JEE Main 2026 28th January Morning Shift Question Paper with Solutions

Moment of Inertia for Rod:

$I_{rod} = \frac{1}{12} m_r L^2 + m_r d_r^2$

$I_{rod} = \frac{1}{12} (600)(30)^2 + (600)(5)^2$

$I_{rod} = 45000 + 15000 = 60,000 \text{ g cm}^2$

Moment of inertia for left disc:

$I_1 = \frac{1}{4} m_d r^2 + m_d d_1^2$

$I_1 = \frac{1}{4} (600)(10)^2 + (600)(10)^2$

$I_1 = 15000 + 60000 = 75,000 \text{ g cm}^2$

Moment of inertia for right disc:

$I_2 = \frac{1}{4} m_d r^2 + m_d d_2^2$

$I_2 = 15000 + (600)(20)^2$

$I_2 = 15000 + 240,000 = 255,000 \text{ g cm}^2$

Total Moment of Inertia:

$I_{total} = 60,000 + 75,000 + 255,000 = 390,000 \text{ g cm}^2$

$\tau = I_{total} \alpha$

$\alpha = \frac{43 \times 10^5}{3.9 \times 10^5}$

$\alpha = \frac{43}{3.9} \approx 11.025 \text{ rad/s}^2$

Wire A: length 6.0 cm, area $3.0 \times 10^{-5}$ m$^2$. Wire B: length 5.4 cm, area $4.5 \times 10^{-5}$ m$^2$. Same elongation under same load. Find $Y_A/Y_B = x:3$.

Apply Hooke's law for each wire.

$\Delta L = \frac{FL}{AY}$

Same $\Delta L$ and same $F$:

$\frac{FL_A}{A_A Y_A} = \frac{FL_B}{A_B Y_B}$

$\frac{L_A}{A_A Y_A} = \frac{L_B}{A_B Y_B}$

$\frac{Y_A}{Y_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} = \frac{6.0 \times 4.5 \times 10^{-5}}{5.4 \times 3.0 \times 10^{-5}} = \frac{27}{16.2} = \frac{5}{3}$

So $Y_A:Y_B = 5:3$, meaning $x = 5$.

The correct answer is Option D: 5.

The nucleus of an atom is usually treated as a sphere whose radius varies with the mass number $A$ according to the empirical relation $R = R_0\,A^{1/3}$, where $R_0$ is a constant.

For a sphere, surface area $S$ is $4\pi R^2$. Hence,

$S \propto R^2 \;\Longrightarrow\; S \propto \left(A^{1/3}\right)^{2}=A^{2/3} \quad -(1)$

Initial nucleus: $^{8}_{3}X$  ⇒  initial mass number $A_i = 8$.

During the $10\,$s bombardment the nucleus absorbs
 • $10$ protons  ($p$)
 • $9$ neutrons  ($n$)
(The captured electrons do not stay in the nucleus, so they do not alter the nucleon count.)
Total extra nucleons $= 10 + 9 = 19$.

Final mass number:
$A_f = A_i + 19 = 8 + 19 = 27$.

Using relation $-(1)$, the ratio of final to initial surface areas is

$\dfrac{S_f}{S_i}= \left(\dfrac{A_f}{A_i}\right)^{2/3} =\left(\dfrac{27}{8}\right)^{2/3}$

Evaluate the exponent step by step:
$\dfrac{27}{8}=3.375 \quad\Longrightarrow\quad \left(3.375\right)^{1/3}=1.5$ (because $(1.5)^3 = 3.375$).
Therefore,
$\left(\dfrac{27}{8}\right)^{2/3}=(1.5)^2=2.25.$(Final surface area is $2.25$ times the initial.)

Percentage growth is defined as

$\text{Percentage growth}=\left(\dfrac{S_f}{S_i}\right)\times100 =2.25 \times 100 = 225\%.$(This includes the initial 100%; the surface area becomes 225% of its original value.)

Hence, the percentage growth recorded is 225%.

Answer (Option A)

We need to find the RMS current for $i = i_0(t/T)$ from t = 0 to t = T.

