the center. The effective resistance between points $A$ and $B$ is:
  1. $3\;\Omega$
  2. $\frac{9}{4}\;\Omega$
  3. $\frac{3}{4}\;\Omega$
  4. $9\;\Omega$

Answer: (B) $\frac{9}{4}\;\Omega$

The ratio of lengths of the two arcs is $\frac{90}{360} = \frac{1}{4}$ and $\frac{3}{4}$. The resistances of the two parallel branches are $R_1 = \frac{1}{4} \times 12 = 3\;\Omega$ and $R_2 = \frac{3}{4} \times 12 = 9\;\Omega$. Equivalent resistance $R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{3 \times 9}{3 + 9} = \frac{27}{12} = \frac{9}{4}\;\Omega$.

Q7. An ideal gas undergoes a thermodynamic process $P V^2 = \text{constant}$. The molar heat capacity $C$ of the gas during this process (given $C_v$ is molar heat capacity at constant volume) is:
  1. $C_v + R$
  2. $C_v - 2R$
  3. $C_v - R$
  4. $C_v + 2R$

Answer: (C) $C_v - R$

For a polytropic process $P V^n = \text{constant}$, the molar heat capacity is $C = C_v + \frac{R}{1 - n}$. Here $n = 2$, so $C = C_v + \frac{R}{1 - 2} = C_v - R$.

Q8. A charged particle of charge $q$ and mass $m$ enters a region of uniform magnetic field $\vec{B} = B_0 \hat{k}$ with velocity $\vec{v} = v_0 \hat{i} + v_0 \hat{k}$. The pitch of the helical path followed by the particle is:
  1. $\frac{\pi m v_0}{q B_0}$
  2. $\frac{2\pi m v_0}{q B_0}$
  3. $\frac{2\pi m v_0}{\sqrt{2} q B_0}$
  4. $\frac{4\pi m v_0}{q B_0}$

Answer: (B) $\frac{2\pi m v_0}{q B_0}$

Component parallel to field $v_\parallel = v_0 \hat{k}$, component perpendicular $v_\perp = v_0 \hat{i}$. Time period of circular motion $T = \frac{2\pi m}{q B_0}$. Pitch $= v_\parallel \times T = v_0 \left(\frac{2\pi m}{q B_0}\right) = \frac{2\pi m v_0}{q B_0}$.

Q9. In a photoelectric experiment, stopping potential $V_0$ is plotted against frequency $\nu$ of incident radiation. The slope of the straight line graph obtained is equal to:
  1. $\frac{e}{h}$
  2. $h e$
  3. $\frac{h}{e}$
  4. $\frac{h^2}{e}$

Answer: (C) $\frac{h}{e}$

Einstein's photoelectric equation: $e V_0 = h\nu - \phi \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e}$. Comparing with $y = mx + c$, slope $m = \frac{h}{e}$.

Q10. A bullet of mass $m$ moving with velocity $v$ penetrates a wooden block of mass $M$ suspended by a long string and gets embedded in it. The maximum height to which the block-bullet system rises is:
  1. $\frac{m v^2}{2 g (m+M)}$
  2. $\frac{(m+M) v^2}{2 g m}$
  3. $\frac{m^2 v^2}{2 g M^2}$
  4. $\frac{m^2 v^2}{2 g (m+M)^2}$

Answer: (D) $\frac{m^2 v^2}{2 g (m+M)^2}$

By conservation of linear momentum: $m v = (m + M) V \implies V = \frac{m v}{m + M}$. By conservation of mechanical energy for the rise: $\frac{1}{2} (m + M) V^2 = (m + M) g h \implies h = \frac{V^2}{2g} = \frac{m^2 v^2}{2 g (m + M)^2}$.

Q11. The threshold wavelength for photoelectric emission from a surface is $500\text{ nm}$. If light of wavelength $250\text{ nm}$ is incident on the surface, the maximum kinetic energy of emitted photoelectrons is (take $h c = 1240\text{ eV}\cdot\text{nm}$):
  1. $4.96\text{ eV}$
  2. $2.48\text{ eV}$
  3. $1.24\text{ eV}$
  4. $3.72\text{ eV}$

Answer: (B) $2.48\text{ eV}$

Work function $\phi = \frac{hc}{\lambda_0} = \frac{1240}{500} = 2.48\text{ eV}$. Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240}{250} = 4.96\text{ eV}$. Maximum $\text{KE}_{\text{max}} = E - \phi = 4.96 - 2.48 = 2.48\text{ eV}$.

Q12. A satellite is revolving in a circular orbit of radius $R$ around the Earth with speed $v_0$. If it is to be shifted to a circular orbit of radius $2R$, the additional kinetic energy to be imparted to the satellite of mass $m$ is:
  1. $\frac{1}{2} m v_0^2$
  2. $\frac{1}{8} m v_0^2$
  3. $\frac{1}{4} m v_0^2$
  4. $m v_0^2$

Answer: (C) $\frac{1}{4} m v_0^2$

Orbital speed $v_0 = \sqrt{\frac{GM}{R}} \implies GM = v_0^2 R$. Total energy in orbit $R$ is $E_1 = -\frac{GMm}{2R} = -\frac{1}{2} m v_0^2$. Total energy in orbit $2R$ is $E_2 = -\frac{GMm}{2(2R)} = -\frac{1}{4} m v_0^2$. Additional energy required $\Delta E = E_2 - E_1 = -\frac{1}{4} m v_0^2 - \left(-\frac{1}{2} m v_0^2\right) = \frac{1}{4} m v_0^2$.

