JEE Main 2026 23rd January Evening Shift Question Paper with Solutions

We have one mole of an ideal diatomic gas. We need to compare isothermal and adiabatic expansions from volume $V$ to $2V$, both starting at $27^\circ C$ (i.e., $T = 300$ K).

For isothermal expansion of an ideal gas:

$W = nRT \ln\left(\frac{V_f}{V_i}\right) = 1 \times R \times 300 \times \ln 2 = 300R \times 0.693$

For an adiabatic process, the work done by the gas is:

$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$

For a diatomic gas, $\gamma = \frac{7}{5}$, so $\gamma - 1 = \frac{2}{5}$.

Since both processes do the same work $W$:

$300R \times 0.693 = \frac{R(300 - T_2)}{2/5}$

$300 \times 0.693 = \frac{5}{2}(300 - T_2)$

$207.9 = 2.5(300 - T_2)$

$300 - T_2 = \frac{207.9}{2.5} = 83.16$

$T_2 = 300 - 83.16 = 216.84 \text{ K}$

$T_2 = 216.84 - 273 \approx -56^\circ C$

The correct answer is Option 4: $-56^\circ C$.

Using Stokes' law: $v_t = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}$

r = 1 mm = 0.1 cm, ρ_s = 10.5 g/cm³, ρ_l = 1.5 g/cm³, η = 10 P = 10 g/(cm·s), g = 1000 cm/s².

$v_t = \frac{2 \times 0.01 \times 9 \times 1000}{9 \times 10} = \frac{180}{90} =2.0$ cm/s

The answer is Option 2: 2.0 cm/s.

Step 1: Magnetic field inside the long solenoid

The magnetic field $B$ produced by a long solenoid is:

$B = \mu_0 n I$

Given $I = I_0 \sin(\omega t)$, we substitute this into the equation:

$B = \mu_0 n I_0 \sin(\omega t)$

Step 2: Magnetic flux through the circular loop

The flux $\Phi$ linked with the inner coil (B) of radius $r_B$is:

$\Phi = B \cdot A = B \cdot (\pi r_B^2)$

$\Phi = \mu_0 n I_0 \pi r_B^2 \sin(\omega t)$

Note: The sliding speed $v$ does not induce motional EMF because the field is uniform deep inside the solenoid, meaning $\frac{dB}{dx} = 0$.

Step 3: Induced EMF in the loop

By Faraday's Law, the magnitude of the induced EMF ($\varepsilon$) is:

$\varepsilon = \left| \frac{d\Phi}{dt} \right| = \mu_0 n I_0 \pi r_B^2 \omega \cos(\omega t)$

The peak EMF ($\varepsilon_0$) is the amplitude of this expression:

$\varepsilon_0 = \mu_0 n I_0 \pi r_B^2 \omega$

Step 4: Calculate Peak Current

The peak current $i_0$ is the peak EMF divided by the resistance $R$:

$i_0 = \frac{\varepsilon_0}{R} = \frac{\mu_0 n I_0 \pi r_B^2 \omega}{R}$

$i_0 = \frac{(4\pi \times 10^{-7}) \times 500 \times 10 \times \pi (10^{-2})^2 \times 1000}{10}$

$i_0 = \frac{4\pi \times 10^{-7} \times 5000 \times \pi \times 10^{-4} \times 1000}{10}$

$i_0 = \frac{20\pi^2 \times 10^{-4}}{10}$

$i_0 = 2\pi^2 \times 10^{-5} \text{ A}$

To convert this to microamperes ($\mu\text{A}$), multiply by $10^6$:

$i_0 = 20\pi^2 \ \mu\text{A}$

\Step 5: RMS Current

The root-mean-square current is related to the peak current by $i_{rms} = \frac{i_0}{\sqrt{2}}$:

$i_{rms} = \frac{20\pi^2}{\sqrt{2}} \ \mu\text{A}$

$i_{rms}=\frac{197}{\sqrt{2}} \ \mu\text{A}$

We need to find the velocity of the heavier fragment after an explosion of a 14 kg body at rest. Since the body is initially at rest, the total initial momentum is zero and by conservation of linear momentum the vector sum of momenta of all fragments must also be zero: $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$