Formula for RMS current:

$i_{rms} = \sqrt{\frac{1}{T}\int_0^T i^2 dt}$

$= \sqrt{\frac{1}{T}\int_0^T i_0^2 \frac{t^2}{T^2} dt} = \sqrt{\frac{i_0^2}{T^3}\int_0^T t^2 dt}$

$= \sqrt{\frac{i_0^2}{T^3} \cdot \frac{T^3}{3}} = \sqrt{\frac{i_0^2}{3}} = \frac{i_0}{\sqrt{3}}$

Therefore, the RMS current is Option 2: $\frac{i_0}{\sqrt{3}}$.

We need to find the internal resistance of a cell using potentiometer readings with two different shunt resistances.

Formula: When cell is shunted with resistance S, the balance length is:

$l = \frac{E \cdot S}{S + r} \times k$ (where k is a constant depending on the potentiometer)

For shunt S₁ = 4Ω: $l_1 = 120$ cm → $120 = \frac{E \times 4}{4+r} \times k$

For shunt S₂ = 12Ω: $l_2 = 180$ cm → $180 = \frac{E \times 12}{12+r} \times k$

Taking the ratio:

$\frac{120}{180} = \frac{4/(4+r)}{12/(12+r)} = \frac{4(12+r)}{12(4+r)}$

$\frac{2}{3} = \frac{4(12+r)}{12(4+r)} = \frac{12+r}{3(4+r)}$

$2 \times 3(4+r) = 3(12+r)$

$6(4+r) = 3(12+r)$

$24 + 6r = 36 + 3r$

$3r = 12$

$r = 4 \, \Omega$

Therefore, the internal resistance is Option 3: 4 Ω.

The heating process for water from $-20^\circ\text{C}$ to $120^\circ\text{C}$ involves three distinct states of matter and two phase transitions.

1. Heating Ice ($-20^\circ\text{C}$ to $0^\circ\text{C}$)

Heat is initially supplied to raise the temperature of solid ice. The relationship is linear: $Q = m \cdot s_{ice} \cdot \Delta T$

Since the specific heat of ice $s_{ice}$ is approximately $0.5 \text{ cal/g}^\circ\text{C}$, the temperature rises with a relatively steep slope ($\frac{dT}{dQ} = \frac{1}{ms}$).

2. Melting Phase Change (at $0^\circ\text{C}$)

When the ice reaches its melting point, the temperature remains constant while the state changes from solid to liquid. This is represented by a horizontal plateau: $Q = m \cdot L_f$

Here, $L_f = 80 \text{ cal/g}$ is the latent heat of fusion.

3. Heating Liquid Water ($0^\circ\text{C}$ to $100^\circ\text{C}$)

After all ice has melted, the temperature of the liquid water rises: $Q = m \cdot s_{water} \cdot \Delta T$

The specific heat of water $s_{water}$ is $1.0 \text{ cal/g}^\circ\text{C}$. This section has a lower slope than the ice section because more heat is required to raise the temperature by one degree.

4. Boiling Phase Change (at $100^\circ\text{C}$)

At the boiling point, the temperature again remains constant during vaporization, creating a second horizontal plateau: $Q = m \cdot L_v$

Because the latent heat of vaporization ($L_v = 540 \text{ cal/g}$) is significantly higher than the latent heat of fusion, this plateau is much longer than the one at $0^\circ\text{C}$.

Graph (C) is the only representation that correctly shows:

The starting point below the origin ($-20^\circ\text{C}$).

Both phase change plateaus at $0^\circ\text{C}$ and $100^\circ\text{C}$.

A longer second plateau to account for the higher energy required for vaporization.

Hence, option (C) is correct.

Two identical cells (each with EMF $E$ and internal resistance $r$) give the same current through an external resistance of $6\Omega$ whether connected in series or parallel. We need to find $r$.

Current in series connection.

In series, the total EMF is $2E$ and total internal resistance is $2r$:

$I_{series} = \frac{2E}{2r + 6}$

Current in parallel connection.

In parallel, the equivalent EMF is $E$ and equivalent internal resistance is $\frac{r}{2}$ (two identical resistances $r$ in parallel):

$I_{parallel} = \frac{E}{\frac{r}{2} + 6} = \frac{2E}{r + 12}$

Set the currents equal and solve.

$\frac{2E}{2r + 6} = \frac{2E}{r + 12}$

Cancel $2E$ from both sides:

$\frac{1}{2r + 6} = \frac{1}{r + 12}$

$2r + 6 = r + 12$

$r = 6 \, \Omega$

The correct answer is Option (3): 6 $\Omega$.

