JEE Main 2026 Jan 24 shift 1

Potential in a spherical region is given by $V = ar^3 + b$. We seek the total charge enclosed \in a sphere of unit radius. Since the electric field is related to the potential by $E = -\frac{dV}{dr}$, it follows that $E = -3ar^2$. The negative sign indicates the field points inward when $a>0$. Using Gauss's law for a sphere of radius $r$, we have $\oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$, which simplifies to $E\cdot 4\pi r^2 = \frac{Q}{\varepsilon_0}$. At $r = 1$, the field becomes $E = -3a$ (with magnitude $|E| = 3|a|$, directed radially). Substituting these values gives $(-3a)\,(4\pi)\,(1)^2 = \frac{Q}{\varepsilon_0}$, and hence $Q = -12\pi a\varepsilon_0$. Comparing this result with the form $Q = \alpha \pi a \varepsilon_0$ shows that $\alpha = -12$. The correct answer is Option (3): $-12$.

Three point masses of 200 kg, 300 kg, and 400 kg originally occupy the vertices of an equilateral triangle of side 20 m, and they are then rearranged to the vertices of a larger equilateral triangle of side 25 m with the same center. We wish to find the work done in this rearrangement. The gravitational potential energy of three point masses at the vertices of a triangle is given by $U = -G\left(\frac{m_1 m_2}{r_{12}} + \frac{m_2 m_3}{r_{23}} + \frac{m_1 m_3}{r_{13}}\right).$ Since the triangle is equilateral, all pairwise distances are equal to a common value $r$, so $U = -\frac{G}{r}\bigl(m_1m_2 + m_2m_3 + m_1m_3\bigr)\,.$ Calculating the \sum of mass products gives $m_1m_2 + m_2m_3 + m_1m_3 = 200\times300 + 300\times400 + 200\times400 = 60000 + 120000 + 80000 = 260000 \text{ kg}^2\,.$ The work done is the change \in potential energy, so $W = U_f - U_i = -\frac{G \cdot 260000}{25} - \Bigl(-\frac{G \cdot 260000}{20}\Bigr).$ This gives $W = G \cdot 260000\Bigl(\frac{1}{20} - \frac{1}{25}\Bigr) = G \cdot 260000 \cdot \frac{1}{100} = G \cdot 2600\,.$ Substituting the gravitational constant $G = 6.7 \times 10^{-11}$ yields $W = 6.7 \times 10^{-11} \times 2600 = 6.7 \times 2.6 \times 10^{-8} = 17.42 \times 10^{-8} = 1.742 \times 10^{-7} \approx 1.74 \times 10^{-7} \text{ J}.\,$ The correct answer is Option (4): $1.74 \times 10^{-7}$.

We need to find the height $H$ of the building where a ball reaches its maximum height \in 2 s and lands on the building \in 3 s. Since the ball is projected at $45°$, its horizontal and vertical components of the initial velocity are equal, namely $u_x = u_y = \frac{u}{\sqrt{2}}$. The time to reach the maximum height is $t_{max} = \frac{u_y}{g} = \frac{u/\sqrt{2}}{10} = 2$, which gives $u = 20\sqrt{2}$ m/s and therefore $u_x = 20$ m/s and $u_y = 20$ m/s. To find the height after 3 s, we use the vertical displacement formula $H = u_y t - \frac{1}{2}gt^2$. Substituting $u_y = 20$ m/s and $t = 3$ s into this expression yields $H = 20(3) - \frac{1}{2}(10)(9) = 60 - 45 = 15$ m. Hence, the height of the building is $H = 15$ m. Therefore, the answer is Option D.

Dipole moment: $\vec{p} = \sum q_i \vec{r}_i = 2q(0,-3a) + 3q(2a,0) + (-4q)(-2a,0) = (0,-6qa) + (6qa,0) + (8qa,0) = (14qa, -6qa) = 2qa(7\hat{i}-3\hat{j})$. The answer is Option 2 : $2qa(7\hat{i}-3\hat{j})$.