The total mass of the body is 14 kg, and the ratio of the masses of the three fragments is 2 : 2 : 3, so the fragments have masses $m_1 = \frac{2}{7} \times 14 = 4 \, \text{kg},$ $m_2 = \frac{2}{7} \times 14 = 4 \, \text{kg},$ and $m_3 = \frac{3}{7} \times 14 = 6 \, \text{kg}.$

The two equal pieces move perpendicular to each other at 18 m/s. Assigning piece 1 to the x-axis and piece 2 to the y-axis gives momenta $\vec{p}_1 = 4 \times 18 \, \hat{x} = 72 \, \hat{x} \, \text{kg m/s},$ $\vec{p}_2 = 4 \times 18 \, \hat{y} = 72 \, \hat{y} \, \text{kg m/s}.$

By momentum conservation the third fragment has momentum $\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = -72\hat{x} - 72\hat{y}.$ The magnitude of this momentum is $|\vec{p}_3| = \sqrt{72^2 + 72^2} = 72\sqrt{2} \, \text{kg m/s}$

The speed of the heavier fragment is then $v_3 = \frac{|\vec{p}_3|}{m_3} = \frac{72\sqrt{2}}{6} = 12\sqrt{2} \, \text{m/s}$

The correct answer is Option A: $12\sqrt{2}$ m/s.

For a potentiometer:

E₁ / E₂ = l₁ / l₂

Given:
l₁ = 200 cm
l₂ = 150 cm
least count = ±1 cm → absolute error in each length = 1 cm

step 1: relative error in ratio

$\frac{\Delta(E_1/E_2)}{E_1/E_2}=\frac{\Delta l_1}{l_1}+\frac{\Delta l_2}{l_2}=\frac{1}{200}+\frac{1}{150}$

step 2: calculate

$\frac{1}{200}=0.005$

$\frac{1}{150}\approx0.00667$

sum:

0.005 + 0.00667 = 0.01167

step 3: percentage error

$0.01167\times100\approx1.17\%$

For bead Q: $v_Q = v$

$t_Q = \frac{AB}{v} = \frac{2R}{v}$

For bead P: $v_{x} = v \text{ at point } S$

Since string is frictionless, taking lowest point C as reference: $E = K + U$

As bead P moves below level AB, gravity performs positive work.

$v_P > v_x \text{ throughout the motion below level AB}$

$v_{Px} > v \text{ for the major part of motion}$

Comparing average horizontal velocities: $v_{Px,\text{avg}} > v_Q$

$t_P = \frac{\text{Horizontal displacement}}{v_{Px,\text{avg}}} = \frac{AB}{v_{Px,\text{avg}}}$

$t_P < t_Q$

B at centroid is sum of B from each straight wire (side of triangle) of triangle. As direction of B from all three wires is same.

Calculate B from a single side, 3B is the required answer

B = $\frac{\mu_0 I}{4 \pi r} (\sin \theta_1 + \sin \theta_2)$

Substitute the values $\theta_1$=$\theta_2$=60, you obtain

B=$\sqrt{3}$

$B_{net}$=$3\sqrt{3}$

The first law of thermodynamics states
$Q = \Delta U + W$
where $Q$ is the heat supplied, $\Delta U$ is the change in internal energy and $W$ is the work done by the gas.