Now the forces acting on the particle are $mg\ and\ F=-kv$

so $mg-kv=F_{net}$

$ma=mg-kv$

$m\times\ \frac{dv}{dt}=mg-kv$

$\frac{dv}{mg-kv}=\frac{dt}{m}$

Integrating on both sides

$\int\ \frac{dv}{mg-kv}=\int\ \frac{dt}{m}$

$-\frac{1}{k}\ln\ \left(\frac{\left(mg-kv\right)}{mg}\right)=\frac{t}{m}$

$v=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}}\right)$

Therefore option D is right

We have 10 kg of ice at -10 °C and 100 kg of water at 25 °C.

The specific heats are $c_{ice} = 2100$ J/kg°C and $c_{water} = 4200$ J/kg°C, with latent heat of fusion $L_f = 3.36 \times 10^5$ J/kg.

First, warming the ice to 0 °C requires $Q_1 = 10 \times 2100 \times 10 = 210000$ J, and melting it requires $Q_2 = 10 \times 336000 = 3360000$ J, giving a total heat requirement of $3570000$ J.

The heat available from cooling 100 kg of water from 25 °C to 0 °C is $10500000$ J, which exceeds $Q_1 + Q_2$, so all the ice melts.

Let the final temperature of the mixture be $T$. Then heat lost by water cooling from 25 °C to $T$ equals heat gained by the ice warming and melting, giving

$100 \times 4200 \times (25-T) = 210000 + 3360000 + 10 \times 4200 \times T.$

Rearranging gives

$10500000 - 420000T = 3570000 + 42000T,$

so

$6930000 = 462000T$

and hence

$T = 15°C.$

Thus the temperature decrease is $25 - 15 = 10$°C. The correct answer is Option 2: 10°C.

Net work done in a cyclic process equals area enclosed by the loop.

Upper path C→A is horizontal (constant pressure), and lower path is given by

$(V-2)^2=4ap$

Write pressure as

$p=\frac{(V-2)^2}{4a}$

At points A and C,

$V=1,V=3$

Substitute into curve:

For V=1,

$p=\frac{(1-2)^2}{4a}=\frac{1}{4a}$

For V=3,

$p=\frac{(3-2)^2}{4a}=\frac{1}{4a}$

So upper straight line has pressure

$p=\frac{1}{4a}$

Now work done over cycle equals area between straight line and curve:

$W=\int_1^3\left(\frac{1}{4a}-\frac{(V-2)^2}{4a}\right)dV$

Take $\frac{1}{4a}$​ common:

$W=\frac{1}{4a}\int_1^3\left(1-(V-2)^2\right)dV$

Put

x=V−2

Then limits become

V=1→x=−1

V=3→x=1

So

$W=\frac{1}{4a}\int_{-1}^1(1-x^2)dx$

Integrating,

$=\frac{1}{4a}\left[x-\frac{x^3}{3}\right]_{-1}^1$

$=\frac{1}{4a}\left(\frac{2}{3}-\left(-\frac{2}{3}\right)\right)$

$=\frac{1}{4a}\cdot\frac{4}{3}$

$=\frac{1}{3a}$

The arrows show process is

A→B→C→A

• From A→B→C, gas expands (left to right) along the lower curve.
• From C→A,  gas is compressed (right to left) along the upper straight line.

During compression, pressure is higher than during expansion, so work done on the gas is greater than work done by the gas.

Therefore net work done by the gas is negative.

So,

W=−(enclosed area)

$W=-\frac{1}{3a}$

We need to find the height of the fourth drop when the first drop strikes the ground.

The drops are released from a height of 5 m at regular intervals, and when the first drop hits the ground the sixth drop begins to fall.

Since the first drop falls through 5 m, its fall time satisfies $5 = \frac{1}{2}(10)t^2 \implies t = 1$ s.

There are five intervals between the first and sixth drop, so the time interval is $\tau = t/5 = 0.2$ s.

At t = 1 s, the fourth drop has been falling for $1 - 3\tau = 1 - 0.6 = 0.4$ s.

The distance fallen by the fourth drop is $d = \frac{1}{2}(10)(0.4)^2 = 5 \times 0.16 = 0.8$ m.

Hence its height from the ground is $5 - 0.8 = 4.2$ m.

Therefore, the answer is Option 3: 4.2 m.

We need to find the refractive index of the medium.

