Indian Navy 10+2 B.Tech Cadet Physics — practice by topic

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.

For total internal reflection at a denser-to-rarer interface, $\mu = \dfrac{1}{\sin C}$. With $C = 42^\circ$, $\sin 42^\circ \approx 0.669$, so $\mu = 1/0.669 \approx 1.49$, which matches typical crown glass. **Option A is water and option D is diamond.**

Faraday's law: $|\varepsilon| = N \dfrac{d\Phi}{dt} = 50 \times 4 \times 10^{-3} = 0.20\,\text{V}$.