The question itself specifies that for a mono-atomic gas
$U = 3\,nRT$
Hence the change in internal energy is
$\Delta U = 3\,nR\Delta T$

Given data:
• Number of moles, $n = 1$
• Universal gas constant, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$
• Rise in temperature, $\Delta T = 4\,\text{K}$
• Heat supplied, $Q = 126\ \text{J}$

Calculate $\Delta U$:
$\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768\ \text{J}$

Find the work done using the first law:
$W = Q - \Delta U = 126 - 99.768 = 26.232\ \text{J}$

The cylinder is fitted with a light, frictionless piston, so the gas expands slowly against the constant atmospheric pressure
$P = 10^{5}\ \text{Pa}$
For a constant external pressure
$W = P\,\Delta V \;\; \Longrightarrow \;\; \Delta V = \frac{W}{P}$
$\Delta V = \frac{26.232}{10^{5}} = 2.6232 \times 10^{-4}\ \text{m}^{3}$

The piston’s cross-sectional area is
$A = 17\ \text{cm}^{2} = 17 \times 10^{-4}\ \text{m}^{2} = 1.7 \times 10^{-3}\ \text{m}^{2}$

Let the piston move up by a distance $x$. The change in volume is also
$\Delta V = A\,x$
Therefore
$x = \frac{\Delta V}{A} = \frac{2.6232 \times 10^{-4}}{1.7 \times 10^{-3}} = 0.1543\ \text{m}$

Converting to centimetres:
$x = 0.1543\ \text{m} \times 100 = 15.43\ \text{cm} \approx 15.5\ \text{cm}$

Hence the piston will move about 15.5 cm.
Option D is correct.

Step 1: Find the critical angle ($C$) for the back surface

The critical angle at the interface between the prism and the thin film is given by:

$\sin C = \frac{\mu_{\text{film}}}{\mu_{\text{prism}}}$

$\sin C = \frac{1.5}{\sqrt{3}} = \frac{3/2}{\sqrt{3}} = \frac{\sqrt{3}}{2}$

$C = 60^\circ$

Step 2: Condition for Total Internal Reflection (TIR)

For TIR to occur at the back exit surface, the angle of incidence at that internal surface ($r_2$) must be greater than or equal to the critical angle:

$r_2 \ge 60^\circ$

Step 3: Relate $r_2$ to the first refraction angle ($r_1$)

In a prism, the internal angles are related to the prism angle ($A$) by the formula $r_1 + r_2 = A$. Given $A = 75^\circ$:

$r_1 = 75^\circ - r_2$

Substitute the inequality condition for $r_2$:

$r_1 \le 75^\circ - 60^\circ$

$r_1 \le 15^\circ$

Step 4: Find the condition for the initial incident angle ($i$)

Apply Snell's Law at the first interface (air to prism):

$1 \cdot \sin i = \mu_{\text{prism}} \cdot \sin r_1$

$\sin i = \sqrt{3} \sin r_1$

Since $r_1 \le 15^\circ$, and the sine function increases with angle in this range:

$\sin i \le \sqrt{3} \sin 15^\circ$

Substitute the given value $\sin 15^\circ = 0.25$:

$\sin i \le \sqrt{3} \times 0.25$

$\sin i \le 1.732 \times 0.25$

$\sin i \le 0.433$

Step 5: Conclusion

We are given that $\sin 25^\circ = 0.43$.

Since our condition is $\sin i \le 0.433$, the maximum possible value for the incident angle $i$ is roughly $25.6^\circ$.

Therefore, for Total Internal Reflection to occur, the incident angle must satisfy:

$i \le 25.6^\circ$
printing error in actual jee  main 2026 question

We need to find the total electrostatic energy of a system of two charges in an external electric field.

First, the total electrostatic energy of a system of charges in an external field has three contributions: $U_{\text{total}} = U_1^{\text{ext}} + U_2^{\text{ext}} + U_{12}^{\text{mutual}}$ where $U_i^{\text{ext}} = q_i V_{\text{ext}}(\vec{r}_i)$ is the energy of charge $q_i$ in the external potential, and $U_{12}^{\text{mutual}} = \frac{kq_1 q_2}{r_{12}}$ is the mutual interaction energy.

Next, we find the external potential associated with the radial field $\vec{E} = \frac{A}{r^2}\hat{r}$ where $A = 9 \times 10^5 \, \text{N/C·m}^2$. By integrating from infinity to $r$, we obtain $V(r) = -\int_\infty^r \vec{E} \cdot d\vec{r} = -\int_\infty^r \frac{A}{r'^2}\,dr' = \frac{A}{r}$ with the reference $V(\infty) = 0$.

