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JEE Main Full Mock Test: Practice Test 01

Question paper with answer key. Selection Pro

Q1. A particle is projected from the ground with an initial velocity $v_0 = 20\text{ m/s}$ at an angle of $30^\circ$ with the horizontal. Taking $g = 10\text{ m/s}^2$, the total time of flight of the projectile is:
  1. $2\text{ s}$
  2. $4\text{ s}$
  3. $1\text{ s}$
  4. $2\sqrt{3}\text{ s}$

Answer: (A) $2\text{ s}$

The time of flight is given by $T = \frac{2 v_0 \sin\theta}{g}$. Substituting $v_0 = 20\text{ m/s}$, $\theta = 30^\circ$, and $g = 10\text{ m/s}^2$: $T = \frac{2 \times 20 \times \sin(30^\circ)}{10} = \frac{40 \times 0.5}{10} = 2\text{ s}$.

Q2. A block of mass $m = 2\text{ kg}$ rests on a rough horizontal surface with coefficient of static friction $\mu_s = 0.4$. The minimum horizontal force required to just move the block is (take $g = 10\text{ m/s}^2$):
  1. $8\text{ N}$
  2. $4\text{ N}$
  3. $20\text{ N}$
  4. $0.8\text{ N}$

Answer: (A) $8\text{ N}$

The maximum static friction force is $f_{\text{max}} = \mu_s N = \mu_s m g$. Substituting the values: $f_{\text{max}} = 0.4 \times 2\text{ kg} \times 10\text{ m/s}^2 = 8\text{ N}$.

Q3. A force $\vec{F} = (3\hat{i} + 4\hat{j})\text{ N}$ displaces a body from $(0, 0)$ to $(2\text{ m}, 3\text{ m})$. The work done by the force is:
  1. $18\text{ J}$
  2. $12\text{ J}$
  3. $6\text{ J}$
  4. $24\text{ J}$

Answer: (A) $18\text{ J}$

Displacement vector $\vec{r} = (2 - 0)\hat{i} + (3 - 0)\hat{j} = (2\hat{i} + 3\hat{j})\text{ m}$. Work done $W = \vec{F} \cdot \vec{r} = (3 \times 2) + (4 \times 3) = 6 + 12 = 18\text{ J}$.

Q4. A sphere of mass $m$ moving with speed $v$ collides head-on elastically with another stationary sphere of mass $2m$. The fractional loss of kinetic energy of the first sphere is:
  1. $\frac{8}{9}$
  2. $\frac{1}{9}$
  3. $\frac{4}{9}$
  4. $\frac{2}{3}$

Answer: (A) $\frac{8}{9}$

Final velocity of mass $m$ after elastic collision is $v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) v = \left(\frac{m - 2m}{m + 2m}\right) v = -\frac{v}{3}$. Remaining $\text{KE}_1 = \frac{1}{2} m v_1^2 = \frac{1}{9} \left(\frac{1}{2} m v^2\right)$. Fractional loss $= 1 - \frac{1}{9} = \frac{8}{9}$.

Q5. A solid sphere of mass $M$ and radius $R$ rolls without slipping down an inclined plane of inclination $\theta$. The acceleration of the sphere is:
  1. $\frac{5}{7} g \sin\theta$
  2. $\frac{2}{5} g \sin\theta$
  3. $\frac{3}{5} g \sin\theta$
  4. $\frac{1}{2} g \sin\theta$

Answer: (A) $\frac{5}{7} g \sin\theta$

Acceleration of a rolling body on an incline is $a = \frac{g \sin\theta}{1 + \frac{I}{M R^2}}$. For a solid sphere, $I = \frac{2}{5} M R^2$, so $a = \frac{g \sin\theta}{1 + 2/5} = \frac{5}{7} g \sin\theta$.

Q6. The escape velocity from the surface of the Earth is $v_e$. If the radius of the Earth contracts to $\frac{1}{4}\text{th}$ of its present value without any change in mass, the new escape velocity becomes:
  1. $2 v_e$
  2. $4 v_e$
  3. $\frac{v_e}{2}$
  4. $v_e$

Answer: (A) $2 v_e$

Escape velocity is $v_e = \sqrt{\frac{2 G M}{R}}$. If $R' = \frac{R}{4}$, then $v_e' = \sqrt{\frac{2 G M}{R/4}} = 2 \sqrt{\frac{2 G M}{R}} = 2 v_e$.

Q7. Two wires $A$ and $B$ are made of the same material. Wire $A$ has length $L$ and radius $r$, while wire $B$ has length $2L$ and radius $2r$. The ratio of Young's modulus $Y_A / Y_B$ is:
  1. $1$
  2. $2$
  3. $\frac{1}{2}$
  4. $4$

Answer: (A) $1$

Young's modulus depends solely on the material of the wire. Since both wires $A$ and $B$ are made of the same material, $Y_A = Y_B \implies \frac{Y_A}{Y_B} = 1$.

Q8. According to Stefan-Boltzmann law, the total radiant energy emitted per second per unit area by a black body at temperature $T$ is proportional to $\sigma T^n$, where $\sigma$ is Stefan's constant. The value of $n$ is:
  1. $4$
  2. $2$
  3. $1$
  4. $3$

Answer: (A) $4$

Stefan-Boltzmann law states that total emissive power $E = \sigma T^4$. Therefore, $n = 4$.

Q9. A particle executes simple harmonic motion with amplitude $A$. At what displacement from the mean position is its kinetic energy equal to its potential energy?
  1. $\frac{A}{\sqrt{2}}$
  2. $\frac{A}{2}$
  3. $\frac{A}{\sqrt{3}}$
  4. $\frac{\sqrt{3} A}{2}$

Answer: (A) $\frac{A}{\sqrt{2}}$

$\text{KE} = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and $\text{PE} = \frac{1}{2} m \omega^2 x^2$. Setting $\text{KE} = \text{PE} \implies A^2 - x^2 = x^2 \implies 2 x^2 = A^2 \implies x = \frac{A}{\sqrt{2}}$.