If the block is pushed down by a small distance x, extra volume submerged is Ax So extra buoyant force upward is $F_b=\rho g(Ax)$ This acts as restoring force: $F=-\rho gAx$ which is of SHM form $F=−kx$ with effective spring constant $k=\rho gA$ For SHM, $T=2\pi\sqrt{\frac{M}{k}}$ Substituting $k=\rho gA$ we get $2\pi \sqrt{\frac{M}{\rho Ag}}$

When the reflected and refracted rays are perpendicular, the angle of incidence is the Brewster angle ($\theta_B$).
According to Brewster's Law:
$\tan \theta_B = \frac{\mu_2}{\mu_1}$
Given $\mu_1 = 2$ and $\mu_2 = 2\sqrt{3}$:
$\tan \theta_B = \frac{2\sqrt{3}}{2} = \sqrt{3}$
$\theta_B = 60^{\circ}$

We need to find the increase in internal energy of 4 kg of water heated from 4 °C to 20 °C. Given that the mass of water is $m = 4$ kg, the density at 4 °C is $\rho_1 = 1000$ kg/m³, the density at 20 °C is $\rho_2 = 998$ kg/m³, the specific heat capacity is $c = 4.2 \times 10^3$ J/(kg·°C), the temperature change is $\Delta T = 20 - 4 = 16\ ^\circ\mathrm{C}$, and the atmospheric pressure is $P = 10^5$ Pa. Using the first law of thermodynamics, the change \in internal energy is given by $\Delta U = Q - W$. Since the heat supplied is $Q = mc\Delta T = 4 \times 4200 \times 16 = 268800\text{ J},$ we have $Q = 268800\text{ J}.$ At constant atmospheric pressure, the work done by the system is $W = P\,\Delta V = P\Bigl(\frac{m}{\rho_2} - \frac{m}{\rho_1}\Bigr).$ Substituting the values gives $W = 10^5\Bigl(\frac{4}{998} - \frac{4}{1000}\Bigr) = 10^5\Bigl(\frac{4\times1000 - 4\times998}{998\times1000}\Bigr) = 10^5 \times \frac{8}{998000} = \frac{8 \times 10^5}{998000} \approx 0.8\text{ J}.$ Therefore, the internal energy change is $\Delta U = Q - W = 268800 - 0.8 = 268799.2\text{ J}.$ Thus, the increase in internal energy is Option 3 : 268799.2 J.

To match List-I with List-II correctly, we analyze each option:

• A: Radio-wave is IV (due to rapid acceleration of electrons).
• B: Micro-wave is III (due to inner shell electrons moving from higher to lower energy levels).
• C: Infrared-wave is I (produced by Magnetron valve).
• D: X-ray is II (due to change in the vibrational mode of atoms).

The correct matches are A-IV, B-III, C-I, D-II, which corresponds to option A, not option C. Thus, the target answer is incorrect.

