TN State Board 12th Physics Question Paper 2026

When the identical conducting balls touch, charge redistributes equally because they are identical. The total charge becomes $q_1+q_2$, so each ball gets $\frac{q_1+q_2}{2}$. The new force is $F'=\frac{1}{4\pi\epsilon_0}\frac{\left(\frac{q_1+q_2}{2}\right)^2}{r^2}$. Since the initial force was $F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$, the force after touching is less than before for positive charges. Hence, the correct option is (b) less than before.

For a wire, $R=\rho \frac{L}{A}=\rho \frac{L}{\pi r^2}$. Since both wires have the same material and length, $R\propto \frac{1}{r^2}$. Given $R_A=3R_B$, we get $\frac{1}{r_A^2}=3\frac{1}{r_B^2}$, so $r_A^2=\frac{r_B^2}{3}$. Therefore, $\frac{r_A}{r_B}=\frac{1}{\sqrt{3}}$. Hence, option (c) is correct.

For a charged ring rotating with angular speed $\omega$, the current is $I=\frac{q}{T}=\frac{q\omega}{2\pi}$. Its magnetic moment is $M=IA=\frac{q\omega}{2\pi}\cdot \pi r^2=\frac{q\omega r^2}{2}$. The angular momentum is $L=m r^2\omega$. Hence, $$\frac{M}{L}=\frac{\frac{q\omega r^2}{2}}{mr^2\omega}=\frac{q}{2m}.$$ So the correct option is (b).

The induced emf is given by $e=L\frac{\Delta I}{\Delta t}$. Here, the current changes from $12$ A to $-2$ A, so $\Delta I=-2-12=-14$ A. Taking magnitude, $|\Delta I|=14$ A and $e=8$ V, $\Delta t=0.05$ s. $$L=\frac{e\,\Delta t}{\Delta I}=\frac{8\times 0.05}{14}\approx 0.0286\ \text{H}$$ This value does not match the listed options, so the intended exam calculation is likely based on a different interpretation or a misprint. Among the given choices, the closest standard answer expected in such problems is not consistent with the data. However, using the usual formula directly gives $L\approx 0.029$ H.

An electromagnetic wave is produced by an accelerating charge. A stationary charge produces only an electric field, and a charge moving with constant velocity does not radiate electromagnetic waves. An uncharged particle cannot produce such radiation directly. Hence, option (b) is correct.