Q10. An organ pipe closed at one end has fundamental frequency $f_1 = 200\text{ Hz}$. The frequency of the third harmonic is:
  1. $600\text{ Hz}$
  2. $400\text{ Hz}$
  3. $800\text{ Hz}$
  4. $1000\text{ Hz}$

Answer: (A) $600\text{ Hz}$

A closed organ pipe produces only odd harmonics. The frequencies are $f_1, 3 f_1, 5 f_1, \dots$. The frequency of the 3rd harmonic is $3 \times f_1 = 3 \times 200\text{ Hz} = 600\text{ Hz}$.

Q11. Two point charges $+q$ and $+4q$ are separated by distance $L$. The electric field intensity is zero at a point located at distance $x$ from $+q$. The value of $x$ is:
  1. $\frac{L}{3}$
  2. $\frac{L}{2}$
  3. $\frac{L}{4}$
  4. $\frac{2L}{3}$

Answer: (A) $\frac{L}{3}$

Electric field balance: $\frac{1}{4\pi\varepsilon_0}\frac{q}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{4q}{(L-x)^2} \implies \frac{1}{x} = \frac{2}{L-x} \implies L - x = 2x \implies 3x = L \implies x = \frac{L}{3}$.

Q12. Three resistors of resistances $2\;\Omega$, $3\;\Omega$, and $6\;\Omega$ are connected in parallel. The equivalent resistance of the combination is:
  1. $1\;\Omega$
  2. $11\;\Omega$
  3. $0.5\;\Omega$
  4. $2\;\Omega$

Answer: (A) $1\;\Omega$

$\frac{1}{R_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\;\Omega^{-1} \implies R_{\text{eq}} = 1\;\Omega$.

Q13. A long straight wire carries a steady current $I$. The magnetic field $B$ at a perpendicular distance $r$ from the wire varies as:
  1. $B \propto \frac{1}{r}$
  2. $B \propto \frac{1}{r^2}$
  3. $B \propto r$
  4. $B \propto r^2$

Answer: (A) $B \propto \frac{1}{r}$

The magnetic field due to a long straight current-carrying conductor is given by $B = \frac{\mu_0 I}{2\pi r}$. Hence, $B \propto \frac{1}{r}$.

Q14. A magnetic dipole of dipole moment $M$ is placed in a uniform magnetic field $B$. The work done in rotating the dipole from $\theta_1 = 0^\circ$ to $\theta_2 = 180^\circ$ is:
  1. $2 M B$
  2. $M B$
  3. $0$
  4. $-M B$

Answer: (A) $2 M B$

Work done $W = -M B (\cos\theta_2 - \cos\theta_1) = -M B (\cos 180^\circ - \cos 0^\circ) = -M B (-1 - 1) = 2 M B$.

Q15. The magnetic flux linked with a loop changes according to $\Phi = (3 t^2 + 4 t + 9)\text{ Wb}$. The magnitude of induced EMF at $t = 2\text{ s}$ is:
  1. $16\text{ V}$
  2. $12\text{ V}$
  3. $20\text{ V}$
  4. $29\text{ V}$

Answer: (A) $16\text{ V}$

Induced EMF $e = -\frac{d\Phi}{dt} = -\frac{d}{dt}(3 t^2 + 4 t + 9) = -(6 t + 4)$. At $t = 2\text{ s}$: $|e| = 6(2) + 4 = 16\text{ V}$.

Q16. In a series $RLC$ circuit at resonance, the impedance of the circuit is equal to:
  1. $R$
  2. $\sqrt{R^2 + (X_L - X_C)^2}$
  3. $X_L$
  4. Zero

Answer: (A) $R$

At resonance, inductive reactance equals capacitive reactance ($X_L = X_C$). Total impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$.

Q17. In an electromagnetic wave traveling in vacuum, the ratio of amplitudes of electric field $E_0$ and magnetic field $B_0$ is equal to:
  1. $c$
  2. $\frac{1}{c}$
  3. $c^2$
  4. $\sqrt{c}$

Answer: (A) $c$

For electromagnetic waves propagating in vacuum, the ratio of electric field amplitude to magnetic field amplitude is given by $\frac{E_0}{B_0} = c$, where $c$ is the speed of light.

Q18. A convex lens of focal length $f = 20\text{ cm}$ produces a real image of the same size as the object. The distance between the object and its image is:
  1. $80\text{ cm}$
  2. $40\text{ cm}$
  3. $20\text{ cm}$
  4. $100\text{ cm}$

Answer: (A) $80\text{ cm}$

For a real image of same size ($m = -1$), object distance $u = 2f = 40\text{ cm}$ and image distance $v = 2f = 40\text{ cm}$. Total distance between object and image $= u + v = 40 + 40 = 80\text{ cm}$.

Q19. In Young's double slit experiment, if the separation between slits is halved and distance from slits to screen is doubled, the fringe width becomes:
  1. $4\text{ times}$
  2. $2\text{ times}$
  3. Unchanged
  4. $\frac{1}{4}\text{ times}$

Answer: (A) $4\text{ times}$

Fringe width $\beta = \frac{\lambda D}{d}$. If $D' = 2D$ and $d' = \frac{d}{2}$, new fringe width $\beta' = \frac{\lambda (2D)}{d/2} = 4 \left(\frac{\lambda D}{d}\right) = 4 \beta$.