We need to find the potentials of three concentric conducting spherical shells A, B, and C with radii $a$, $b$, and $c$ (where $c > b > a$) carrying charges $q_1$, $q_2$, and $q_3$ respectively. Since the key principle is that the potential at any point due to a conducting spherical shell of radius R carrying charge q is $\frac{q}{4\pi\epsilon_0 R}$ when the point is on the shell or inside it ($r \le R$), and $\frac{q}{4\pi\epsilon_0 r}$ when the point is outside ($r > R$). For sphere A at radius $a$, the contribution from $q_1$ on A itself is $V_1 = \frac{q_1}{4\pi\epsilon_0 a}$ since the point lies on A, from $q_2$ on B is $V_2 = \frac{q_2}{4\pi\epsilon_0 b}$ as the point is inside B, and from $q_3$ on C is $V_3 = \frac{q_3}{4\pi\epsilon_0 c}$ because the point is inside C. Therefore, the total potential at A is $V_A = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{a} + \frac{q_2}{b} + \frac{q_3}{c}\right)$. Turning to sphere B at radius $b$, the potential due to $q_1$ on A is $\frac{q_1}{4\pi\epsilon_0 b}$ since B lies outside A, the potential from $q_2$ on B itself is $\frac{q_2}{4\pi\epsilon_0 b}$, and the potential from $q_3$ on C is $\frac{q_3}{4\pi\epsilon_0 c}$ because B is inside C. Hence, the total potential at B is $V_B = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1 + q_2}{b} + \frac{q_3}{c}\right)$. Finally, for sphere C at radius $c$, both A and B lie inside C, so the contributions from $q_1$ and $q_2$ are $\frac{q_1}{4\pi\epsilon_0 c}$ and $\frac{q_2}{4\pi\epsilon_0 c}$ respectively, and the self-potential from $q_3$ is $\frac{q_3}{4\pi\epsilon_0 c}$. Thus, the total potential at C is $V_C = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1 + q_2 + q_3}{c}\right)$. Comparing these results with the given options shows that they correspond to Option 2.

We need to find the force constant of the smaller piece when a spring of force constant 15 N/m is cut in the ratio 1:3. Since the force constant k of a spring of length L is inversely proportional to its length, we have $k \times L = k_1 \times L_1 = k_2 \times L_2 = \text{constant}$ Given that k = 15 N/m and the spring is cut \in the ratio 1:3, if the total length is L then the smaller piece has length $L_1 = \frac{L}{4}$ and the larger piece has length $L_2 = \frac{3L}{4}$. Substituting into the relation gives $k_1 = \frac{k \times L}{L_1} = \frac{15 \times L}{\frac{L}{4}} = 15 \times 4 = 60 \text{ N/m}$ Therefore, the force constant of the smaller piece is 60 N/m (Option 3).

We need to find the temperature to which a brass wire must be cooled to increase its tension from $T$ to $1.4T$. Since when a wire held between rigid supports is cooled it tries to contract but cannot, a thermal stress develops, producing a tension given by $T = YA\alpha\Delta\theta$, where $Y$ = Young's modulus, $A$ = cross-sectional area, $\alpha$ = coefficient of linear expansion, and $\Delta\theta$ = temperature change. Initially the wire is at 27°C and is cooled to -43°C, so $\Delta\theta_1 = 27 - (-43) = 70°C$; substituting into the tension expression gives $T = YA\alpha \times 70$. For the increased tension $1.4T$ we therefore have $1.4T = YA\alpha \times \Delta\theta_2$. Dividing these equations yields $\frac{1.4T}{T} = \frac{\Delta\theta_2}{70}$ and hence $\Delta\theta_2 = 1.4 \times 70 = 98°C$. This means the wire must be cooled from 27°C by 98°C, giving $\theta = 27 - 98 = -71°C$. Therefore, the required temperature is Option 4 : -71°C.

In an LCR series circuit, the impedance $Z$ can be calculated as:

$Z = R + j(X_L - X_C)$

Here, $R = 60 \, \Omega$, $X_L = 70 \, \Omega$, and $X_C = 150 \, \Omega$. Therefore,

$X = X_L - X_C = 70 - 150 = -80 \, \Omega$

The total impedance thus becomes:

$Z = 60 - j80$

The magnitude is:

$|Z| = \sqrt{R^2 + X^2} = \sqrt{60^2 + (-80)^2} = \sqrt{3600 + 6400} = 100 \, \Omega$

The power factor $\cos \phi$ is given by:

$\cos \phi = \frac{R}{|Z|} = \frac{60}{100} = 0.6$

Since the power factor is $\frac{\alpha}{10}$, we find:

$\frac{\alpha}{10} = 0.6 \implies \alpha = 6$

Thus, the value of $\alpha$ is $6$.