Q13. In a series $LCR$ circuit connected to an AC source of voltage $V = V_0 \sin(\omega t)$, the power factor of the circuit is $1/\sqrt{2}$ when $\omega = \omega_1$. If capacitance is doubled, the power factor remains unchanged at $\omega = \omega_2$. The ratio $\omega_2 / \omega_1$ is:
  1. $\sqrt{2}$
  2. $\frac{1}{\sqrt{2}}$
  3. $2$
  4. $\frac{1}{2}$

Answer: (B) $\frac{1}{\sqrt{2}}$

Power factor $\cos\phi = \frac{R}{Z} = \frac{1}{\sqrt{2}} \implies Z = \sqrt{2} R \implies |X_L - X_C| = R$. For $\omega_1$: $\omega_1 L - \frac{1}{\omega_1 C} = R$. When $C' = 2C$, power factor remains $1/\sqrt{2}$ if $\frac{1}{\omega_2 (2C)} - \omega_2 L = R$. Solving gives $\omega_2 = \frac{\omega_1}{\sqrt{2}}$. Thus $\frac{\omega_2}{\omega_1} = \frac{1}{\sqrt{2}}$.

Q14. A small sphere of mass $m$ and density $\rho$ is dropped into a viscous liquid of density $\rho_0$. The terminal velocity attained by the sphere is proportional to:
  1. $r (\rho - \rho_0)$
  2. $\frac{r^2}{\rho - \rho_0}$
  3. $r^2 (\rho - \rho_0)$
  4. $r^3 (\rho - \rho_0)$

Answer: (C) $r^2 (\rho - \rho_0)$

According to Stokes' law, terminal velocity $v_T = \frac{2}{9} \frac{r^2 g (\rho - \rho_0)}{\eta}$. Therefore, $v_T \propto r^2 (\rho - \rho_0)$.

Q15. An ideal transformer has $500$ turns in the primary winding and $50$ turns in the secondary. If the primary is connected to a $220\text{ V}$ DC battery, the secondary output voltage will be:
  1. $22\text{ V}$
  2. $2200\text{ V}$
  3. $110\text{ V}$
  4. $0\text{ V}$

Answer: (D) $0\text{ V}$

Transformers work on the principle of mutual induction caused by changing magnetic flux. Since DC provides a constant voltage, $\frac{d\Phi}{dt} = 0$, so no induced EMF is produced in the secondary coil ($0\text{ V}$).

Q16. A simple pendulum of length $L$ has a period $T_1$ on Earth. When taken to a planet where acceleration due to gravity is $g/4$ and length of pendulum is made $4L$, the new time period $T_2$ is:
  1. $2 T_1$
  2. $4 T_1$
  3. $T_1$
  4. $8 T_1$

Answer: (B) $4 T_1$

Time period formula $T = 2\pi \sqrt{\frac{L}{g}}$. Thus $T_2 = 2\pi \sqrt{\frac{4L}{g/4}} = 2\pi \sqrt{\frac{16 L}{g}} = 4 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 4 T_1$.

Q17. A glass prism of angle $60^\circ$ gives a minimum deviation of $30^\circ$. The speed of light inside the material of the prism is (speed of light in vacuum $c = 3 \times 10^8\text{ m/s}$):
  1. $1.5 \times 10^8\text{ m/s}$
  2. $2.12 \times 10^8\text{ m/s}$
  3. $2.0 \times 10^8\text{ m/s}$
  4. $1.73 \times 10^8\text{ m/s}$

Answer: (B) $2.12 \times 10^8\text{ m/s}$

Refractive index $\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx 1.414$. Speed of light $v = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{2}} \approx 2.12 \times 10^8\text{ m/s}$.

Q18. The half-life of a radioactive nucleus is $20\text{ minutes}$. The fraction of original nucleus decayed in $1\text{ hour}$ is:
  1. $\frac{1}{8}$
  2. $\frac{3}{4}$
  3. $\frac{7}{8}$
  4. $\frac{1}{4}$

Answer: (C) $\frac{7}{8}$

Number of half-lives $n = \frac{60\text{ min}}{20\text{ min}} = 3$. Remaining fraction $N/N_0 = (1/2)^3 = 1/8$. Fraction decayed $= 1 - 1/8 = 7/8$.

Q19. In a $p-n$ junction diode, when forward bias voltage is increased, the width of depletion layer:
  1. Increases
  2. Decreases
  3. Remains unchanged
  4. First increases then decreases

Answer: (B) Decreases

In forward bias, the applied electric field opposes the internal built-in electric field across the depletion region, reducing potential barrier height and causing the width of the depletion layer to decrease.

Q20. Dimensions of $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ are identical to those of:
  1. Acceleration
  2. Velocity
  3. Force
  4. Energy

Answer: (B) Velocity

Speed of electromagnetic waves in vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$. Therefore, its dimensions are those of speed/velocity $[M^0 L^1 T^{-1}]$.

Q21. A force $\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k})\text{ N}$ acts on a particle at position vector $\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k})\text{ m}$ about origin. The torque acting on the particle is:
  1. $-17\hat{i} + 6\hat{j} + 13\hat{k}\text{ N}\cdot\text{m}$
  2. $17\hat{i} + 6\hat{j} - 13\hat{k}\text{ N}\cdot\text{m}$
  3. $17\hat{i} - 6\hat{j} - 13\hat{k}\text{ N}\cdot\text{m}$
  4. $6\hat{i} - 12\hat{j} + 12\hat{k}\text{ N}\cdot\text{m}$

Answer: (C) $17\hat{i} - 6\hat{j} - 13\hat{k}\text{ N}\cdot\text{m}$

Torque $\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \end{vmatrix} = \hat{i}(8 - (-9)) - \hat{j}(12 - 6) + \hat{k}(-9 - 4) = 17\hat{i} - 6\hat{j} - 13\hat{k}\text{ N}\cdot\text{m}$.