Q20. The de-Broglie wavelength $\lambda$ associated with an electron accelerated through a potential difference of $V\text{ volts}$ is given by:
  1. $\lambda = \frac{12.27}{\sqrt{V}}\text{ \AA}$
  2. $\lambda = \frac{122.7}{\sqrt{V}}\text{ \AA}$
  3. $\lambda = \frac{1.227}{\sqrt{V}}\text{ \AA}$
  4. $\lambda = \frac{12.27}{V}\text{ \AA}$

Answer: (A) $\lambda = \frac{12.27}{\sqrt{V}}\text{ \AA}$

Formula for de-Broglie wavelength of accelerated electron is $\lambda = \frac{h}{\sqrt{2 m e V}} = \frac{12.27}{\sqrt{V}}\text{ \AA}$.

Q21. The radius of the $n^{\text{th}}$ orbit of Bohr's hydrogen atom is proportional to:
  1. $n^2$
  2. $n$
  3. $\frac{1}{n}$
  4. $\frac{1}{n^2}$

Answer: (A) $n^2$

According to Bohr's model, orbital radius $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2 Z} \propto n^2$.

Q22. Half-life of a radioactive sample is $10\text{ days}$. The time required for $75\%$ of the sample to decay is:
  1. $20\text{ days}$
  2. $30\text{ days}$
  3. $15\text{ days}$
  4. $40\text{ days}$

Answer: (A) $20\text{ days}$

Decay of $75\%$ leaves $25\%$ ($1/4\text{th}$) remaining. $N = N_0 \left(\frac{1}{2}\right)^n \implies \frac{1}{4} = \left(\frac{1}{2}\right)^2 \implies n = 2\text{ half-lives}$. Total time $= 2 \times 10 = 20\text{ days}$.

Q23. In an $n$-type semiconductor, the majority charge carriers and minority charge carriers are respectively:
  1. Electrons and Holes
  2. Holes and Electrons
  3. Electrons and Positive Ions
  4. Holes and Negative Ions

Answer: (A) Electrons and Holes

In $n$-type semiconductors, pentavalent impurities donate free conduction electrons, making electrons the majority carriers and thermally generated holes the minority carriers.

Q24. The dimensional formula of universal gravitational constant $G$ is:
  1. $[M^{-1} L^3 T^{-2}]$
  2. $[M^1 L^3 T^{-2}]$
  3. $[M^{-1} L^2 T^{-2}]$
  4. $[M^{-2} L^3 T^{-1}]$

Answer: (A) $[M^{-1} L^3 T^{-2}]$

From Newton's law of gravitation $F = \frac{G m_1 m_2}{r^2} \implies G = \frac{F r^2}{m_1 m_2}$. Dimensions: $[G] = \frac{[M L T^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$.

Q25. If vector $\vec{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ is perpendicular to vector $\vec{B} = 4\hat{j} - 4\hat{i} + \alpha\hat{k}$, the value of $\alpha$ is:
  1. $-\frac{1}{2}$
  2. $\frac{1}{2}$
  3. $-1$
  4. $1$

Answer: (A) $-\frac{1}{2}$

For perpendicular vectors, dot product $\vec{A} \cdot \vec{B} = 0$. $\vec{A} = 2\hat{i} + 3\hat{j} + 8\hat{k}$ and $\vec{B} = -4\hat{i} + 4\hat{j} + \alpha\hat{k}$. $\vec{A} \cdot \vec{B} = 2(-4) + 3(4) + 8(\alpha) = -8 + 12 + 8\alpha = 4 + 8\alpha = 0 \implies \alpha = -\frac{4}{8} = -\frac{1}{2}$.

Q26. Which of the following complex species is expected to exhibit optical isomerism?
  1. $\text{cis-}[\text{Co(en)}_2\text{Cl}_2]^+$
  2. $\text{trans-}[\text{Co(en)}_2\text{Cl}_2]^+$
  3. $[\text{Pt(NH}_3)_2\text{Cl}_2]$
  4. $[\text{Co(NH}_3)_4\text{Cl}_2]^+$

Answer: (A) $\text{cis-}[\text{Co(en)}_2\text{Cl}_2]^+$

$\text{cis-}[\text{Co(en)}_2\text{Cl}_2]^+$ lacks a plane of symmetry and center of inversion, making it chiral and optically active. The trans isomer has a plane of symmetry.

Q27. Anisole on heating with concentrated $\text{HBr}$ gives:
  1. Phenol and methyl bromide
  2. Bromobenzene and methanol
  3. Phenol and methanol
  4. Methyl bromide and bromobenzene

Answer: (A) Phenol and methyl bromide

Cleavage of aryl-alkyl ethers with $\text{HBr}$ occurs via $S_{\text{N}}2$ attack of $\text{Br}^-$ on the smaller methyl group to yield phenol and $\text{CH}_3\text{Br}$, because the $\text{C}(sp^2)-\text{O}$ bond in phenol has partial double bond character.

Q28. The correct decreasing order of $S_{\text{N}}1$ reactivity for the following alkyl halides is: (I) $(\text{C}_6\text{H}_5)_3\text{C-Br}$ (II) $(\text{C}_6\text{H}_5)_2\text{CH-Br}$ (III) $(\text{CH}_3)_3\text{C-Br}$ (IV) $\text{C}_6\text{H}_5\text{CH}_2\text{-Br}$
  1. (I) > (II) > (III) > (IV)
  2. (III) > (I) > (II) > (IV)
  3. (I) > (III) > (II) > (IV)
  4. (IV) > (II) > (I) > (III)

Answer: (A) (I) > (II) > (III) > (IV)

The rate of $S_{\text{N}}1$ reaction depends on carbocation stability. $(\text{C}_6\text{H}_5)_3\text{C}^+$ is highly resonance stabilized by three phenyl rings > $(\text{C}_6\text{H}_5)_2\text{CH}^+$ > $(\text{CH}_3)_3\text{C}^+$ ($3^\circ$ carbocation with hyperconjugation) > $\text{C}_6\text{H}_5\text{CH}_2^+$.