Step 1: initial balance condition Let the balancing length from X be l. Then, 2 / 3 = l / (100 − l) Solve: 2(100 − l) = 3l 200 − 2l = 3l 200 = 5l l = 40 cm So initial null point is at 40 cm. Step 2: after adding unknown resistor in parallel with 3Ω Let equivalent resistance of (3Ω || R) be: R' = (3R)/(3 + R) Now new balance length shifts 22.5 cm toward Y → length increases: new length = 40 + 22.5 = 62.5 cm Step 3: new balance condition 2 / R' = 62.5 / (100 − 62.5) 2 / R' = 62.5 / 37.5 2 / R' = 5 / 3 So, R' = 6 / 5 = 1.2 Ω Step 4: solve for unknown R (3R)/(3 + R) = 1.2 3R = 1.2(3 + R) 3R = 3.6 + 1.2R 3R − 1.2R = 3.6 1.8R = 3.6 R = 2 Ω

We need to find the energy states (quantum numbers) of electrons in two hydrogen-like atoms given their speeds and that the orbital radii are nearly the same. Key formulas for hydrogen-like atoms: Speed of electron in nth orbit: $v_n = \frac{Ze^2}{2\epsilon_0 n h}$, which gives $v \propto \frac{Z}{n}$ Radius of nth orbit: $r_n = \frac{n^2 a_0}{Z}$, which gives $r \propto \frac{n^2}{Z}$ $v_1 = 3 \times 10^5$ m/s, $v_2 = 2.5 \times 10^5$ m/s, and $r_1 \approx r_2$ Since radii are approximately equal: $\frac{n_1^2}{Z_1} = \frac{n_2^2}{Z_2}$ $\frac{Z_1}{Z_2} = \frac{n_1^2}{n_2^2} \quad \cdots (1)$ From the velocity relation: $\frac{v_1}{v_2} = \frac{Z_1/n_1}{Z_2/n_2} = \frac{Z_1 n_2}{Z_2 n_1}$ Substituting from equation (1): $\frac{Z_1}{Z_2} = \frac{n_1^2}{n_2^2}$ $\frac{v_1}{v_2} = \frac{n_1^2}{n_2^2} \times \frac{n_2}{n_1} = \frac{n_1}{n_2}$ $\frac{3 \times 10^5}{2.5 \times 10^5} = \frac{n_1}{n_2}$ $\frac{n_1}{n_2} = \frac{3}{2.5} = \frac{6}{5}$ So $n_1 = 6$ and $n_2 = 5$ (taking the simplest integer ratio). Therefore, the possible energy states are Option 3 : 6 and 5.

Magnetic induction BBB: From Lorentz force: $F=qvB\Rightarrow B=\frac{F}{qv}$ $[B]=\frac{MLT^{-2}}{(AT)(LT^{-1})}=[MT^{-2}A^{-1}]$ So, (A) matches with (III) Magnetic flux $\Phi$: $\Phi=B\cdot A$ $[\Phi]=\$[MT^{-2}A^{-1}]\$\cdot\$[L^2]\$=[ML^2T^{-2}A^{-1}]$ So, (B) matches with (IV) Magnetic permeability \mu \mu\mu : $B=\mu\frac{I}{L}\Rightarrow\mu=\frac{BL}{I}$ $[\mu]=\frac{(MT^{-2}A^{-1})\cdot L}{A}=[MLT^{-2}A^{-2}]$ So, (C) matches with (I) Self inductance L: $\Phi=LI\Rightarrow L=\frac{\Phi}{I}$ $[L]=\frac{ML^2T^{-2}A^{-1}}{A}=[ML^2T^{-2}A^{-2}]$ So, (D) matches with (II) Final matching: (A)−(III), (B)−(IV), (C)−(I), (D)−(II)

For total internal reflection to occur at the prism exit surface coated with a material of refractive index $\frac{n}{2}$, the critical angle $\theta_c$ must satisfy $\sin \theta_c = \frac{n/2}{n} = \frac{1}{2}$, giving $\theta_c = 30^{\circ}$. At minimum angle of deviation, the prism angle $A$ is twice the critical angle, thus $A = 2 \times 30^{\circ} = 60^{\circ}$. Therefore, the prism angle is $60^{\circ}$.

