A metallic rod of $10 \mathrm{~cm}$ is rotated with a frequency 100 revolution per second about an axis perpendicular to its length and passing through its one end in uniform transverse magnetic field of strength $1 \mathrm{~T}$. The emf developed across its ends is:

The induced electromotive force (emf) across the ends of a conducting rod of length $l$ rotating with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation is given by the formula:

$\varepsilon = \dfrac{1}{2} B \omega l^2$

Given values are:
Length $l = 10 \text{ cm} = 0.1 \text{ m}$
Frequency $f = 100 \text{ rev/s}$
Magnetic field $B = 1 \text{ T}$

The angular velocity $\omega$ is related to frequency $f$ by:
$\omega = 2 \pi f = 2 \times \pi \times 100 = 200 \pi \text{ rad/s}$

Substituting these values into the emf formula:
$\varepsilon = \dfrac{1}{2} \times 1 \times (200 \pi) \times (0.1)^2$

$\varepsilon = \dfrac{1}{2} \times 200 \pi \times 0.01$

$\varepsilon = 100 \pi \times 0.01$

$\varepsilon = \pi \text{ V}$

Using $\pi \approx 3.14$, we get:
$\varepsilon = 3.14 \text{ V}$

Answer: 3.14 V