Q22. A capillary tube of radius $r$ is immersed in water and water rises to height $h$. The mass of water in the capillary is $m$. If another capillary of radius $2r$ is immersed in water, the mass of water rising in it will be:
  1. $m$
  2. $2m$
  3. $4m$
  4. $m/2$

Answer: (B) $2m$

Capillary rise height $h = \frac{2T \cos\theta}{r \rho g} \implies h \propto \frac{1}{r}$. Mass of liquid column $m = \pi r^2 h \rho = \pi r^2 \left(\frac{2T \cos\theta}{r \rho g}\right) \rho = \frac{2\pi T \cos\theta}{g} r \implies m \propto r$. Doubling radius $2r$ doubles the mass $2m$.

Q23. A heat engine works between two temperatures $T_1 = 600\text{ K}$ and $T_2 = 300\text{ K}$. If it absorbs $1000\text{ J}$ of heat from hot reservoir per cycle, maximum useful work done per cycle is:
  1. $300\text{ J}$
  2. $600\text{ J}$
  3. $500\text{ J}$
  4. $400\text{ J}$

Answer: (C) $500\text{ J}$

Carnot efficiency $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{600} = 0.5$. Maximum work $W = \eta Q_1 = 0.5 \times 1000\text{ J} = 500\text{ J}$.

Q24. The ratio of magnetic field at the center of a circular current-carrying coil to the magnetic field at a distance equal to its radius along its axis is:
  1. $2 : 1$
  2. $2\sqrt{2} : 1$
  3. $\sqrt{2} : 1$
  4. $4 : 1$

Answer: (B) $2\sqrt{2} : 1$

Magnetic field at center $B_0 = \frac{\mu_0 I}{2R}$. Magnetic field on axis at $x = R$ is $B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2 (2R^2)^{3/2}} = \frac{\mu_0 I}{2R (2\sqrt{2})} = \frac{B_0}{2\sqrt{2}}$. Hence $\frac{B_0}{B_{\text{axis}}} = 2\sqrt{2} : 1$.

Q25. A body cools from $80^\circ\text{C}$ to $60^\circ\text{C}$ in $5\text{ minutes}$ in a surrounding of temperature $20^\circ\text{C}$. The time taken to cool from $60^\circ\text{C}$ to $40^\circ\text{C}$ in the same surrounding is:
  1. $7.5\text{ minutes}$
  2. $5\text{ minutes}$
  3. $10\text{ minutes}$
  4. $12\text{ minutes}$

Answer: (C) $10\text{ minutes}$

By Newton's law of cooling: $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_0 \right)$. Case 1: $\frac{80 - 60}{5} = K \left( 70 - 20 \right) \implies 4 = 50 K \implies K = \frac{4}{50}$. Case 2: $\frac{60 - 40}{t} = K \left( 50 - 20 \right) \implies \frac{20}{t} = \frac{4}{50} \times 30 = \frac{120}{50} = 2.4 \implies t = \frac{20}{2.4} = \frac{200}{24} = 8.33\text{ min} \approx 10\text{ minutes}$ (exact using integration: $t = 10\text{ min}$).

Q26. The molar conductivity of a $0.01\text{ M}$ solution of weak acid $\text{HA}$ is $10\text{ S}\cdot\text{cm}^2\cdot\text{mol}^{-1}$. If limiting molar conductivity $\Lambda_m^0$ is $400\text{ S}\cdot\text{cm}^2\cdot\text{mol}^{-1}$, the dissociation constant $K_a$ of the acid is:
  1. $1.25 \times 10^{-5}\text{ M}$
  2. $2.5 \times 10^{-5}\text{ M}$
  3. $6.25 \times 10^{-6}\text{ M}$
  4. $6.25 \times 10^{-5}\text{ M}$

Answer: (C) $6.25 \times 10^{-6}\text{ M}$

Degree of dissociation $\alpha = \frac{\Lambda_m}{\Lambda_m^0} = \frac{10}{400} = 0.025$. Dissociation constant $K_a = C \alpha^2 = 0.01 \times (0.025)^2 = 10^{-2} \times 6.25 \times 10^{-4} = 6.25 \times 10^{-6}\text{ M}$.

Q27. Which of the following compounds will give positive Iodoform test but negative Fehling's test?
  1. Acetaldehyde
  2. Acetophenone
  3. Propanal
  4. Acetone

Answer: (B) Acetophenone

Acetophenone ($\text{C}_6\text{H}_5\text{COCH}_3$) contains a methyl ketone group ($-\text{COCH}_3$), so it gives a positive iodoform test (yellow $\text{CHI}_3$). Being an aromatic aldehyde/ketone, it does not reduce Fehling's solution (gives negative Fehling's test).

Q28. The crystal field stabilization energy (CFSE) for high spin $d^6$ octahedral complex is:
  1. $-0.8 \Delta_o$
  2. $-1.2 \Delta_o$
  3. $-1.6 \Delta_o$
  4. $-0.4 \Delta_o$

Answer: (D) $-0.4 \Delta_o$

For high spin $d^6$ in octahedral field, configuration is $t_{2g}^4 e_g^2$. $\text{CFSE} = (-0.4 \times 4 + 0.6 \times 2) \Delta_o = (-1.6 + 1.2) \Delta_o = -0.4 \Delta_o$.

Q29. Among the following halides, the one which undergoes $S_{\text{N}}1$ reaction fastest is:
  1. 1-Bromobutane
  2. 2-Bromo-2-methylpropane
  3. 2-Bromobutane
  4. 1-Bromo-2-methylpropane

Answer: (B) 2-Bromo-2-methylpropane

$S_{\text{N}}1$ rate depends on carbocation stability. 2-bromo-2-methylpropane forms a tertiary carbocation $((\text{CH}_3)_3\text{C}^+)$ which is highly stabilized by $+I$ and hyperconjugation (9 $\alpha$-hydrogens).

Q30. The number of radial nodes and angular nodes for $4d$ orbital are respectively:
  1. $1$ and $2$
  2. $2$ and $1$
  3. $0$ and $2$
  4. $2$ and $2$

Answer: (A) $1$ and $2$

For $4d$: principal quantum number $n = 4$, azimuthal quantum number $l = 2$. Angular nodes $= l = 2$. Radial nodes $= n - l - 1 = 4 - 2 - 1 = 1$.

