CG PET Physics — practice by topic

Inside any closed conducting shell with all charge residing on its outer surface, Gauss's law applied to a sphere of radius $r < R$ encloses zero charge so $E = 0$ everywhere inside the cavity.

Standard result for a solid sphere about a central axis: $I = \dfrac{2}{5}MR^2$. Compare with a hollow (thin-shell) sphere where $I = \dfrac{2}{3}MR^2$ — that's option D and a common trap.

Lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$. Sign convention: $u = -20$ cm (object on the left), $v = +60$ cm (real image on the right). We have $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{-20}$. That is $\dfrac{1}{f} = \dfrac{1}{60} + \dfrac{3}{60}$. That is $\dfrac{1}{f} = \dfrac{4}{60}$. So $\dfrac{1}{f} = \dfrac{1}{15}$. So $f = 15$ cm. **Trap.** Forgetting the sign on $u$ gives $\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{20} = -\dfrac{1}{30}$, leading to a negative $f$ — that's how option B (12) or A (10) creeps in.