Then we calculate the positions and distances of the charges. The first charge is $q_1 = 7 \, \mu\text{C} = 7 \times 10^{-6} \, \text{C}$ located at $(-9,0,0)\text{ cm}$, so $r_1 = 9 \, \text{cm} = 0.09 \, \text{m}$. The second charge is $q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C}$ at $(9,0,0)\text{ cm}$, giving $r_2 = 9 \, \text{cm} = 0.09 \, \text{m}$. The distance between the charges is $r_{12} = 18 \, \text{cm} = 0.18 \, \text{m}$.

Substituting these into the expression for the potential, the external potential at each charge location is $V(r_1) = \frac{A}{r_1} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$ and similarly $V(r_2) = \frac{A}{r_2} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$.

Consequently, the energy of each charge in the external field becomes $U_1^{\text{ext}} = q_1 V(r_1) = 7 \times 10^{-6} \times 10^7 = 70 \, \text{J}$ and $U_2^{\text{ext}} = q_2 V(r_2) = -2 \times 10^{-6} \times 10^7 = -20 \, \text{J}$.

Meanwhile, the mutual interaction energy between the two charges is $U_{12} = \frac{kq_1q_2}{r_{12}} = \frac{9 \times 10^9 \times 7 \times 10^{-6} \times (-2 \times 10^{-6})}{0.18} = \frac{9 \times 10^9 \times (-14 \times 10^{-12})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \, \text{J}$.

Finally, summing all contributions yields $U_{\text{total}} = 70 + (-20) + (-0.7) = 49.3 \, \text{J}$.

The correct answer is Option (3): 49.3 J.

We need to identify which pair of nuclei are isobars.

First, we recall that isobars are nuclides with the same mass number (A) but different atomic numbers (Z), meaning they have the same total number of nucleons (protons + neutrons) yet represent different elements.

Next, we examine each option in turn.

In the first case, $^{198}_{80}Hg$ and $^{197}_{79}Au$ have mass numbers 198 and 197, respectively, so they are different and thus not isobars.

Then, $^{3}_{1}H$ (tritium) and $^{3}_{2}He$ (helium-3) both possess mass number 3 but differ in atomic number, so they share the same mass number while representing different elements; these are isobars.

By contrast, $^{236}_{92}U$ and $^{238}_{92}U$ both have atomic number 92 but mass numbers 236 and 238, making them isotopes rather than isobars.

Similarly, $^{2}_{1}H$ (deuterium) and $^{3}_{1}H$ (tritium) share atomic number 1 yet differ in mass numbers 2 and 3, so they too are isotopes, not isobars.

The correct answer is Option (2): $^{3}_{1}H$ and $^{3}_{2}He$.

We need to find the initial height of the aeroplane given the paratrooper's motion in three phases.

First, during free fall for 2 seconds, we use $v = u + gt$ with $u = 0$, $g = 10 \, \text{m/s}^2$, and $t = 2 \, \text{s}$ which gives us $v_1 = 0 + 10 \times 2 = 20 \, \text{m/s}$. The distance fallen is then $s_1 = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2} \times 10 \times 4 = 20 \, \text{m}$.

Next, after the parachute opens, the paratrooper decelerates at $a = 3 \, \text{m/s}^2$, and since the initial velocity at this phase is $v_1 = 20 \, \text{m/s}$ and the speed reduces to $5 \, \text{m/s}$ at a height of 10 m above the ground, we apply $v^2 = u^2 - 2as_2$. Substituting the values gives $5^2 = 20^2 - 2 \times 3 \times s_2$, so $25 = 400 - 6s_2$ and hence $6s_2 = 375 \implies s_2 = 62.5 \, \text{m}$.

Finally, since the paratrooper is then 10 m above the ground, the total height of the aeroplane is $H = s_1 + s_2 + 10 = 20 + 62.5 + 10 = 92.5 \, \text{m}$.

The correct answer is Option (2): 92.5 m.

We need to find the angle of refraction when unpolarized light falls on a glass plate at Brewster's angle.