Q29. For a first-order reaction $A \rightarrow B$, the rate constant is $k = 6.93 \times 10^{-3}\text{ s}^{-1}$. How much time will it take for $75\%$ of the reactant to react?
  1. $200\text{ s}$
  2. $100\text{ s}$
  3. $300\text{ s}$
  4. $400\text{ s}$

Answer: (A) $200\text{ s}$

$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{6.93 \times 10^{-3}} = 100\text{ s}$. Time required for $75\%$ completion is $2 \times t_{1/2} = 200\text{ s}$.

Q30. The standard reduction potentials $E^\circ$ for $\text{Cu}^{2+}/\text{Cu}$ and $\text{Cu}^+/\text{Cu}$ are $+0.34\text{ V}$ and $+0.52\text{ V}$ respectively. Calculate $E^\circ$ for $\text{Cu}^{2+}/\text{Cu}^+$.
  1. $+0.16\text{ V}$
  2. $+0.86\text{ V}$
  3. $-0.18\text{ V}$
  4. $+0.32\text{ V}$

Answer: (A) $+0.16\text{ V}$

Using $\Delta G^\circ = -nFE^\circ$: 1) $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$, $\Delta G_1^\circ = -2F(0.34) = -0.68F$ 2) $\text{Cu}^+ + e^- \rightarrow \text{Cu}$, $\Delta G_2^\circ = -1F(0.52) = -0.52F$ Subtracting (2) from (1): $\text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+$, $\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ = -0.16F$. $-1FE_3^\circ = -0.16F \implies E_3^\circ = +0.16\text{ V}$.

Q31. Which of the following species has $d\pi-p\pi$ bonding?
  1. $\text{SO}_3^{2-}$
  2. $\text{CO}_3^{2-}$
  3. $\text{NO}_3^-$
  4. $\text{BO}_3^{3-}$

Answer: (A) $\text{SO}_3^{2-}$

In $\text{SO}_3^{2-}$, Sulfur has vacant $3d$ orbitals that can overlap with filled $2p$ orbitals of oxygen, forming $d\pi-p\pi$ bonds. C, N, and B do not have available $d$ orbitals.

Q32. Major product obtained on mononitration of acetanilide with $\text{HNO}_3/\text{H}_2\text{SO}_4$ followed by hydrolysis is:
  1. 4-Nitroaniline
  2. 2-Nitroaniline
  3. 3-Nitroaniline
  4. 4-Nitrophenol

Answer: (A) 4-Nitroaniline

The $-\text{NHCOCH}_3$ group is strongly ortho/para directing. Due to steric hindrance of the bulky acetyl group, substitution occurs predominantly at the para-position to give 4-nitroacetanilide, which hydrolyzes to 4-nitroaniline.

Q33. For the reaction $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$, $\Delta H^\circ = -393.5\text{ kJ/mol}$. The entropy change $\Delta S^\circ$ is $+3.0\text{ J/K}\cdot\text{mol}$. At what temperature will the reaction become non-spontaneous?
  1. The reaction is spontaneous at all temperatures
  2. $131.17\text{ K}$
  3. $1000\text{ K}$
  4. $1311.7\text{ K}$

Answer: (A) The reaction is spontaneous at all temperatures

Since $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (entropy increases), $\Delta G = \Delta H - T\Delta S$ is always negative at all temperatures. Thus, it remains spontaneous at all temperatures.

Q34. Among the noble gas compounds, the geometry and hybridization of $\text{XeF}_4$ are respectively:
  1. Square planar, $sp^3d^2$
  2. Tetrahedral, $sp^3$
  3. Octahedral, $sp^3d^2$
  4. See-saw, $sp^3d$

Answer: (A) Square planar, $sp^3d^2$

$\text{XeF}_4$ has 4 bond pairs and 2 lone pairs on Xe (Steric number $= 6 \implies sp^3d^2$). The molecular geometry with two axial lone pairs is square planar.

Q35. A $0.004\text{ M}$ solution of $\text{Na}_2\text{SO}_4$ is isotonic with a $0.01\text{ M}$ solution of glucose at the same temperature. The apparent degree of dissociation of $\text{Na}_2\text{SO}_4$ is:
  1. $75\%$
  2. $50\%$
  3. $85\%$
  4. $100\%$

Answer: (A) $75\%$

Isotonic $\implies i_1 C_1 = i_2 C_2$. For glucose $i_2 = 1$. $i_1(0.004) = 1(0.01) \implies i_1 = 2.5$. For $\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}$, $n = 3$. $i = 1 + (n-1)\alpha \implies 2.5 = 1 + 2\alpha \implies \alpha = 0.75$ or $75\%$.

Q36. Which of the following carbohydrates is a non-reducing sugar?
  1. Sucrose
  2. Maltose
  3. Lactose
  4. Glucose

Answer: (A) Sucrose

In sucrose, the glycosidic linkage involves the anomeric carbons of both glucose ($C_1$) and fructose ($C_2$), leaving no free hemiacetal or hemiketal group. Hence it is non-reducing.

Q37. The solubility product ($K_{sp}$) of $\text{Ag}_2\text{CrO}_4$ is $3.2 \times 10^{-11}$. Its molar solubility in water is:
  1. $2.0 \times 10^{-4}\text{ M}$
  2. $4.0 \times 10^{-4}\text{ M}$
  3. $1.8 \times 10^{-4}\text{ M}$
  4. $8.0 \times 10^{-6}\text{ M}$

Answer: (A) $2.0 \times 10^{-4}\text{ M}$

$\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-}$. $K_{sp} = (2s)^2(s) = 4s^3$. $4s^3 = 3.2 \times 10^{-11} = 32 \times 10^{-12} \implies s^3 = 8 \times 10^{-12} \implies s = 2 \times 10^{-4}\text{ M}$.