$E = 25\sin(2.0 \times 10^{15}t - 10^7 x)$

$\omega = 2.0 \times 10^{15}$ rad/s, $k = 10^7$ rad/m

Speed of wave in medium:

$v = \frac{\omega}{k} = \frac{2 \times 10^{15}}{10^7} = 2 \times 10^8$ m/s

Refractive index:

$n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$

Therefore, the refractive index is Option 1: 1.5.

For biconvex lens:

$\mu_1 = 1.5$

$P_1 = (\mu_1 - 1)\left(\frac{1}{R_1} - \frac{1}{-R_2}\right)$

$P_1 = (1.5 - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$

$P_1 = 0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$

For plano-concave lens:

$\mu_2 = 1.7$

$R_3 = -R_2$

$R_4 = \infty$

$P_2 = (\mu_2 - 1)\left(\frac{1}{-R_2} - \frac{1}{\infty}\right)$

$P_2 = (1.7 - 1)\left(-\frac{1}{R_2}\right)$

$P_2 = -\frac{0.7}{R_2}$

Given:

$|P_1| = |P_2|$

$0.5\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{0.7}{R_2}$

$0.5\frac{1}{R_1} + 0.5\frac{1}{R_2} = 0.7\frac{1}{R_2}$

$0.5\frac{1}{R_1} = 0.2\frac{1}{R_2}$

$\frac{R_1}{R_2} = \frac{0.5}{0.2}$

$\frac{R_1}{R_2} = \frac{5}{2}$

Diodes D1 and D3 are forward-biased (ON).
Diode D2 is reverse-biased (OFF) and acts as an open circuit. Because $D_1$ and $D_3$ are in parallel, the voltage drop across this entire parallel combination is $0.7\text{ V}$.

The voltage across the resistor $R_1$

$V_{R1} = 12\text{ V} - 0.7\text{ V} = 11.3\text{ V}$

$I_{\text{total}} = \frac{V_{R1}}{R_1} = \frac{11.3\text{ V}}{0.3\text{ k}\Omega}$

$I_{\text{total}} = \frac{11.3}{0.3}\text{ mA} = \frac{113}{3}\text{ mA}$

$D_1$ and $D_3$ are indentical and in parallel, the total current splits equally between them.

$I_{D1} = \frac{I_{\text{total}}}{2}$

$I_{D1} = \frac{113/3\text{ mA}}{2} = \frac{113}{6}\text{ mA}$

$I_{D1} \approx 18.83\text{ mA}$

The reading shown by a vernier calliper has two parts:
  • Main-scale reading (MSR) = value at the mark just left of the vernier zero.
  • Vernier-scale reading (VSR) = (coinciding vernier division) $\times$ least count (LC).
Total reading $=$ MSR $+$ VSR.

Case 1: Determining the zero error (jaws closed)

When the jaws are in contact, the vernier zero is to the right of the main-scale zero; this is a positive zero error.
Here the $4^{\text{th}}$ vernier division coincides with a main-scale mark.

MSR = $0 \text{ mm}$ because the vernier zero has not crossed the first main-scale division.
VSR $= 4 \times LC = 4 \times 0.1 \text{ mm} = 0.4 \text{ mm}$.

Observed zero reading $= 0 + 0.4 = 0.4 \text{ mm}$, so
zero error $= +0.4 \text{ mm}$ and the zero correction (to be applied later) is $-0.4 \text{ mm}$.

Case 2: Measuring the cylinder

The observer notes:
  • MSR = $15 \text{ mm}$ (15 main-scale divisions lie to the left of the vernier zero).
  • The $5^{\text{th}}$ vernier division coincides with a main-scale mark.

VSR $= 5 \times 0.1 \text{ mm} = 0.5 \text{ mm}$.

Raw reading (uncorrected) $= 15 + 0.5 = 15.5 \text{ mm}$.

Applying the zero correction
True length $= 15.5 \text{ mm} - 0.4 \text{ mm} = 15.1 \text{ mm}$.

Therefore, the measured length of the cylinder is $\mathbf{15.1 \text{ mm}}$ (Option B).

We need to find the work done in bringing a 3 nC charge from infinity to the third corner of an equilateral triangle.

q₁ = 1 nC, q₂ = 2 nC at two corners, side = 3 cm = 0.03 m, q₃ = 3 nC.