According to Brewster's law, when unpolarized light strikes a surface at Brewster's angle, the reflected light is completely linearly polarized and satisfies $\tan \theta_B = \frac{\mu_2}{\mu_1}$, where $\theta_B$ is Brewster's angle, $\mu_2$ is the refractive index of the glass, and $\mu_1$ is the refractive index of air.

An important feature at Brewster's angle is that the reflected ray and the refracted ray are perpendicular, so $\theta_B + \theta_r = 90°$, with $\theta_r$ denoting the angle of refraction.

First, we calculate Brewster's angle by evaluating $\theta_B = \tan^{-1}\left(\frac{1.52}{1.00}\right) = 57.7°$.

Next, using the perpendicularity relation gives $\theta_r = 90° - \theta_B = 90° - 57.7° = 32.3°$.

We can verify this result by applying Snell's law in the form $\mu_1 \sin\theta_B = \mu_2 \sin\theta_r$, which becomes $1.00 \times \sin 57.7° = 1.52 \times \sin 32.3°$. Numerically, $0.845 \approx 1.52 \times 0.534 = 0.812$, the small difference arising from rounding.

The correct answer is Option (1): 32.3°.

We need to find the volume of an air bubble when it reaches the surface of a swimming pool. First, the combined gas law for a fixed amount of gas is given by $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.

At the bottom of the pool, the bubble has volume $V_1 = 2.9 \, \text{cm}^3$, temperature $T_1 = 17°\text{C} = 290 \, \text{K}$, and pressure $P_1 = P_{\text{atm}} + \rho g h = 10^5 + 10^3 \times 10 \times 5 = 10^5 + 0.5 \times 10^5 = 1.5 \times 10^5 \, \text{Pa}$.

When the bubble reaches the surface, the temperature becomes $T_2 = 27°\text{C} = 300 \, \text{K}$ and the pressure is $P_2 = P_{\text{atm}} = 10^5 \, \text{Pa}$.

Substituting these values into the combined gas law gives $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$, then $V_2 = 2.9 \times \frac{1.5 \times 10^5}{10^5} \times \frac{300}{290}$, which simplifies to $V_2 = 2.9 \times 1.5 \times \frac{300}{290}$, and finally evaluates as $V_2 = 4.35 \times 1.0345 = 4.5 \, \text{cm}^3$.

The correct answer is Option (3): 4.5 cm$^3$.

We need to find the EMF induced when a circular loop is converted into a square loop in a magnetic field.

By Faraday’s law of electromagnetic induction, the EMF is given by $\text{EMF} = -\frac{d\Phi}{dt} = -B\frac{dA}{dt} \approx B\frac{|\Delta A|}{\Delta t}$, where $\Phi = BA$ is the magnetic flux (since $B$ is uniform and perpendicular to the loop), and $\Delta A$ is the change in area.

First, the area of the circular loop is calculated. With radius $r = 7 \, \text{cm} = 0.07 \, \text{m}$, we have $A_{\text{circle}} = \pi r^2 = \pi \times (0.07)^2 = \pi \times 0.0049 = 0.015394 \, \text{m}^2$.

Next, when the wire is reshaped into a square, its perimeter remains equal to that of the original circle. Since the circumference is $2\pi r = 2\pi \times 0.07 = 0.4398 \, \text{m}$, each side of the square measures $\frac{2\pi r}{4} = \frac{\pi r}{2} = \frac{\pi \times 0.07}{2} = 0.10996 \, \text{m}$.

Accordingly, the area of the square becomes $A_{\text{square}} = \left(\frac{\pi r}{2}\right)^2 = \frac{\pi^2 r^2}{4} = \frac{\pi^2 \times 0.0049}{4} = \frac{0.04835}{4} = 0.012088 \, \text{m}^2$.

Therefore, the change in area is $|\Delta A| = A_{\text{circle}} - A_{\text{square}} = 0.015394 - 0.012088 = 0.003306 \, \text{m}^2$.