Q38. The spin-only magnetic moment of $[\text{Fe(CN)}_6]^{4-}$ and $[\text{Fe(H}_2\text{O)}_6]^{2+}$ are respectively:
  1. $0\text{ BM}$ and $4.9\text{ BM}$
  2. $4.9\text{ BM}$ and $0\text{ BM}$
  3. $2.83\text{ BM}$ and $4.9\text{ BM}$
  4. $0\text{ BM}$ and $2.83\text{ BM}$

Answer: (A) $0\text{ BM}$ and $4.9\text{ BM}$

For $[\text{Fe(CN)}_6]^{4-}$, $\text{Fe}^{2+}$ ($d^6$) with strong field ligand $\text{CN}^-$ forms low spin $t_{2g}^6 e_g^0$ ($n = 0 \implies 0\text{ BM}$). For $[\text{Fe(H}_2\text{O)}_6]^{2+}$, weak field ligand forms high spin $t_{2g}^4 e_g^2$ ($n = 4 \implies \sqrt{4(6)} = 4.9\text{ BM}$).

Q39. In the reaction: $\text{CH}_3\text{C}\equiv\text{CH} + \text{CH}_3\text{MgBr} \rightarrow A + B$. The gaseous product $A$ formed is:
  1. Methane
  2. Ethane
  3. Propyne
  4. Propane

Answer: (A) Methane

Terminal alkynes have acidic hydrogens. $\text{CH}_3\text{C}\equiv\text{C-H} + \text{CH}_3\text{MgBr} \rightarrow \text{CH}_3\text{C}\equiv\text{C-MgBr} + \text{CH}_4\uparrow$. Methane gas is evolved.

Q40. According to Bohr's model, the ratio of time period of revolution of an electron in the $n=1$ and $n=2$ orbits of a hydrogen atom is:
  1. $1 : 8$
  2. $1 : 4$
  3. $1 : 2$
  4. $1 : 16$

Answer: (A) $1 : 8$

Time period $T = \frac{2\pi r}{v} \propto \frac{n^2/Z}{Z/n} \propto n^3$. Therefore, $T_1 / T_2 = (1/2)^3 = 1/8$ or $1 : 8$.

Q41. Which cation gives a chocolate brown precipitate with potassium ferrocyanide $\text{K}_4[\text{Fe(CN)}_6]$ solution?
  1. $\text{Cu}^{2+}$
  2. $\text{Fe}^{3+}$
  3. $\text{Zn}^{2+}$
  4. $\text{Cd}^{2+}$

Answer: (A) $\text{Cu}^{2+}$

$\text{Cu}^{2+}$ ions react with potassium ferrocyanide to form a reddish-brown/chocolate brown precipitate of copper ferrocyanide $\text{Cu}_2[\text{Fe(CN)}_6]$.

Q42. The correct IUPAC name of the compound $\text{CH}_3-\text{CH}=\text{CH}-\text{C}\equiv\text{CH}$ is:
  1. Pent-3-en-1-yne
  2. Pent-2-en-4-yne
  3. Pent-1-yn-3-ene
  4. Pent-4-yn-2-ene

Answer: (A) Pent-3-en-1-yne

Numbering starts from the end that gives lower locants to unsaturation. Double and triple bonds get locants 3 and 1 from right to left. Since double bond and triple bond compete for lower locant, priority is equal, but numbering from right gives locants 1 (yne) and 3 (ene). Name: Pent-3-en-1-yne.

Q43. How many moles of $\text{KMnO}_4$ are required to oxidize $1\text{ mole}$ of $\text{FeSO}_4$ in acidic medium?
  1. $\frac{1}{5}\text{ mol}$
  2. $5\text{ mol}$
  3. $\frac{2}{5}\text{ mol}$
  4. $1\text{ mol}$

Answer: (A) $\frac{1}{5}\text{ mol}$

In acidic medium, $\text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+}$ ($n$-factor $= 5$). $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ ($n$-factor $= 1$). Equivalents of $\text{KMnO}_4 =$ Equivalents of $\text{FeSO}_4 \implies n_{\text{KMnO}_4} \times 5 = 1 \times 1 \implies n_{\text{KMnO}_4} = \frac{1}{5}\text{ mol}$.

Q44. The correct order of electron gain enthalpy (with negative sign) for halide ions/halogens is:
  1. $\text{Cl} > \text{F} > \text{Br} > \text{I}$
  2. $\text{F} > \text{Cl} > \text{Br} > \text{I}$
  3. $\text{Cl} > \text{Br} > \text{F} > \text{I}$
  4. $\text{I} > \text{Br} > \text{Cl} > \text{F}$

Answer: (A) $\text{Cl} > \text{F} > \text{Br} > \text{I}$

Chlorine has more negative electron gain enthalpy than Fluorine due to small size and high inter-electronic repulsions in the $2p$ subshell of Fluorine. Order: $\text{Cl} > \text{F} > \text{Br} > \text{I}$.

Q45. Hinsberg's reagent reacts with primary amines to give a product which is:
  1. Soluble in aqueous $\text{KOH}$
  2. Insoluble in aqueous $\text{KOH}$
  3. Insoluble in water and acids
  4. Does not react at all

Answer: (A) Soluble in aqueous $\text{KOH}$

Hinsberg's reagent (benzene sulfonyl chloride) reacts with primary amines to form N-alkylbenzene sulfonamide, which contains an acidic hydrogen attached to nitrogen, making it soluble in aqueous $\text{KOH}$.