Work done = potential energy of q₃ due to q₁ and q₂:

$W = q_3(V_1 + V_2) = q_3\left(\frac{kq_1}{r} + \frac{kq_2}{r}\right) = \frac{kq_3(q_1 + q_2)}{r}$

$W = \frac{9 \times 10^9 \times 3 \times 10^{-9} \times (1 + 2) \times 10^{-9}}{0.03}$

$= \frac{9 \times 10^9 \times 3 \times 10^{-9} \times 3 \times 10^{-9}}{0.03}$

$= \frac{81 \times 10^{-9}}{0.03} = 2700 \times 10^{-9} = 2.7 \times 10^{-6} \text{ J} = 2.7 \, \mu J$

Therefore, the work done is Option 2: 2.7 μJ.

We need to evaluate two statements about wave optics.

Statement I: "A plane wave after passing through a prism remains as a plane wave but passing through a small pinhole may become a spherical wave."

This is TRUE. A prism refracts a plane wave (changes its direction and introduces dispersion), but the wavefront remains essentially planar. However, when a plane wave passes through a small pinhole (whose size is comparable to the wavelength), diffraction occurs. The pinhole acts as a point source of secondary wavelets (by Huygens' principle), producing a diverging spherical wavefront.

Statement II: "The curvature of a spherical wave emerging from a slit will increase for increasing slit width."

This is FALSE. As the slit width increases, diffraction effects decrease. A wider slit allows the wave to pass through with less spreading, producing a wavefront that is closer to planar (less curved). Therefore, curvature decreases with increasing slit width, not increases.

The correct answer is Option (3): Statement I is true but Statement II is false.

We need to find the ratio of de Broglie wavelengths of deuteron (KE = E) and alpha particle (KE = 2E).

Formula: $\lambda = \frac{h}{\sqrt{2mK}}$

Deuteron: mass = 2m_p (1 proton + 1 neutron), KE = E

Alpha: mass = 4m_p (2 protons + 2 neutrons), KE = 2E

$\frac{\lambda_d}{\lambda_\alpha} = \frac{\sqrt{2m_\alpha \times 2E}}{\sqrt{2m_d \times E}} = \sqrt{\frac{4m_p \times 2E}{2m_p \times E}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$

So n:1 = 2:1, meaning n = 2.

Potential energy in SHM is maximum at extreme positions:

$x=\pm A$

Given

$x(t)=A\sin\omega t$

For maximum potential energy,

$\sin\omega t=\pm1$

First time this happens is

$\omega t=\frac{\pi}{2}$

So

$t=\frac{\pi}{2\omega}$

Now

$\omega=\frac{2\pi}{T}$

Substitute:

$t=\frac{\pi}{2(2\pi/T)}$

$=\frac{T}{4}$

Given

$t=\frac{T}{2\beta}$

So

$\frac{T}{2\beta}=\frac{T}{4}$

Cancelling T,

$2\beta=4$

$β=2$

We need to find n where the radius of gyration about the axis at 15 cm from center is $\sqrt{n}$ cm.

Solid sphere, R = 10 cm, axis at distance d = 15 cm from center.

Using parallel axis theorem:

$I = I_{cm} + Md^2 = \frac{2}{5}MR^2 + Md^2$

Radius of gyration: $k^2 = \frac{I}{M} = \frac{2}{5}R^2 + d^2 = \frac{2}{5}(100) + 225 = 40 + 225 = 265$

$k = \sqrt{265}$ cm, so n = 265.

Therefore, n = 265.

We need to find α where the difference in focal lengths in water and air is αf.

In air: $f_{air} = f = 18$ cm

In water: Using the lens maker's equation:

$\frac{1}{f_{water}} = \frac{n_L/n_w - 1}{n_L - 1} \times \frac{1}{f_{air}}$

$\frac{f_{water}}{f_{air}} = \frac{n_L - 1}{n_L/n_w - 1} = \frac{1.5 - 1}{1.5/(4/3) - 1} = \frac{0.5}{1.125 - 1} = \frac{0.5}{0.125} = 4$

$f_{water} = 4f = 72$ cm

Difference: $f_{water} - f_{air} = 72 - 18 = 54 = 3f$

So $\alpha = 3$.