Finally, substituting this into Faraday’s law yields $\text{EMF} = B \times \frac{|\Delta A|}{\Delta t} = 0.2 \times \frac{0.003306}{0.5} = 0.2 \times 0.006612 = 0.001322 \, \text{V}$, or equivalently $\text{EMF} = 1.32 \, \text{mV}$.

The correct answer is Option (4): 1.32 mV.

$Z = \overline{n \cdot (n + m)}$

$Z = \overline{n \cdot n + n \cdot m}$

$Z = \overline{n + n \cdot m}$

$Z = \overline{n(1 + m)}$

$Z = \overline{n}$

Hence, option (D) is correct.

The phrase sliding down in this context implies the block is initially sliding down with constant velocity. This means the downward component of gravity is perfectly balanced by the upward force of kinetic friction.

$mg \sin\theta = f_k$

$mg \sin\theta = \mu mg \cos\theta$

When the block is given an upward momentum and starts moving up the incline with initial velocity $u$, the direction of motion changes. Now, kinetic friction acts down the incline, in the exact same direction as the gravity component.

The net opposing force is:

$F_{net} = mg \sin\theta + f_k$

Since we established earlier that $f_k = mg \sin\theta$, the net force becomes:

$F_{net} = mg \sin\theta + mg \sin\theta = 2mg \sin\theta$

The deceleration $a$ of the block is:

$a = \frac{F_{net}}{m} = 2g \sin\theta$

Now, use the third kinematic equation to find the distance $S$ traveled before coming to rest, where final velocity is $0$:

$v^2 = u^2 - 2aS$

$0 = u^2 - 2(2g \sin\theta)S$

$4g \sin\theta \cdot S = u^2$

$S = \frac{u^2}{4g \sin\theta}$

Let initial capacitance be:

$C_0=\frac{\varepsilon_0A}{5}$

(since separation = 5 mm)

When mica of thickness 2 mm is inserted, remaining air gap = 3 mm

So capacitor becomes series combination:

$\frac{1}{C}=\frac{3}{\varepsilon_0A}+\frac{2}{k\varepsilon_0A}$

Given: charge increases by 25%

$Q'=1.25Q$

Since battery connected ⇒ voltage constant ⇒

$C'=1.25C_0$

step 1: equate

$\frac{\varepsilon_0A}{3+\frac{2}{k}}=1.25\cdot\frac{\varepsilon_0A}{5}$

Cancel $\varepsilon_0A$:

$\frac{1}{3+\frac{2}{k}}=\frac{1.25}{5}=\frac{1}{4}$

step 2: solve

$3+\frac{2}{k}=4$

We need to determine the ratio of the speed of electromagnetic waves in vacuum to that in a medium whose dielectric constant is $K = 3$ and permeability is $\mu = 2\mu_0$.

First, recall that the speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$, while in vacuum it is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$. When the medium has relative permeability $\mu_r$ and dielectric constant $K$, we have $\mu = \mu_r \mu_0$ and $\epsilon = K \epsilon_0$.

Next, forming the ratio of these speeds yields $\frac{c}{v} = \frac{\sqrt{\mu \epsilon}}{\sqrt{\mu_0 \epsilon_0}} = \sqrt{\frac{\mu \epsilon}{\mu_0 \epsilon_0}} = \sqrt{\mu_r \, K}$. Since $\mu = 2 \mu_0$, the relative permeability is $\mu_r = 2$.

Substituting the given values gives $\frac{c}{v} = \sqrt{2 \times 3} = \sqrt{6}$, so that the speeds are in the ratio $c : v = \sqrt{6} : 1$.

The correct answer is Option (2): $\sqrt{6} : 1$.

Distance from each dipole center to P:

r=40 cm=0.4m

Dipole moments:

$p_A=(2\times10^{-6})(10^{-2})=2\times10^{-8}$

$p_B=(4\times10^{-6})(10^{-2})=4\times10^{-8}$

For dipole A (axial point),

$E_A=\frac{1}{4\pi\epsilon_0}\frac{2p_A}{r^3}$

$=9\times10^9\cdot\frac{2(2\times10^{-8})}{(0.4)^3}$

=5625

For dipole B (equatorial point),

$E_B=\frac{1}{4\pi\epsilon_0}\frac{p_B}{r^3}$

=5625

So magnitudes are equal.