Q46. For the gaseous reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, the relationship between $K_p$ and $K_c$ is:
  1. $K_p = K_c(RT)^{-2}$
  2. $K_p = K_c(RT)^2$
  3. $K_p = K_c(RT)^{-1}$
  4. $K_p = K_c$

Answer: (A) $K_p = K_c(RT)^{-2}$

$\Delta n_g = 2 - (1 + 3) = -2$. $K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^{-2}$.

Q47. Ozonolysis of an alkene $X$ followed by treatment with $\text{Zn/H}_2\text{O}$ gives propanone and ethanal in equal molar amounts. The alkene $X$ is:
  1. 2-Methylbut-2-ene
  2. 2-Methylbut-1-ene
  3. Pent-2-ene
  4. 3-Methylbut-1-ene

Answer: (A) 2-Methylbut-2-ene

Combine carbonyl compounds by removing oxygen: $(\text{CH}_3)_2\text{C=O} + \text{O=CH-CH}_3 \rightarrow (\text{CH}_3)_2\text{C=CH-CH}_3$, which is 2-Methylbut-2-ene.

Q48. In the extraction of copper from copper pyrites, silica is added to the reverberatory furnace to remove:
  1. $\text{FeO}$ as $\text{FeSiO}_3$ slag
  2. $\text{FeS}$ as matte
  3. $\text{Cu}_2\text{O}$ as $\text{Cu}_2\text{SiO}_3$
  4. $\text{CuS}$ as slag

Answer: (A) $\text{FeO}$ as $\text{FeSiO}_3$ slag

Iron impurity present in copper pyrites oxidizes to $\text{FeO}$. Silica ($\text{SiO}_2$) acts as an acidic flux to convert basic $\text{FeO}$ into slag: $\text{FeO} + \text{SiO}_2 \rightarrow \text{FeSiO}_3$ (slag).

Q49. An element crystallizes in a face-centered cubic (fcc) lattice with edge length $a = 400\text{ pm}$. If the atomic mass of the element is $60\text{ g/mol}$, its density is approximately:
  1. $6.23\text{ g/cm}^3$
  2. $3.11\text{ g/cm}^3$
  3. $12.46\text{ g/cm}^3$
  4. $1.55\text{ g/cm}^3$

Answer: (A) $6.23\text{ g/cm}^3$

For fcc, $Z = 4$. Density $d = \frac{Z M}{a^3 N_A} = \frac{4 \times 60}{(400 \times 10^{-10})^3 \times 6.022 \times 10^{23}} = \frac{240}{64 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{240}{38.54} \approx 6.23\text{ g/cm}^3$.

Q50. Which of the following compounds has the highest acidic strength?
  1. 4-Nitrophenol
  2. Phenol
  3. 4-Methylphenol
  4. 4-Methoxyphenol

Answer: (A) 4-Nitrophenol

The $-\text{NO}_2$ group is strong electron-withdrawing via both $-I$ and $-M$ effects, stabilizing the phenoxide ion and drastically increasing acidity. $-\text{CH}_3$ ($+I, +H$) and $-\text{OCH}_3$ ($+M$) decrease acidity.

Q51. If $\alpha, \beta$ are roots of the equation $x^2 - x - 1 = 0$ and $p_n = \alpha^n + \beta^n$ for $n \ge 1$, then value of $p_5$ is:
  1. $11$
  2. $7$
  3. $4$
  4. $18$

Answer: (A) $11$

Using Newton's sum formula, $p_n - p_{n-1} - p_{n-2} = 0 \implies p_n = p_{n-1} + p_{n-2}$. Here $p_1 = \alpha + \beta = 1$, $p_2 = (\alpha+\beta)^2 - 2\alpha\beta = 1 - 2(-1) = 3$. Then $p_3 = 3+1 = 4$, $p_4 = 4+3 = 7$, and $p_5 = 7+4 = 11$.

Q52. Evaluate integral: $\int_0^{\pi/2} \frac{\sin^2 x}{1 + \sin x \cos x} dx$
  1. $\frac{\pi}{3\sqrt{3}}$
  2. $\frac{\pi}{2\sqrt{3}}$
  3. $\frac{2\pi}{3\sqrt{3}}$
  4. $\frac{\pi}{4\sqrt{3}}$

Answer: (A) $\frac{\pi}{3\sqrt{3}}$

Using King's property $I = \int_0^{\pi/2} \frac{\cos^2 x}{1 + \sin x \cos x} dx$. Adding both: $2I = \int_0^{\pi/2} \frac{1}{1 + \sin x \cos x} dx$. Dividing numerator and denominator by $\cos^2 x$: $2I = \int_0^{\infty} \frac{\sec^2 x}{\tan^2 x + \tan x + 1} dx = \frac{2\pi}{3\sqrt{3}} \implies I = \frac{\pi}{3\sqrt{3}}$.

Q53. Find the area bounded by the curves $y = |x - 1|$ and $y = 3 - |x|$.
  1. $4\text{ sq units}$
  2. $3\text{ sq units}$
  3. $4.5\text{ sq units}$
  4. $5\text{ sq units}$

Answer: (A) $4\text{ sq units}$

Intersections occur at $x = 2$ ($y=1$) and $x = -1$ ($y=2$). The region forms a quadrilateral with vertices $(-1, 2)$, $(1, 0)$, $(2, 1)$, and $(0, 3)$. Calculating area by integrating top curve minus bottom curve gives $4\text{ sq units}$.

Q54. If matrix $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, then $A^n$ for any positive integer $n$ is equal to:
  1. $\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$
  2. $\begin{bmatrix} n & 1 \\ 0 & n \end{bmatrix}$
  3. $\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$
  4. $\begin{bmatrix} n & n \\ 0 & n \end{bmatrix}$

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

By mathematical induction or direct multiplication, $A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$, $A^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$, so $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.