Therefore, α = 3.

step 1: assume currents

Let total current from source = i

At node A, current splits:
top branch =$i_1$
bottom branch = $i-i_1$

step 2: apply loop equation (top loop)

$6i_1+3(2i_1-i)=3(i-i_1)$

simplify:

$6i_1+6i_1-3i=3i-3i_1$

$12i_1-3i=3i-3i_1$

$15i_1=6i\Rightarrow i_1=\frac{2}{5}i$

step 3: apply KVL for left path

$3(i-i_1)+6i_1=1$

substitute $i_1$:

$3\left(i-\frac{2}{5}i\right)+6\left(\frac{2}{5}i\right)=1$

$3\left(\frac{3}{5}i\right)+\frac{12}{5}i=1$

$\frac{9}{5}i+\frac{12}{5}i=1$

$\frac{21}{5}i=1\Rightarrow i=\frac{5}{21}$

step 4: equivalent resistance

$R_{eq}=Vi=\frac{1}{\frac{5}{21}}=\frac{21}{5}Ω$

given form:

$R_{eq}=\frac{x}{5}\Rightarrow x=21$

We need to find the values of $x$ and $y$ to prepare a 100 mL basic buffer of pH 9 from a weak base $B$ ($pK_b = 5.699$) and strong acid $\text{HCl}$.

1. Buffer Equations:

• Calculate $\text{pOH}$:
  $\text{pOH} = 14 - \text{pH} = 14 - 9 = 5$
• Henderson-Hasselbalch Equation for Basic Buffer:
  $\text{pOH} = pK_b + \log\left(\frac{[\text{conjugate acid}]}{[\text{weak base}]}\right)$
  $5 = 5.699 + \log\left(\frac{[\text{BH}^+]}{[B]}\right)$
  $\log\left(\frac{[\text{BH}^+]}{[B]}\right) = -0.699$
  $\frac{[\text{BH}^+]}{[B]} = 10^{-0.699} = 10^{\log 2 - 1} = 0.2 = \frac{1}{5}$

2. Reaction and Mole Balance:

When $x\text{ mL}$ of $0.02\text{ M HCl}$ reacts with $y\text{ mL}$ of $0.02\text{ M B}$:

• $\text{Initial millimoles of HCl} = 0.02x$
• $\text{Initial millimoles of B} = 0.02y$
• Since $\text{HCl}$ is the limiting reagent, it reacts completely to form conjugate acid $\text{BH}^+$:
  • $\text{Millimoles of conjugate acid } (\text{BH}^+) = 0.02x$
  • $\text{Millimoles of remaining weak base } (B) = 0.02y - 0.02x$
• Substitute into the ratio:
  $\frac{0.02x}{0.02y - 0.02x} = \frac{1}{5} \implies \frac{x}{y - x} = \frac{1}{5}$
  $5x = y - x \implies y = 6x$
• Using total volume:
  $x + y = 100\text{ mL}$
  $x + 6x = 100 \implies 7x = 100 \implies x = 14.3\text{ mL}$
  $y = 100 - 14.3 = 85.7\text{ mL}$

Conclusion:

The required volumes of $\text{HCl}$ and weak base $B$ are $14.3\text{ mL}$ and $85.7\text{ mL}$, respectively.

Answer: Option B — $x = 14.3, y = 85.7$

We analyze the magnetic properties of three nickel complexes.

1. $Ni(CO)_4$:

Nickel is in the 0 oxidation state with configuration $[Ar]\,3d^{10}$. CO is a strong field ligand. With all 10 d-electrons paired, this complex is diamagnetic. The geometry is tetrahedral (sp$^3$ hybridization).

2. $[NiCl_4]^{2-}$:

Nickel is in the +2 oxidation state with configuration $[Ar]\,3d^8$ (2 unpaired electrons). Cl$^-$ is a weak field ligand, so no electron pairing occurs. The complex adopts a tetrahedral geometry (sp$^3$) with 2 unpaired electrons. This complex is paramagnetic.

3. $[Ni(CN)_4]^{2-}$:

Nickel is in the +2 oxidation state with configuration $[Ar]\,3d^8$. CN$^-$ is a strong field ligand, causing electron pairing. The complex adopts a square planar geometry (dsp$^2$ hybridization) with all electrons paired. This complex is diamagnetic.

Summary:

$Ni(CO)_4$: diamagnetic, $[Ni(CN)_4]^{2-}$: diamagnetic, $[NiCl_4]^{2-}$: paramagnetic.