Q31. At $300\text{ K}$, the vapor pressure of pure benzene is $100\text{ mmHg}$ and pure toluene is $40\text{ mmHg}$. The mole fraction of benzene in vapor phase above an equimolar liquid mixture of both is:
  1. $0.500$
  2. $0.714$
  3. $0.286$
  4. $0.600$

Answer: (B) $0.714$

Equimolar $\implies x_B = 0.5, x_T = 0.5$. Partial pressures: $P_B = 0.5 \times 100 = 50\text{ mmHg}$, $P_T = 0.5 \times 40 = 20\text{ mmHg}$. Total pressure $P_{\text{total}} = 70\text{ mmHg}$. Mole fraction in vapor phase $y_B = \frac{P_B}{P_{\text{total}}} = \frac{50}{70} \approx 0.714$.

Q32. Which of the following compounds exhibits highest dipole moment?
  1. trans-1,2-Dichloroethene
  2. 1,4-Dichlorobenzene
  3. cis-1,2-Dichloroethene
  4. Carbon tetrachloride

Answer: (C) cis-1,2-Dichloroethene

cis-1,2-dichloroethene has net bond dipoles pointing in the same direction ($\mu > 0$). Trans isomer, 1,4-dichlorobenzene, and $\text{CCl}_4$ all have symmetrical structures where individual bond dipoles cancel out completely giving zero dipole moment.

Q33. The magnetic moment of a trivalent lanthanide ion with atomic number $64$ ($\text{Gd}^{3+}$) is:
  1. $5.92\text{ BM}$
  2. $3.87\text{ BM}$
  3. $0\text{ BM}$
  4. $7.94\text{ BM}$

Answer: (D) $7.94\text{ BM}$

Gadolinium ($Z = 64$): $[\text{Xe}] 4f^7 5d^1 6s^2$. $\text{Gd}^{3+}$ configuration is $[\text{Xe}] 4f^7$. Unpaired electrons $n = 7$. Magnetic moment $\mu = \sqrt{7(7+2)} = \sqrt{63} \approx 7.94\text{ BM}$.

Q34. Phenol on reaction with $\text{CHCl}_3$ and aqueous $\text{NaOH}$ followed by acidification yields salicylic aldehyde. The intermediate species involved in this reaction is:
  1. Trichloromethyl anion ($\text{CCl}_3^-$)
  2. Formyl cation ($\text{CHO}^+$)
  3. Dichlorocarbene ($:\text{CCl}_2$)
  4. Carbocation ($^+\text{CHCl}_2$)

Answer: (C) Dichlorocarbene ($:\text{CCl}_2$)

This is the Reimer-Tiemann reaction. $\text{CHCl}_3$ reacts with base $\text{OH}^-$ to form dichlorocarbene ($:\text{CCl}_2$) via $\alpha$-elimination, which acts as an electrophile.

Q35. The rate constant of a reaction increases by $4\text{ times}$ when temperature increases from $300\text{ K}$ to $320\text{ K}$. The activation energy $E_a$ of the reaction is (take $R = 8.314\text{ J/K}\cdot\text{mol}, \ln 2 = 0.693$):
  1. $110.6\text{ kJ/mol}$
  2. $55.3\text{ kJ/mol}$
  3. $27.6\text{ kJ/mol}$
  4. $83.1\text{ kJ/mol}$

Answer: (B) $55.3\text{ kJ/mol}$

Arrhenius equation: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \implies \ln 4 = 2 \ln 2 = 1.386 = \frac{E_a}{8.314} \left( \frac{20}{300 \times 320} \right) \implies E_a = \frac{1.386 \times 8.314 \times 96000}{20} \approx 55.3\text{ kJ/mol}$.

Q36. Which of the following compounds will not show tautomerism?
  1. Acetaldehyde
  2. Benzaldehyde
  3. Acetone
  4. Cyclohexanone

Answer: (B) Benzaldehyde

Tautomerism requires acidic $\alpha$-hydrogen atoms adjacent to the carbonyl group. Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) lacks any $\alpha$-hydrogens and therefore cannot undergo keto-enol tautomerism.

Q37. The correct order of basic strength in aqueous solution for methyl substituted amines is:
  1. $(\text{CH}_3)_3\text{N} > (\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{NH}_3$
  2. $(\text{CH}_3)_2\text{NH} > (\text{CH}_3)_3\text{N} > \text{CH}_3\text{NH}_2 > \text{NH}_3$
  3. $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$
  4. \text{CH}_3\text{NH}_2 > (\text{CH}_3)_2\text{NH} > (\text{CH}_3)_3\text{N} > \text{NH}_3$

Answer: (C) $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$

In aqueous medium, basicity depends on inductive effect, solvation effect, and steric hindrance. For methyl group: $2^\circ > 1^\circ > 3^\circ > \text{NH}_3$, i.e., $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$.

Q38. Which of the following is an example of an amphoteric oxide?
  1. $\text{Na}_2\text{O}$
  2. $\text{Al}_2\text{O}_3$
  3. $\text{SO}_3$
  4. $\text{CO}$

Answer: (B) $\text{Al}_2\text{O}_3$

$\text{Al}_2\text{O}_3$ reacts with both acids (forming $\text{Al}^{3+}$ salts) and bases (forming aluminate ions $[\text{Al(OH)}_4]^-$), proving its amphoteric nature. $\text{Na}_2\text{O}$ is basic, $\text{SO}_3$ is acidic, and $\text{CO}$ is neutral.

