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(B) Given: Inclination $	heta = 30^{\circ}$, length $l = 60 \,cm = 0.6 \,m$, acceleration due to gravity $g = 10 \,m/s^2$.Using the law of conservati

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(B) Given: Inclination $	heta = 30^{\circ}$, length $l = 60 \,cm = 0.6 \,m$, acceleration due to gravity $g = 10 \,m/s^2$.Using the law of conservation of mechanical energy, the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Here, $h = l \sin 	heta = 0.6 	imes \sin 30^{\circ} = 0.6 	imes 0.5 = 0.3 \,m$.For a solid cylinder, the moment of inertia $I = \frac{1}{2}mr^2$ and for rolling without slipping, $\omega = \frac{v}{r}$.Substituting these values:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$$gh = \frac{3}{4}v^2 \Rightarrow v = \sqrt{\frac{4gh}{3}}$$v = \sqrt{\frac{4 	imes 10 	imes 0.3}{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2 \,m/s$.

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 \,cm$. If the cylinder rolls without slipping, then the speed when it reaches the bottom is (in $\,m/s$)

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