JEE Main 2025 Physics Question Paper with Answer and Solution

(D) The radius of a nucleus is given by the relation $R = R_0 A^{1/3}$,where $R_0$ is a constant (approximately $1.2 \times 10^{-15} \ m$).
The mass of a nucleus is approximately $M = A \times m_p$,where $m_p$ is the mass of a nucleon (proton or neutron).
The volume of the nucleus is $V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
The mass density $\rho$ is given by $\rho = \frac{M}{V} = \frac{A \cdot m_p}{\frac{4}{3} \pi R_0^3 A}$.
Canceling $A$ from the numerator and denominator,we get $\rho = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.
Since $m_p$,$\pi$,and $R_0$ are constants,the density $\rho$ is independent of the mass number $A$.

(A) The electric field $E$ due to an infinite non-conducting sheet is $E = \frac{\sigma}{2\varepsilon_0}$.
Since the electron is released from rest,the force on it is $F = eE = \frac{e\sigma}{2\varepsilon_0}$.
The acceleration $a$ of the electron is $a = \frac{F}{m} = \frac{e\sigma}{2m\varepsilon_0}$,which is constant.
The velocity of the electron at time $t$ is $v = at = \left(\frac{e\sigma}{2m\varepsilon_0}\right)t$.
The momentum of the electron is $p = mv = m(at) = \left(\frac{e\sigma}{2\varepsilon_0}\right)t$.
The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{(e\sigma / 2\varepsilon_0)t}$.
The rate of change of wavelength is $\frac{d\lambda}{dt} = \frac{d}{dt} \left[ \frac{2h\varepsilon_0}{e\sigma t} \right] = -\frac{2h\varepsilon_0}{e\sigma} \cdot \frac{1}{t^2}$.
Thus,the magnitude of the rate of change of wavelength varies inversely as $t^2$,so $n = 2$.

(C) The capacitor is divided into two capacitors in series,each with plate separation $d' = d/2 = 0.885 \ mm = 0.885 \times 10^{-3} \ m$ and area $A = 4 \ cm^2 = 4 \times 10^{-4} \ m^2$.
The capacitance of the first part is $C_1 = \frac{k_1 \varepsilon_0 A}{d'} = \frac{5 \times 8.85 \times 10^{-12} \times 4 \times 10^{-4}}{0.885 \times 10^{-3}} = 20 \ pF$.
The capacitance of the second part is $C_2 = \frac{k_2 \varepsilon_0 A}{d'} = \frac{3 \times 8.85 \times 10^{-12} \times 4 \times 10^{-4}}{0.885 \times 10^{-3}} = 12 \ pF$.
Since these are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \implies C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{20 \times 12}{20 + 12} = \frac{240}{32} = 7.5 \ pF$.
This capacitor is connected in parallel with another capacitor of $7.5 \ pF$. The total effective capacitance is $C_{total} = C_{eq} + 7.5 \ pF = 7.5 + 7.5 = 15 \ pF$.