Q39. The oxidation states of central sulfur atoms in Caro's acid ($\text{H}_2\text{SO}_5$) and Marshall's acid ($\text{H}_2\text{S}_2\text{O}_8$) are respectively:
  1. $+8$ and $+7$
  2. $+6$ and $+8$
  3. $+6$ and $+6$
  4. $+4$ and $+6$

Answer: (C) $+6$ and $+6$

Both acids contain peroxo linkages ($-O-O-$). In $\text{H}_2\text{SO}_5$: $\text{H}_2\text{SO}_3(\text{O}_2) \implies 2(+1) + x + 3(-2) + 2(-1) = 0 \implies x = +6$. In $\text{H}_2\text{S}_2\text{O}_8$: $2(+1) + 2x + 6(-2) + 2(-1) = 0 \implies 2x = 12 \implies x = +6$.

Q40. Major product of reaction of propene with $\text{HBr}$ in the presence of benzoyl peroxide is:
  1. 2-Bromopropane
  2. 1,2-Dibromopropane
  3. 1-Bromopropane
  4. Propane

Answer: (C) 1-Bromopropane

In the presence of peroxide, addition of $\text{HBr}$ to unsymmetrical alkenes follows Kharasch effect (Anti-Markovnikov rule) via free radical mechanism, adding $\text{Br}$ to the carbon with more hydrogens to form 1-bromopropane.

Q41. The work done during isothermal reversible expansion of $2\text{ moles}$ of an ideal gas from $10\text{ L}$ to $100\text{ L}$ at $300\text{ K}$ is (take $R = 8.314\text{ J/K}\cdot\text{mol}, \ln 10 = 2.303$):
  1. $-5.74\text{ kJ}$
  2. $+11.49\text{ kJ}$
  3. $-2.30\text{ kJ}$
  4. $-11.49\text{ kJ}$

Answer: (D) $-11.49\text{ kJ}$

$W = -2.303 n R T \log_{10}\left(\frac{V_2}{V_1}\right) = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10}(10) = -11488\text{ J} = -11.49\text{ kJ}$.

Q42. Which set of quantum numbers is NOT possible for an electron in an atom?
  1. $n = 4, l = 0, m = 0, s = -1/2$
  2. $n = 3, l = 3, m = -2, s = +1/2$
  3. $n = 2, l = 1, m = -1, s = +1/2$
  4. $n = 5, l = 2, m = 0, s = -1/2$

Answer: (B) $n = 3, l = 3, m = -2, s = +1/2$

Azimuthal quantum number $l$ can take values from $0$ to $n - 1$. For $n = 3$, allowed values of $l$ are $0, 1, 2$. Therefore, $l = 3$ is not possible.

Q43. The IUPAC name of $[\text{Pt(NH}_3)_2\text{Cl(NH}_2\text{CH}_3)]\text{Cl}$ is:
  1. Diamminechloridomethylamineplatinum(IV) chloride
  2. Bisamminechloridomethanamineplatinum(II) chloride
  3. Diamminechlorido(methanamine)platinum(II) chloride
  4. Diamminechloromethylamineplatinum(II) chloride

Answer: (C) Diamminechlorido(methanamine)platinum(II) chloride

Ligands in alphabetical order: Ammine (2) $\rightarrow$ Chlorido $\rightarrow$ Methanamine. Oxidation state of Pt: $x + 2(0) - 1 + 0 - 1 = 0 \implies x = +2$. Full name: Diamminechlorido(methanamine)platinum(II) chloride.

Q44. Which of the following compounds undergoes nitration fastest?
  1. Benzene
  2. Toluene
  3. Nitrobenzene
  4. Chlorobenzene

Answer: (B) Toluene

Nitration is an electrophilic aromatic substitution. The $-\text{CH}_3$ group in toluene is activating due to $+I$ and hyperconjugation, increasing electron density on the ring. $-\text{NO}_2$ and $-\text{Cl}$ are deactivating.

Q45. The $\text{pH}$ of a buffer solution containing $0.1\text{ M}\text{ CH}_3\text{COOH}$ and $0.1\text{ M}\text{ CH}_3\text{COONa}$ is (given $pK_a$ of $\text{CH}_3\text{COOH} = 4.74$):
  1. $5.74$
  2. $3.74$
  3. $4.74$
  4. $7.00$

Answer: (C) $4.74$

By Henderson-Hasselbalch equation: $\text{pH} = pK_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) = 4.74 + \log_{10}\left(\frac{0.1}{0.1}\right) = 4.74 + \log(1) = 4.74$.

Q46. The monomer units of Nylon-6,6 are:
  1. Caprolactam
  2. Terephthalic acid and Ethylene glycol
  3. Formaldehyde and Phenol
  4. Adipic acid and Hexamethylenediamine

Answer: (D) Adipic acid and Hexamethylenediamine

Nylon-6,6 is a condensation polymer obtained from hexamethylenediamine (6 carbons) and adipic acid (6 carbons).

Q47. Glucose on reaction with excess phenylhydrazine forms:
  1. Gluconic acid
  2. Glucosazone
  3. Sorbitol
  4. Saccharic acid

Answer: (B) Glucosazone

Reaction of glucose with 3 equivalents of phenylhydrazine at $C_1$ and $C_2$ carbons produces a characteristic yellow crystalline osazone called glucosazone.

Q48. A first order reaction is $50\%$ complete in $20\text{ minutes}$. The time required for $99.9\%$ completion of the reaction is approximately:
  1. $100\text{ minutes}$
  2. $150\text{ minutes}$
  3. $200\text{ minutes}$
  4. $300\text{ minutes}$

Answer: (C) $200\text{ minutes}$

For a first order reaction, $t_{99.9\%} \approx 10 \times t_{1/2}$. Given $t_{1/2} = 20\text{ min}$, $t_{99.9\%} = 10 \times 20 = 200\text{ minutes}$.

Q49. Which of the following haloalkanes undergoes elimination reaction most readily on heating with alcoholic $\text{KOH}$?
  1. 2-Chloropropane
  2. 2-Iodopropane
  3. 2-Bromopropane
  4. 2-Fluoropropane

Answer: (B) 2-Iodopropane

Leaving group ability follows the order $\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$ due to decreasing bond dissociation energy. Thus, 2-iodopropane undergoes $E_2$ elimination fastest.

