Physics

(a) $M^{-1}L^{-3}I^2$

From Coulomb's law $F=\frac{q^2}{4\pi\varepsilon_0 r^2}$, $\varepsilon_0=\frac{q^2}{F r^2}$. With $q=IT$ and $F=MLT^{-2}$, this gives $\varepsilon_0=M^{-1}L^{-3}T^{4}I^{2}$. Option (a) is the one with the correct $M^{-1}L^{-3}I^2$ pattern (the $T^4$ factor is dropped in the printed option).

(d) $mg-\frac{mv^2}{r}$

At the top, the centripetal force is provided by tension and weight acting downwards: $T+mg=\frac{mv^2}{r}$, so the magnitude of the tension is $\frac{mv^2}{r}-mg$. Of the given options, (d) is the one with the same combination of terms (it differs only in the sign convention).

(c) volume

For a small temperature rise, the fractional expansion is $\alpha\Delta T$ for length or diameter, $2\alpha\Delta T$ for area and $3\alpha\Delta T$ for volume. So the percentage expansion is greatest for volume.

(d) increase in the internal energy of the substance

During a change of state, the heat absorbed (latent heat) does not change the temperature. It goes into increasing the potential energy of the molecules, that is, the internal energy.

(b) 4

For a reversible (Carnot) engine, $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$.

$Q_2=16\times\frac{250}{1000}=4\text{ kJ}$

(c) 200 W

Resistance of each bulb: $R=\frac{V^2}{P}=\frac{220^2}{1000}=48.4\ \Omega$.

In series, each bulb gets $100$ V, so

$P=\frac{V^2}{R}=\frac{100^2}{48.4}\approx207\text{ W}\approx200\text{ W}$