We need to find the voltmeter reading when a 400 ohm voltmeter is connected across one of two 100 ohm resistors in series with a 9V battery. Two resistors: $R_1 = R_2 = 100 \, \Omega$ \in series Battery voltage: $V = 9$ V Voltmeter resistance: $R_V = 400 \, \Omega$ The voltmeter (400 ohm) is connected \in parallel with one 100 ohm resistor: $R_{parallel} = \frac{R_1 \times R_V}{R_1 + R_V} = \frac{100 \times 400}{100 + 400} = \frac{40000}{500} = 80 \, \Omega$ $R_{total} = R_{parallel} + R_2 = 80 + 100 = 180 \, \Omega$ $I = \frac{V}{R_{total}} = \frac{9}{180} = 0.05 \text{ A}$ $V_{reading} = I \times R_{parallel} = 0.05 \times 80 = 4 \text{ V}$ Therefore, the voltmeter reading is Option 1 : 4 V.

We need to evaluate two statements about nuclear binding energy. Statement I: "For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon." This statement is FALSE . The binding energy per nucleon does not increase monotonically with mass number. It increases up to iron (Fe-56) and then decreases for heavier nuclei. For example, uranium has a lower binding energy per nucleon than iron. Statement II: "For all elements, nuclei with less binding energy per nucleon transforms to nuclei with greater binding energy per nucleon." This statement is TRUE . Nuclear reactions (both fission and fusion) proceed in the direction of increasing binding energy per nucleon, as this leads to more stable configurations. Light nuclei undergo fusion and heavy nuclei undergo fission, both leading to products with higher binding energy per nucleon. Therefore, Statement I is false and Statement II is true, which corresponds to Option 4 .

We are asked to find the rotational inertia of a heavy pulley given that a 400 g mass falls 81 cm in 9 s when connected to a 350 g mass via a string over the pulley. Here, $m_1 = 400\text{ g} = 0.4\text{ kg}$, $m_2 = 350\text{ g} = 0.35\text{ kg}$, the pulley radius is $R = 2\text{ cm} = 0.02\text{ m}$, the distance fallen is $s = 81\text{ cm} = 0.81\text{ m}$ \in time $t = 9\text{ s}$, and $g = 9.8\text{ m/s}^2$. Starting from rest, the acceleration can be found from the kinematic relation $s = \tfrac12 a t^2$, which gives $0.81 = \frac{1}{2} \times a \times 81$ $a = \frac{2 \times 0.81}{81} = 0.02\text{ m/s}^2.$ Applying Newton's second law to each mass and the torque equation for the pulley yields For mass $m_1$ (heavier, descending): $m_1 g - T_1 = m_1 a \quad (1)$ For mass $m_2$ (lighter, ascending): $T_2 - m_2 g = m_2 a \quad (2)$ For the pulley (with angular acceleration $\alpha = a/R$): $(T_1 - T_2)R = I \alpha = I \frac{a}{R} \quad (3)$ From equation (1), $T_1 = m_1(g - a) = 0.4(9.8 - 0.02) = 0.4 \times 9.78 = 3.912\text{ N}.$ From equation (2), $T_2 = m_2(g + a) = 0.35(9.8 + 0.02) = 0.35 \times 9.82 = 3.437\text{ N}.$ Substituting these into equation (3) and solving for $I$ gives $I = \frac{(T_1 - T_2)R^2}{a} = \frac{(3.912 - 3.437)\times (0.02)^2}{0.02} = \frac{0.475 \times 0.0004}{0.02} = \frac{0.00019}{0.02} = 0.0095\text{ kg·m}^2 = 9.5 \times 10^{-3}\text{ kg·m}^2.$ Therefore, the rotational inertia of the pulley is Option 1: $9.5 \times 10^{-3}\text{ kg·m}^2.$