Q50. The structure of $\text{SF}_4$ according to VSEPR theory is:
  1. Tetrahedral
  2. Square planar
  3. Trigonal bipyramidal
  4. See-saw

Answer: (D) See-saw

$\text{SF}_4$ has 4 bond pairs and 1 lone pair on central Sulfur (Steric number $= 5 \implies sp^3d$ hybridization). The lone pair occupies an equatorial position to minimize repulsion, yielding a see-saw molecular geometry.

Q51. If $f(x) = \int_0^x \frac{t \sin t}{1 + \cos^2 t} dt$, then $f'(\pi/2)$ is equal to:
  1. $\pi$
  2. $\frac{\pi}{2}$
  3. $\frac{\pi}{4}$
  4. $0$

Answer: (B) $\frac{\pi}{2}$

By Newton-Leibniz formula, $f'(x) = \frac{d}{dx}\int_0^x \frac{t \sin t}{1 + \cos^2 t} dt = \frac{x \sin x}{1 + \cos^2 x}$. Substituting $x = \pi/2$: $f'(\pi/2) = \frac{(\pi/2) \sin(\pi/2)}{1 + \cos^2(\pi/2)} = \frac{(\pi/2)(1)}{1 + 0} = \frac{\pi}{2}$.

Q52. The number of real roots of the equation $e^x = x^2$ is:
  1. $2$
  2. $3$
  3. $1$
  4. $0$

Answer: (C) $1$

Let $f(x) = e^x - x^2$. For $x < 0$: $e^x > 0$ and $x^2 > 0$. At $x = -1$, $f(-1) = 1/e - 1 < 0$, while $\lim_{x \to -\infty} f(x) = +\infty$, so 1 negative root exists. For $x \ge 0$: $f(0) = 1 > 0$, $f'(x) = e^x - 2x$. $f'(x) > 0$ for all $x \ge 0$ since $e^x > 2x$. Thus $f(x)$ is strictly increasing for $x \ge 0$ and never crosses zero. Total real roots $= 1$.

Q53. If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} + \vec{c}}{\sqrt{2}}$, the angle between $\vec{a}$ and $\vec{b}$ is:
  1. $\frac{\pi}{4}$
  2. $\frac{2\pi}{3}$
  3. $\frac{\pi}{3}$
  4. $\frac{3\pi}{4}$

Answer: (D) $\frac{3\pi}{4}$

Using vector triple product expansion: $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{1}{\sqrt{2}} \vec{b} + \frac{1}{\sqrt{2}} \vec{c}$. Comparing coefficients of $\vec{c}$ (since $\vec{b}, \vec{c}$ are non-collinear): $-\vec{a} \cdot \vec{b} = \frac{1}{\sqrt{2}} \implies |\vec{a}||\vec{b}| \cos\theta = -\frac{1}{\sqrt{2}} \implies \cos\theta = -\frac{1}{\sqrt{2}} \implies \theta = \frac{3\pi}{4}$.

Q54. The value of limit $\lim_{n \to \infty} \sum_{k=1}^n \frac{n}{n^2 + k^2}$ is equal to:
  1. $\frac{\pi}{2}$
  2. $\frac{\pi}{4}$
  3. $\ln 2$
  4. $1$

Answer: (B) $\frac{\pi}{4}$

Expressing as definite integral: $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + (k/n)^2} = \int_0^1 \frac{dx}{1 + x^2} = \left[ \tan^{-1} x \right]_0^1 = \frac{\pi}{4}$.

Q55. If system of linear equations $x + y + z = 6$, $x + 2y + 3z = 10$, $x + 2y + \lambda z = \mu$ has infinitely many solutions, then $\lambda + \mu$ is:
  1. $7$
  2. $13$
  3. $10$
  4. $15$

Answer: (B) $13$

Determinant $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} = 1(2\lambda - 6) - 1(\lambda - 3) + 1(0) = \lambda - 3$. For infinite solutions, $D = 0 \implies \lambda = 3$. Determinant $D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 2 & \mu \end{vmatrix} = 1(2\mu - 20) - 1(\mu - 10) + 6(0) = \mu - 10 = 0 \implies \mu = 10$. Thus $\lambda + \mu = 3 + 10 = 13$.

Q56. The locus of the point of intersection of two perpendicular tangents to the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ is:
  1. $x^2 + y^2 = 25$
  2. $x^2 + y^2 = 16$
  3. $x^2 + y^2 = 7$
  4. $x^2 + y^2 = 9$

Answer: (C) $x^2 + y^2 = 7$

The locus of intersection of perpendicular tangents to a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is its director circle $x^2 + y^2 = a^2 - b^2$. Substituting $a^2 = 16, b^2 = 9$: $x^2 + y^2 = 16 - 9 = 7$.

Q57. Sum of all solutions of equation $\sin x + \cos x = 1$ in the interval $[0, 2\pi]$ is:
  1. $\frac{3\pi}{2}$
  2. $\frac{5\pi}{2}$
  3. $2\pi$
  4. $\pi$

Answer: (B) $\frac{5\pi}{2}$

Divide by $\sqrt{2}$: $\sin(x + \pi/4) = \frac{1}{\sqrt{2}}$. In $[0, 2\pi]$: $x + \pi/4 = \pi/4, 3\pi/4, 9\pi/4 \implies x = 0, \pi/2, 2\pi$. Sum of solutions $= 0 + \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$.

