A solid cylinder rolls down on an inclined plane of height ' $h$ ' and inclination ' $\theta$ '. The speed…
A solid cylinder rolls down on an inclined plane of height ' $h$ ' and inclination ' $\theta$ '. The speed of the cylinder at the bottom is
$\sqrt{\frac{g h}{2}}$
$\sqrt{\frac{3 g h}{2}}$
$\sqrt{2 g h}$
$\sqrt{\frac{4 g h}{3}}$
Solution
The speed of cylinder at the bottom of inclined plane is
$v=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{K}^2}{\mathrm{R}^2}}}=\sqrt{\frac{2 \mathrm{gh}}{\left(1+\frac{1}{2}\right)}}=\sqrt{\frac{4 \mathrm{gh}}{3}}$