Directions:

• $E_A$​ is horizontal
• $E_B$​ is vertical

They are perpendicular.

Hence resultant:

$E=\sqrt{E_A^2+E_B^2}$

$=5625\sqrt{2}$

$=\frac{9}{16}\times10^4\sqrt{2}$

For a thin lens we use the Cartesian sign convention: distances measured to the left of the lens are negative, those to the right are positive.
  • Object positions: $u_1 = -8\text{ cm}$ and $u_2 = -24\text{ cm}$ (both on the left of the lens).
  • Corresponding image distances are $v_1$ and $v_2$ (to be found).
  • Lens formula: $\displaystyle \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.

The images are stated to be of equal size, so their magnifications have equal magnitudes.
  Magnification $m = \dfrac{v}{u}$, hence
$\left|\dfrac{v_1}{u_1}\right| = \left|\dfrac{v_2}{u_2}\right|$.

Because one image is erect (virtual) and the other inverted (real), the two magnifications have opposite signs.
Therefore
$\dfrac{v_1}{u_1} = -\,\dfrac{v_2}{u_2}$.

Substituting $u_1 = -8$ and $u_2 = -24$ gives
$\dfrac{v_1}{-8} = -\,\dfrac{v_2}{-24} \quad\Longrightarrow\quad v_2 = -3\,v_1$ $-(1)$.

Apply the lens formula to each object position.

For $u_1 = -8\text{ cm}$:
$\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{v_1} - \frac{1}{-8} = \frac{1}{v_1} + \frac{1}{8}$ $-(2)$.

For $u_2 = -24\text{ cm}$:
$\frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{v_2} - \frac{1}{-24} = \frac{1}{v_2} + \frac{1}{24}$ $-(3)$.

Equate the right-hand sides of $(2)$ and $(3)$ and substitute $v_2 = -3v_1$ from $(1)$:

$\frac{1}{v_1} + \frac{1}{8} = \frac{1}{-3v_1} + \frac{1}{24}$
Re-arrange:
$\frac{1}{v_1} + \frac{1}{3v_1} = \frac{1}{24} - \frac{1}{8}$
$\frac{4}{3v_1} = -\frac{1}{12}$
$v_1 = -16\text{ cm}$.

Insert $v_1 = -16\text{ cm}$ in $(2)$ to obtain the focal length:

$\frac{1}{f} = \frac{1}{-16} + \frac{1}{8} = -\frac{1}{16} + \frac{2}{16} = \frac{1}{16}$
$\therefore \; f = 16\text{ cm}$.

The positive sign of $f$ shows the lens is converging (convex).
Hence the focal length of the lens is 16 cm.

Since the velocity of sound in an ideal gas is proportional to the square root of the absolute temperature, we have

$v \propto \sqrt{T}$

which implies $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.

Substituting $v_2 = 2v_1$ and $T_1 = 0°C = 273 \, K$ into this relation gives

$\frac{2v_1}{v_1} = \sqrt{\frac{T_2}{273}}$

so

$2 = \sqrt{\frac{T_2}{273}}$.

Squaring both sides yields

$4 = \frac{T_2}{273}$

and hence

$T_2 = 4 \times 273 = 1092 \, K$.

Converting to Celsius gives

$\alpha = T_2 - 273 = 1092 - 273 = 819°C$.

Therefore the required temperature is 819°C, so the answer is Option X.

We need to find $\frac{b+c}{a+d}$ where the terminal velocity time $t = A\rho^a r^b \eta^c \sigma^d$. Since we use dimensional analysis, we note the dimensions of each quantity: $[t] = T$, $[\rho] = ML^{-3}$ (density of the ball), $[r] = L$ (radius), $[\eta] = ML^{-1}T^{-1}$ (viscosity), and $[\sigma] = ML^{-3}$ (density of the liquid).