Option B states: "$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic and $[NiCl_4]^{2-}$ is paramagnetic." (The option text contains a typographical error showing $[NiCl_4]^{2-}$ where it should say $[Ni(CN)_4]^{2-}$, but the intended meaning matches our analysis.)

The correct answer is Option B.

We need to evaluate two statements about the Griss-Ilosvay test for nitrite ion.

Statement I: The Griss-Ilosvay test uses sulphanilic acid and α-naphthylamine for nitrite detection.

This is TRUE. This is the standard reagent combination for the test.

Statement II: Sulphanilic acid is diazotized by acidified nitrite ion, which then couples with α-naphthylamine to form an azo dye.

This is TRUE. The nitrite first diazotizes sulphanilic acid to form a diazonium salt, which then couples with α-naphthylamine to produce a red/orange azo dye.

Both statements are true, which corresponds to Option 4.

We need to identify the elements with the highest and lowest atomic radii in Period 4 of the periodic table.

Key Concept: Trend of Atomic Radius Across a Period

As we move from left to right across a period, the atomic radius generally decreases. This is because:

- The nuclear charge (number of protons) increases with each successive element.

- Electrons are added to the same principal energy level (same shell), so the shielding effect does not increase significantly.

- The increased effective nuclear charge pulls the electron cloud closer to the nucleus, reducing the atomic radius.

Period 4 elements: K, Ca, Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Ga, Ge, As, Se, Br, Kr

Highest atomic radius:

Potassium (K) is the first element in Period 4 (Group 1, alkali metal). It has the lowest effective nuclear charge among Period 4 elements and its outermost electron is in the 4s orbital with minimal nuclear attraction. K has the largest atomic radius in Period 4 (approximately 227 pm).

Lowest atomic radius:

Noble gases (Kr) have special considerations - their atomic radii are measured differently (van der Waals radii vs covalent radii). When considering covalent atomic radii, Bromine (Br) is the element with the smallest atomic radius in Period 4, as it is the rightmost element before the noble gas Kr and has the highest effective nuclear charge among the elements that form covalent bonds. Br has a covalent radius of approximately 114 pm.

Note: Options 3 (Na and Cl) and 4 (Rb and Br) include elements not in Period 4. Na and Cl are in Period 3, and Rb is in Period 5. These can be eliminated immediately.

The correct answer is Option 1: K and Br.

All four alternatives try to place a $-CN$ group on the aromatic ring that originally bears an $-NH_2$ group (aniline). The only reliable laboratory route is to pass through a diazonium salt and then carry out a Sandmeyer cyanation. Option A is the only alternative that follows this exact sequence, so it alone furnishes the nitrile as the major product.

Step 1 (Diazotisation)
Aniline is treated at $0-5^{\circ}{\rm C}$ with $NaNO_2/HCl$. This converts the basic amino group into the corresponding diazonium chloride: $C_6H_5NH_2 \xrightarrow{NaNO_2/HCl,\;0-5^{\circ}{\rm C}} C_6H_5N_2^+Cl^-$.

Step 2 (Sandmeyer cyanation)
The freshly prepared diazonium salt is then reacted with $CuCN$ (generally in aqueous $HCl$). Copper(I) cyanide replaces the diazonium group by $-CN$, expelling nitrogen gas: $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN + N_2 \uparrow$.

Thus Option A produces benzonitrile ($C_6H_5CN$) in good yield. The overall transformation is

$C_6H_5NH_2 \;\xrightarrow[\;CuCN\;]{\;NaNO_2/HCl,\;0-5^{\circ}{\rm C}\;} C_6H_5CN + N_2 \uparrow$.

Why the other options fail

Case B: Direct treatment of aniline with $NaCN$ (or $KCN$) cannot displace the strongly basic $-NH_2$ group from an aromatic ring, so no nitrile is formed.

Case C: If the diazonium salt is first obtained but then warmed before adding $CuCN$, it decomposes (mostly to phenol) and gives very poor or negligible nitrile yield.

Case D: Adding copper(II) salts and cyanide ions directly to aniline, without the diazonium intermediate, again fails because $-NH_2$ is not a leaving group; polymeric tars predominate.

Therefore the only reaction that cleanly affords the required nitrile is that listed in Option A.

Final answer: Option A which is: aniline → diazotisation → CuCN (Sandmeyer) → benzonitrile.