Q55. If complex number $z = x + iy$ satisfies $|z - 2| = |z + 2|$, then the locus of $z$ is:
  1. y-axis
  2. x-axis
  3. Circle of radius $2$
  4. Ellipse

Answer: (A) y-axis

$|z - 2| = |z + 2| \implies (x-2)^2 + y^2 = (x+2)^2 + y^2 \implies -4x = 4x \implies 8x = 0 \implies x = 0$, which represents the y-axis.

Q56. Evaluate limit: $\lim_{x \to 0} \frac{e^{x^2} - \cos x}{x^2}$
  1. $\frac{3}{2}$
  2. $\frac{1}{2}$
  3. $1$
  4. $2$

Answer: (A) $\frac{3}{2}$

Using standard expansion: $e^{x^2} = 1 + x^2 + O(x^4)$ and $\cos x = 1 - \frac{x^2}{2} + O(x^4)$. $\lim_{x \to 0} \frac{(1 + x^2) - (1 - x^2/2)}{x^2} = \lim_{x \to 0} \frac{\frac{3}{2}x^2}{x^2} = \frac{3}{2}$.

Q57. Two dice are thrown. What is the conditional probability that the sum of the numbers is $8$ given that at least one die shows a $5$?
  1. $\frac{2}{11}$
  2. $\frac{1}{11}$
  3. $\frac{3}{11}$
  4. $\frac{5}{36}$

Answer: (A) $\frac{2}{11}$

Event $B$ (at least one 5): $11$ outcomes. Event $A$ (sum is 8): $(2,6), (3,5), (4,4), (5,3), (6,2)$. Intersection $A \cap B$: $(3,5)$ and $(5,3)$ ($2$ outcomes). $P(A|B) = \frac{2}{11}$.

Q58. The general solution of differential equation $\frac{dy}{dx} + y = e^{-x}$ with initial condition $y(0) = 1$ is:
  1. $y = (x+1)e^{-x}$
  2. $y = xe^{-x}$
  3. $y = e^{-x} + x$
  4. $y = (1-x)e^{-x}$

Answer: (A) $y = (x+1)e^{-x}$

Integrating factor $\text{I.F.} = e^{\int 1 dx} = e^x$. Solution: $y e^x = \int e^x e^{-x} dx = \int 1 dx = x + C \implies y = (x+C)e^{-x}$. Given $y(0)=1 \implies 1 = (0+C)1 \implies C=1$. So $y = (x+1)e^{-x}$.

Q59. Sum of the infinite series $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \dots$ is:
  1. $\frac{9}{4}$
  2. $\frac{3}{2}$
  3. $\frac{4}{3}$
  4. $\frac{9}{2}$

Answer: (A) $\frac{9}{4}$

$S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \dots$ ; $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \dots$. Subtracting gives $\frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots = \frac{1}{1 - 1/3} = \frac{3}{2}$. Thus $S = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}$.

Q60. If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$, projection of vector $\vec{a}$ on vector $\vec{b}$ is:
  1. $\frac{4}{\sqrt{14}}$
  2. $\frac{4}{3}$
  3. $\frac{2}{\sqrt{14}}$
  4. $\frac{4}{\sqrt{3}}$

Answer: (A) $\frac{4}{\sqrt{14}}$

Projection $= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{(1)(2) + (1)(-1) + (1)(3)}{\sqrt{2^2 + (-1)^2 + 3^2}} = \frac{2 - 1 + 3}{\sqrt{4 + 1 + 9}} = \frac{4}{\sqrt{14}}$.

Q61. The perpendicular distance of point $(1, 2, 3)$ from the plane $x - 2y + 2z = 9$ is:
  1. $2$
  2. $3$
  3. $1$
  4. $4$

Answer: (A) $2$

Perpendicular distance $d = \frac{|1(1) - 2(2) + 2(3) - 9|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{|1 - 4 + 6 - 9|}{\sqrt{9}} = \frac{|-6|}{3} = 2$.

Q62. The term independent of $x$ in the expansion of $\left(2x^2 - \frac{1}{x}\right)^9$ is:
  1. $672$
  2. $-672$
  3. $84$
  4. $-84$

Answer: (A) $672$

General term $T_{r+1} = \binom{9}{r}(2x^2)^{9-r}\left(-\frac{1}{x}\right)^r = \binom{9}{r}2^{9-r}(-1)^r x^{18-3r}$. For independent term, $18 - 3r = 0 \implies r = 6$. Term $= \binom{9}{6} 2^3 (-1)^6 = 84 \times 8 = 672$.

Q63. The value of $\tan\left(2 \tan^{-1}\left(\frac{1}{5}\right) - \frac{\pi}{4}\right)$ is equal to:
  1. $-\frac{7}{17}$
  2. $\frac{7}{17}$
  3. $-\frac{17}{7}$
  4. $1$

Answer: (A) $-\frac{7}{17}$

$2\tan^{-1}(1/5) = \tan^{-1}\left(\frac{2/5}{1 - 1/25}\right) = \tan^{-1}(5/12)$. Then $\tan\left(\tan^{-1}(5/12) - \frac{\pi}{4}\right) = \frac{5/12 - 1}{1 + 5/12} = \frac{-7/12}{17/12} = -\frac{7}{17}$.

Q64. Total number of 4-digit numbers formed using digits 1, 2, 3, 4, 5 without repetition that are divisible by 4 is:
  1. $24$
  2. $18$
  3. $12$
  4. $36$

Answer: (A) $24$

Divisible by 4 means last two digits must be divisible by 4. Possible last two digits from {1,2,3,4,5}: 12, 24, 32, 52 ($4$ choices). For each choice, remaining 2 places are filled by remaining 3 digits in $^3P_2 = 6$ ways. Total $= 4 \times 6 = 24$.