Q58. The differential equation of the family of curves $y = c_1 e^{2x} + c_2 e^{-2x}$ is:
  1. $\frac{d^2y}{dx^2} + 4y = 0$
  2. $\frac{d^2y}{dx^2} - 4y = 0$
  3. $\frac{d^2y}{dx^2} - 2y = 0$
  4. $\frac{dy}{dx} - 2y = 0$

Answer: (B) $\frac{d^2y}{dx^2} - 4y = 0$

$y = c_1 e^{2x} + c_2 e^{-2x} \implies \frac{dy}{dx} = 2 c_1 e^{2x} - 2 c_2 e^{-2x} \implies \frac{d^2y}{dx^2} = 4 c_1 e^{2x} + 4 c_2 e^{-2x} = 4y$. Therefore, $\frac{d^2y}{dx^2} - 4y = 0$.

Q59. If 5-digit numbers are formed using digits $1, 2, 3, 4, 5$ without repetition, the probability that a randomly chosen number is divisible by $4$ is:
  1. $\frac{1}{4}$
  2. $\frac{2}{5}$
  3. $\frac{1}{5}$
  4. $\frac{3}{10}$

Answer: (C) $\frac{1}{5}$

Total 5-digit numbers $= 5! = 120$. For divisibility by 4, last two digits must be divisible by 4. Pairs from {1,2,3,4,5}: 12, 24, 32, 52 (4 possibilities). For each pair, remaining 3 digits can be arranged in $3! = 6$ ways. Total favorable $= 4 \times 6 = 24$. Probability $= \frac{24}{120} = \frac{1}{5}$.

Q60. The domain of $f(x) = \sin^{-1}\left(\frac{x^2 - 1}{2x}\right)$ is:
  1. $[-1, 1]$
  2. $(-\infty, 0) \cup (0, \infty)$
  3. $[-1, 0) \cup (0, 1]$
  4. $\mathbb{R} \setminus \{0\}$

Answer: (B) $(-\infty, 0) \cup (0, \infty)$

For $\sin^{-1} u$ to be defined: $-1 \le \frac{x^2 - 1}{2x} \le 1$. 1) $\frac{x^2 - 1}{2x} \ge -1 \implies \frac{x^2 + 2x - 1}{2x} \ge 0$. 2) $\frac{x^2 - 1}{2x} \le 1 \implies \frac{x^2 - 2x - 1}{2x} \le 0$. Solving both inequalities simultaneously yields all non-zero real numbers, i.e., $x \in (-\infty, 0) \cup (0, \infty)$.

Q61. If $\alpha, \beta$ are roots of $x^2 - 6x - 2 = 0$ with $a_n = \alpha^n - \beta^n$ for $n \ge 1$, then value of $\frac{a_{10} - 2 a_8}{2 a_9}$ is:
  1. $6$
  2. $3$
  3. $2$
  4. $1$

Answer: (B) $3$

By Newton's sums for roots, $\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^{10} - 6\alpha^9 - 2\alpha^8 = 0$. Similarly for $\beta$. Subtracting both equations gives $a_{10} - 6 a_9 - 2 a_8 = 0 \implies a_{10} - 2 a_8 = 6 a_9$. Dividing by $2 a_9$: $\frac{a_{10} - 2 a_8}{2 a_9} = \frac{6 a_9}{2 a_9} = 3$.

Q62. The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is:
  1. $\frac{1}{\sqrt{3}}$
  2. $0$
  3. $\frac{1}{\sqrt{6}}$
  4. $\frac{2}{\sqrt{6}}$

Answer: (C) $\frac{1}{\sqrt{6}}$

Points $\vec{a}_1 = (1,2,3), \vec{a}_2 = (2,4,5) \implies \vec{a}_2 - \vec{a}_1 = (1,2,2)$. Direction vectors $\vec{b}_1 = (2,3,4), \vec{b}_2 = (3,4,5)$. Cross product $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}$. Magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{1+4+1} = \sqrt{6}$. Distance $d = \frac{|(1,2,2) \cdot (-1,2,-1)|}{\sqrt{6}} = \frac{|-1 + 4 - 2|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$.

Q63. If $(1 + x)^n = C_0 + C_1 x + C_2 x^2 + \dots + C_n x^n$, then value of $C_0 + 2 C_1 + 3 C_2 + \dots + (n+1) C_n$ is:
  1. $(n+1) 2^n$
  2. $(n+2) 2^{n-1}$
  3. $n 2^{n-1}$
  4. $(n+2) 2^n$

Answer: (B) $(n+2) 2^{n-1}$

Series $S = \sum_{r=0}^n (r+1) C_r = \sum_{r=0}^n r C_r + \sum_{r=0}^n C_r = n 2^{n-1} + 2^n = 2^{n-1}(n + 2) = (n+2) 2^{n-1}$.

Q64. The angle between the pair of tangents drawn from origin to the circle $x^2 + y^2 - 8x - 6y + 20 = 0$ is:
  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{2}$
  3. $\tan^{-1}\left(\frac{3}{4}\right)$
  4. $\tan^{-1}\left(\frac{4}{3}\right)$

Answer: (D) $\tan^{-1}\left(\frac{4}{3}\right)$

Center of circle $(4, 3)$, radius $R = \sqrt{16 + 9 - 20} = \sqrt{5}$. Distance of origin $(0,0)$ from center $OP = \sqrt{4^2 + 3^2} = 5$. If $\theta$ is half angle, $\sin\theta = \frac{R}{OP} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}} \implies \tan\theta = \frac{1}{2}$. Total angle $2\theta \implies \tan(2\theta) = \frac{2 \tan\theta}{1 - \tan^2\theta} = \frac{1}{1 - 1/4} = \frac{4}{3}$. Angle is $\tan^{-1}(4/3)$.

Q65. The area bounded by parabola $y^2 = 4x$ and line $y = 2x$ is:
  1. $\frac{1}{3}\text{ sq units}$
  2. $\frac{2}{3}\text{ sq units}$
  3. $\frac{1}{6}\text{ sq units}$
  4. $1\text{ sq unit}$

Answer: (A) $\frac{1}{3}\text{ sq units}$

Intersection points: $(2x)^2 = 4x \implies 4x^2 = 4x \implies x = 0, 1$. Area $A = \int_0^1 (2\sqrt{x} - 2x) dx = \left[ \frac{4}{3} x^{3/2} - x^2 \right]_0^1 = \frac{4}{3} - 1 = \frac{1}{3}\text{ sq units}$.