A compound microscope has tube length $L = 10$ cm, objective focal length $f_o = 2$ cm, and eyepiece focal length $f_e = 5$ cm. We wish to find the exponent $k$ such that the total magnification is $(5)^k$. In normal adjustment (image at infinity), the total magnification is given by $M = \frac{L \times D}{f_o \times f_e},$ where $D = 25$ cm is the least distance of distinct vision. Substituting the given values, $M = \frac{10 \times 25}{2 \times 5} = \frac{250}{10} = 25 = 5^2.$ Since $5^k = 25 = 5^2,$ it follows that $k = 2$. Therefore, the correct answer is Option A: $k = 2$.

To calculate the force $F$ required to slide block $C$ with constant speed, we need to consider the total weight of the blocks above it and the coefficient of friction.

The normal force $N$ on block $C$ is the weight of blocks $A$ and $B$:

The frictional force $f$ is given by:

Since it takes this much force to overcome friction and slide block $C$, the required force $F$ is $50 \, \text{N}$, not $210 \, \text{N}$.

Gas at 3.23 kPa and 50°C is heated to double its temperature in a closed cylinder. The initial temperature of the gas is $T_1 = 50°C = 323$ K, and doubling the Celsius temperature gives $T_2 = 2 \times 50 = 100°C = 373$ K. At constant volume the pressure and temperature are related by $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ so that $P_2 = P_1 \times \frac{T_2}{T_1} = 3.23 \times \frac{373}{323} = 3.23 \times 1.1548 = 3.73$ kPa. Converting to pascals gives $P_2 = 3.73 \times 1000 = 3730$ Pa. The correct answer is $\boxed{3730}$ Pa.

We need to find the work done in moving a short bar magnet from the most stable to the most unstable position. Angle with external field: $\theta = 30°$ External field: $B = 800$ Gauss = 0.08 T Torque: $\tau = 0.016$ N.m Torque on a magnetic dipole: $\tau = mB\sin\theta$ $0.016 = m \times 0.08 \times \sin 30°$ $0.016 = m \times 0.08 \times 0.5 = 0.04m$ $m = \frac{0.016}{0.04} = 0.4 \text{ A.m}^2$ The most stable position is when $\theta = 0°$ (magnet aligned with the field). The most unstable position is when $\theta = 180°$ (magnet anti-parallel to the field). Work done is the change \in potential energy: $W = U_{final} - U_{initial} = (-mB\cos 180°) - (-mB\cos 0°)$ $W = mB + mB = 2mB$ $W = 2 \times 0.4 \times 0.08 = 0.064 \text{ J} = 64 \times 10^{-3} \text{ J}$ Therefore, $\alpha =$ 64 .

We need to find the terminal velocity of a bigger drop formed by the coalescence of 64 small raindrops. Number of small drops: $n = 64$ Radius of each small drop: $r = 1$ mm Terminal velocity of each small drop: $v = 10$ cm/s $n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$ $R^3 = n \times r^3 = 64 r^3$ $R = 4r$ Terminal velocity is proportional to the square of the radius: $v_t \propto r^2$ (This comes from Stokes' law: $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$) $\frac{V}{v} = \frac{R^2}{r^2} = \frac{(4r)^2}{r^2} = 16$ $V = 16 \times v = 16 \times 10 = 160 \text{ cm/s}$ Therefore, the terminal velocity of the bigger drop is 160 cm/s .

