A rod of length $80 \mathrm{~cm}$ rotates about its mid point with a frequency of $10 \mathrm{rev} / \mathrm{s}$. The potential difference (in volts) between two ends of the rod due to a magnetic field, $B=0.5 \mathrm{~T}$ directed perpendicular to the rod is

Given, frequency of revolution of the rod, $f=10 \mathrm{rev} / \mathrm{s}$, length of rod, $L=80 \mathrm{~cm}=0.8 \mathrm{~m}$ and magnetic field $B=0.5 \mathrm{~T}$ Since, rod is rotated about its mid point, hence radius of circular revolution, $$ r=\frac{L}{2}=\frac{0.8}{2}=0.4 \mathrm{~m} $$ $\therefore$ Potential difference between centre of rod and one end of rod due to revolution in magnetic is given by $$ \begin{aligned} V^{\prime} & =\text { induced emf }=\frac{\text { change in flux }}{\text { time taken }} \\ \frac{B . \pi r^2}{T} & =B . \pi r^2 . f \quad\left[\because f=\frac{1}{T}\right] \\ & =0.5 \times \pi \times(0.4)^2 \times 10 \\ & =0.8 \pi \end{aligned} $$ $\therefore$ Potential difference between two ends of rod due to revolution in the magnetic field, $V=2 V^{\prime}=2 \times 0.8 \pi=1.6 \pi$