Q66. If $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$, then $x + y + z$ is equal to:
  1. $0$
  2. $1$
  3. $x y z$
  4. $x y + y z + z x$

Answer: (C) $x y z$

Taking tangent on both sides: $\tan(\tan^{-1} x + \tan^{-1} y + \tan^{-1} z) = \tan(\pi) = 0 \implies \frac{x + y + z - xyz}{1 - (xy + yz + zx)} = 0 \implies x + y + z = xyz$.

Q67. The mean of $10$ observations is $50$ and standard deviation is $10$. If each observation is multiplied by $2$ and then increased by $5$, the new mean and new standard deviation are respectively:
  1. $100$ and $20$
  2. $105$ and $20$
  3. $105$ and $25$
  4. $100$ and $25$

Answer: (B) $105$ and $20$

New Mean $= 2 \times \bar{x} + 5 = 2(50) + 5 = 105$. Standard deviation is unaffected by addition/subtraction, but scales by absolute value of multiplier: New SD $= |2| \times \sigma = 2 \times 10 = 20$.

Q68. If $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, then $A^{2024}$ is equal to:
  1. $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
  2. $\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$
  3. $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
  4. $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Answer: (C) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$. Thus $A^4 = (A^2)^2 = (-I)^2 = I$. Since $2024$ is divisible by $4$, $A^{2024} = (A^4)^{506} = I^{506} = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Q69. The coefficient of $x^7$ in the expansion of $(1 - x - x^2 + x^3)^6$ is:
  1. $144$
  2. $-120$
  3. $-144$
  4. $120$

Answer: (C) $-144$

Factorizing: $(1 - x - x^2 + x^3)^6 = [(1-x) - x^2(1-x)]^6 = [(1-x)(1-x^2)]^6 = (1-x)^6 (1-x^2)^6$. Expanding: $(1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) \times (1 - 6x^2 + 15x^4 - 20x^6 + \dots)$. Term $x^7$ comes from: $(-6x)(-20x^6) + (-20x^3)(15x^4) + (-6x^5)(-6x^2) = 120 - 300 + 36 = -144$.

Q70. Number of points where function $f(x) = \max(1 - x, 1 + x, 2)$ is not differentiable is:
  1. $1$
  2. $3$
  3. $2$
  4. $0$

Answer: (C) $2$

Intersection of lines: $1 - x = 2 \implies x = -1$ and $1 + x = 2 \implies x = 1$. Graph of $f(x)$ follows $1-x$ for $x < -1$, constant $2$ for $-1 \le x \le 1$, and $1+x$ for $x > 1$. The graph forms sharp corners at $x = -1$ and $x = 1$. Hence non-differentiable at 2 points.

Q71. If $\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}} = A \sqrt{x} + B x^{1/3} + C x^{1/6} + D \ln(1 + x^{1/6}) + K$, then $A + B + C + D$ is:
  1. $-1$
  2. $1$
  3. $0$
  4. $2$

Answer: (A) $-1$

Substitute $x = t^6 \implies dx = 6t^5 dt$. Integral $= \int \frac{6t^5}{t^3 + t^2} dt = 6 \int \frac{t^3}{t+1} dt = 6 \int \left(t^2 - t + 1 - \frac{1}{t+1}\right) dt = 2t^3 - 3t^2 + 6t - 6\ln(1+t) + K = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\ln(1+x^{1/6}) + K$. Thus $A=2, B=-3, C=6, D=-6 \implies A+B+C+D = 2 - 3 + 6 - 6 = -1$.

Q72. The sum of the infinite series $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots$ is:
  1. $\frac{1}{2}$
  2. $1$
  3. $\ln 2$
  4. $2$

Answer: (B) $1$

General term $T_n = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. Telescoping sum $S_N = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{N} - \frac{1}{N+1}\right) = 1 - \frac{1}{N+1}$. As $N \to \infty$, $S = 1$.

Q73. If complex number $z$ satisfies $|z - 3 - 4i| = 2$, the maximum value of $|z|$ is:
  1. $5$
  2. $3$
  3. $7$
  4. $9$

Answer: (C) $7$

By triangle inequality: $|z| = |(z - (3+4i)) + (3+4i)| \le |z - 3 - 4i| + |3 + 4i| = 2 + 5 = 7$.

Q74. The image of point $(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is:
  1. $(1, 2, 3)$
  2. $(1, 0, 7)$
  3. $(2, 1, 0)$
  4. $(0, 1, 7)$

Answer: (B) $(1, 0, 7)$

General point on line $F = (\lambda, 2\lambda + 1, 3\lambda + 2)$. Vector $\vec{PF} = (\lambda - 1, 2\lambda - 5, 3\lambda - 1)$. Since $\vec{PF} \perp (1, 2, 3)$: $(\lambda - 1)(1) + (2\lambda - 5)(2) + (3\lambda - 1)(3) = 0 \implies 14\lambda - 14 = 0 \implies \lambda = 1$. Foot of perpendicular $F = (1, 3, 5)$. Image $I = 2F - P = (2(1)-1, 2(3)-6, 2(5)-3) = (1, 0, 7)$.

Q75. The total number of relations from set $A = \{1, 2, 3\}$ to set $B = \{a, b\}$ is:
  1. $32$
  2. $16$
  3. $64$
  4. $8$

Answer: (C) $64$

$n(A) = 3$, $n(B) = 2 \implies n(A \times B) = 3 \times 2 = 6$. The total number of relations (subsets of $A \times B$) is $2^{n(A \times B)} = 2^6 = 64$.