A simple pendulum made of a mass of 10 g and | vedclass

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(B) The induced $EMF$ in a rotating rod of length $\ell$ with angular velocity $\omega$ in a magnetic field $B$ is given by $\varepsilon = \frac{B \om

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) The induced $EMF$ in a rotating rod of length $\ell$ with angular velocity $\omega$ in a magnetic field $B$ is given by $\varepsilon = \frac{B \omega \ell^2}{2}$.To find the maximum angular velocity $\omega_{\max}$,we use the principle of conservation of energy.The potential energy at the release angle $	heta = 60^{\circ}$ is converted into kinetic energy at the lowest point.$mg\ell(1 - \cos 60^{\circ}) = \frac{1}{2} I \omega_{\max}^2$,where $I = m\ell^2$.$mg\ell(1 - 0.5) = \frac{1}{2} m\ell^2 \omega_{\max}^2$.$g(0.5) = \frac{1}{2} \ell \omega_{\max}^2$.$\omega_{\max} = \sqrt{\frac{g}{\ell}} = \sqrt{\frac{10}{0.1}} = \sqrt{100} = 10 \ rad/s$.Now,substitute the values into the $EMF$ formula:$\varepsilon_{\max} = \frac{2 	imes 10 	imes (0.1)^2}{2} = 10 	imes 0.01 = 0.1 \ V$.Converting to $mV$: $0.1 \ V = 100 \ mV$.

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