A magnetic field $4 \times 10^{-2} \mathrm{~T}$ acts at right angles to a coil of area $100 \mathrm{~cm}^2$ with 50 turns. The average e.m.f. induced in the coil is 0.1 V , when it is removed from the field in time ' $t$ '. The value of ' $t$ ' is

Faraday's law gives the average induced emf as $\varepsilon = N \left| \frac{\Delta \Phi_B}{\Delta t} \right|$. The magnetic flux through a coil perpendicular to a uniform magnetic field is $\Phi_B = B \cdot A$.

The area is $A = 100 \mathrm{cm}^2 = 10^{-2} \mathrm{m}^2$. The initial flux is $\Phi_i = (4 \times 10^{-2}) \cdot (10^{-2}) = 4 \times 10^{-4} \mathrm{Wb}$ and final flux is zero, so $\Delta \Phi_B = -4 \times 10^{-4} \mathrm{Wb}$.

Substituting $N = 50$ and $\varepsilon = 0.1 \mathrm{V}$, we get $0.1 = 50 \cdot \frac{4 \times 10^{-4}}{t}$.

Solving for $t$: $t = \frac{50 \cdot 4 \times 10^{-4}}{0.1} = \frac{2 \times 10^{-2}}{10^{-1}} = 2 \times 10^{-1} = 0.2 \mathrm{s}$.

The correct answer is $\boxed{0.2}$ second.