Q65. The function $f(x) = |x| + |x - 1|$ is non-differentiable at:
  1. $x = 0$ and $x = 1$
  2. $x = 0$ only
  3. $x = 1$ only
  4. Everywhere differentiable

Answer: (A) $x = 0$ and $x = 1$

Absolute value functions $|x - a|$ have sharp corners at $x = a$. Therefore, $f(x)$ is continuous everywhere but non-differentiable at corner points $x = 0$ and $x = 1$.

Q66. The minimum value of $f(x) = x^2 + \frac{250}{x}$ for $x > 0$ is:
  1. $75$
  2. $50$
  3. $100$
  4. $25$

Answer: (A) $75$

$f'(x) = 2x - \frac{250}{x^2} = 0 \implies 2x^3 = 250 \implies x^3 = 125 \implies x = 5$. Minimum value $f(5) = 5^2 + \frac{250}{5} = 25 + 50 = 75$.

Q67. The equation of tangent to the circle $x^2 + y^2 = 25$ at point $(3, 4)$ is:
  1. $3x + 4y = 25$
  2. $4x + 3y = 25$
  3. $3x - 4y = 25$
  4. $4x - 3y = 25$

Answer: (A) $3x + 4y = 25$

Equation of tangent at $(x_1, y_1)$ is $x x_1 + y y_1 = r^2$. Substituting $(3, 4)$: $3x + 4y = 25$.

Q68. Length of the latus rectum of parabola $y^2 - 4y - 8x + 12 = 0$ is:
  1. $8$
  2. $4$
  3. $2$
  4. $16$

Answer: (A) $8$

Rewrite equation as $(y - 2)^2 - 4 - 8x + 12 = 0 \implies (y - 2)^2 = 8(x - 1)$. Comparing with $(y - k)^2 = 4a(x - h)$, $4a = 8$. Latus rectum length $= 8$.

Q69. Eccentricity of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is:
  1. $\frac{4}{5}$
  2. $\frac{3}{5}$
  3. $\frac{16}{25}$
  4. $\frac{2}{5}$

Answer: (A) $\frac{4}{5}$

Here $a^2 = 25$, $b^2 = 9$. Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Q70. If $A$ and $B$ are two sets such that $n(A) = 20$, $n(B) = 28$, and $n(A \cup B) = 36$, then $n(A \cap B)$ is equal to:
  1. $12$
  2. $8$
  3. $16$
  4. $14$

Answer: (A) $12$

By Inclusion-Exclusion principle: $n(A \cap B) = n(A) + n(B) - n(A \cup B) = 20 + 28 - 36 = 12$.

Q71. The value of $\cos 20^\circ \cos 40^\circ \cos 80^\circ$ is equal to:
  1. $\frac{1}{8}$
  2. $\frac{1}{4}$
  3. $\frac{1}{2}$
  4. $\frac{1}{16}$

Answer: (A) $\frac{1}{8}$

Using identity $\cos A \cos 2A \cos 4A = \frac{\sin(2^3 A)}{2^3 \sin A}$: $\frac{\sin(160^\circ)}{8 \sin 20^\circ} = \frac{\sin(180^\circ - 20^\circ)}{8 \sin 20^\circ} = \frac{\sin 20^\circ}{8 \sin 20^\circ} = \frac{1}{8}$.

Q72. The domain of function $f(x) = \sqrt{\log_{10}\left(\frac{5x - x^2}{4}\right)}$ is:
  1. $[1, 4]$
  2. $(0, 5)$
  3. $[0, 5]$
  4. $(1, 4)$

Answer: (A) $[1, 4]$

For square root and logarithm to be valid: $\log_{10}\left(\frac{5x - x^2}{4}\right) \ge 0 \implies \frac{5x - x^2}{4} \ge 1 \implies 5x - x^2 \ge 4 \implies x^2 - 5x + 4 \le 0 \implies (x-1)(x-4) \le 0 \implies x \in [1, 4]$.

Q73. Evaluate integral: $\int \frac{dx}{x(x^n + 1)}$
  1. $\frac{1}{n}\ln\left|\frac{x^n}{x^n + 1}\right| + C$
  2. $\frac{1}{n}\ln\left|\frac{x^n + 1}{x^n}\right| + C$
  3. $\ln\left|\frac{x^n}{x^n + 1}\right| + C$
  4. $\frac{1}{n}\ln|x^n + 1| + C$

Answer: (A) $\frac{1}{n}\ln\left|\frac{x^n}{x^n + 1}\right| + C$

Multiply numerator and denominator by $x^{n-1}$: $\int \frac{x^{n-1}}{x^n(x^n + 1)} dx$. Put $x^n = t \implies n x^{n-1} dx = dt$. Integral $= \frac{1}{n}\int \frac{dt}{t(t+1)} = \frac{1}{n}\int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt = \frac{1}{n}\ln\left|\frac{t}{t+1}\right| + C = \frac{1}{n}\ln\left|\frac{x^n}{x^n + 1}\right| + C$.

Q74. The eccentricity of a rectangular hyperbola is always equal to:
  1. $\sqrt{2}$
  2. $2$
  3. $\frac{1}{\sqrt{2}}$
  4. $\sqrt{3}$

Answer: (A) $\sqrt{2}$

For a rectangular hyperbola $a = b$. Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2}$.

Q75. If the variance of observations $x_1, x_2, \dots, x_n$ is $\sigma^2$, then the variance of $2x_1 + 3, 2x_2 + 3, \dots, 2x_n + 3$ is:
  1. $4\sigma^2$
  2. $2\sigma^2$
  3. $4\sigma^2 + 3$
  4. $2\sigma^2 + 3$

Answer: (A) $4\sigma^2$

Properties of variance: $\text{Var}(aX + b) = a^2 \text{Var}(X)$. Here $a = 2$, $b = 3$. New variance $= 2^2 \sigma^2 = 4\sigma^2$.