We need to find the protective resistance in a Zener diode voltage regulating circuit. Zener breakdown voltage: $V_Z = 10$ V Maximum power dissipation of Zener: $P_{max} = 0.4$ W Supply voltage: $V_S = 15$ V $I_Z = \frac{P_{max}}{V_Z} = \frac{0.4}{10} = 0.04 \text{ A} = 40 \text{ mA}$ The voltage drop across the protective resistance: $V_R = V_S - V_Z = 15 - 10 = 5 \text{ V}$ The protective resistance must limit the current through the Zener to its maximum value (assuming no load current, worst case): $R = \frac{V_R}{I_Z} = \frac{5}{0.04} = 125 \, \Omega$ Therefore, the approximate value of the protective resistance is 125 ohm .

We need to evaluate two statements about coordination compounds. Statement I: $K_3[Co(CO_3)_3]$ has hybridisation $sp^3d^2$, octahedral shape and spin-only magnetic moment of 4.9 BM. In $K_3[Co(CO_3)_3]$: The charge on the complex ion $[Co(CO_3)_3]^{3-}$: Since $CO_3^{2-}$ is bidentate, each carbonate acts as a bidentate ligand. With 3 carbonate ions giving a charge of -6, and the complex having a charge of -3, Co has an oxidation state of +3. Co³⁺ has configuration: $[Ar] 3d^6$ $CO_3^{2-}$ is a weak field ligand, so the electrons remain unpaired. With 6 donor atoms (3 bidentate carbonates), the geometry is octahedral. For weak field: 4 unpaired electrons, $\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.9$ BM Hybridisation: $sp^3d^2$ (outer orbital complex for weak field ligand) Statement I is TRUE . Statement II: $[Ni(CN)_4]^{2-}$: Ni²⁺ (3d⁸), CN⁻ is a strong field ligand, forces pairing. Hybridisation: $dsp^2$, square planar, 0 unpaired electrons, $\mu = 0$ BM. ✓ $[MnBr_4]^{2-}$: Mn²⁺ (3d⁵), Br⁻ is a weak field ligand. Hybridisation: $sp^3$, tetrahedral, 5 unpaired electrons, $\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.9$ BM. ✓ $[CoF_6]^{3-}$: Co³⁺ (3d⁶), F⁻ is a weak field ligand. Hybridisation: $sp^3d^2$, octahedral, 4 unpaired electrons, $\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.9$ BM. ✓ Statement II is TRUE . Both statements are true, which corresponds to Option 2 .

The stability of carbanions is influenced by the electronegativity and resonance effects of substituents. Typically, more electronegative groups (like carbonyls) stabilize the negative charge through resonance.

In this case:

• IV (p-CHO) > II (C₆H₅-CH₂) > I (p-Br) > V (p-CH₃) > III (C₆H₃O)

Thus, the correct order of stability is D: IV > I > II > V > III.

We need to identify three metal chlorides based on their solubility properties. x: Water soluble at room temperature y: Sparingly soluble in water at room temperature z: Soluble in hot water Option 2: $CuCl_2, AgCl, PbCl_2$ - $CuCl_2$: Copper(II) chloride is readily soluble \in water at room temperature. ✓ (matches x) - $AgCl$: Silver chloride is sparingly soluble (nearly insoluble) \in water at room temperature. ✓ (matches y) - $PbCl_2$: Lead(II) chloride is sparingly soluble \in cold water but dissolves readily \in hot water. ✓ (matches z) This perfectly matches all three conditions. Checking other options: - Option 1: $AlCl_3$ is soluble, $AgCl$ is sparingly soluble, but $AlCl_3$ doesn't specifically need hot water to dissolve. - Option 3: $BaCl_2$ is freely soluble \in water at room temperature, not specifically requiring hot water. - Option 4: The notation $AlCl$ is unusual, and $Hg_2Cl_2$ is insoluble even \in hot water. Therefore, the correct answer is Option 2 : $CuCl_2, AgCl, PbCl_2$.

We need to find the logarithm of the ratio of rate constants for catalysed and uncatalysed reactions. Temperature: $T = 27°C = 300$ K Reduction \in activation energy: $\Delta E_a = 10$ kJ/mol = 10000 J/mol Frequency factor is the same for both reactions. $k = A e^{-E_a/RT}$ For the ratio: $\frac{k_{cat}}{k_{uncat}} = \frac{A e^{-E_{a,cat}/RT}}{A e^{-E_{a,uncat}/RT}} = e^{(E_{a,uncat} - E_{a,cat})/RT} = e^{\Delta E_a/RT}$ $\log\left(\frac{k_{cat}}{k_{uncat}}\right) = \frac{\Delta E_a}{2.303 \times RT}$ $\log\left(\frac{k_{cat}}{k_{uncat}}\right) = \frac{10000}{2.303 \times 8.314 \times 300}$ $= \frac{10000}{2.303 \times 2494.2} = \frac{10000}{5744.14} \approx 1.741$ Therefore, the logarithm of the ratio is Option 4 : 1.741.

We need to find the number of hydroxyl groups in compound X. Molar mass of compound X: 122 g/mol Molar mass of acetylated product: 290 g/mol During acetylation with acetic anhydride, each -OH group is converted to -OCOCH₃. In this process: Each -OH (mass = 17) is replaced by -OCOCH₃ (mass = 59), so the increase in molar mass per -OH group is: $\Delta M = 59 - 17 = 42 \text{ g/mol per OH group}$ Total increase \in molar mass: $290 - 122 = 168 \text{ g/mol}$ Number of OH groups: $n = \frac{168}{42} = 4$ Therefore, the number of hydroxyl groups present in compound X is Option 2 : 4.

Alkene stability generally follows the order of substitution: more substituted alkenes are more stable due to hyperconjugation and the inductive effect.

1. I has four substituents (tetrasubstituted).
2. III has three substituents (trisubstituted).
3. II has two substituents (disubstituted).
4. IV is monosubstituted.

Thus, the order of stability is: I > III > II > IV, which aligns with option C.

Statement I is true because the C-Cl bond in the allylic compound $CH_{2}=CH-Cl$ is generally weaker due to resonance stabilization compared to the saturated compound $CH_{3}-CH_{2}-Cl$. Statement II is false since the described optically active molecule on hydrolysis does not result in a racemic mixture, which would not rotate plane-polarized light. Thus, the correct answer is A.

A mixture X contains 0.4 mol [Co(NH$_3$)$_5$SO$_4$]Br and 0.4 mol [Co(NH$_3$)$_5$Br]SO$_4$ \in 4 L of water. [Co(NH$_3$)$_5$SO$_4$]Br ionizes as: [Co(NH$_3$)$_5$SO$_4$]$^+$ + Br$^-$. Similarly, [Co(NH$_3$)$_5$Br]SO$_4$ ionizes as: [Co(NH$_3$)$_5$Br]$^{2+}$ + SO$_4^{2-}$. In 4 L total there are 0.4 mol Br$^-$ and 0.4 mol SO$_4^{2-}$. When 2 L of this mixture is treated with excess AgNO$_3$, the 0.2 mol Br$^-$ ions precipitate as AgBr (Y) according to AgNO$_3$ + Br$^-$ $\to$ AgBr, so moles of Y = 0.2 mol. Treating 2 L with excess BaCl$_2$ precipitates the 0.2 mol SO$_4^{2-}$ ions as BaSO$_4$ (Z) according to BaCl$_2$ + SO$_4^{2-}$ $\to$ BaSO$_4$, giving moles of Z = 0.2 mol. Option A: 0.1 mol of Y - Incorrect (0.2 mol formed). Option B: 0.2 mol of Z - Correct (0.2 mol BaSO$_4$). Option C: 0.4 mol of Z - Incorrect. Option D: Y is BaSO$_4$ and Z is AgBr - Incorrect (Y is AgBr, Z is BaSO$_4$). The correct answer is Option B.