Question

A disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in figure. If the velocity of the disc is $v$, height to which the disc will rise will be

• A. $\frac{3v^2}{2g}$
• B. $\frac{3v^2}{4g}$
• C. $\frac{v^2}{4g}$
• D. $\frac{v^2}{2g}$

Answer 1

The correct option is B $\frac{3v^2}{4g}$

$=\frac{1}{2}m{R}^2$

$\frac{1}{2}M{V}^2+\frac{1}{2}I{w}^2=Mgh$

$\frac{1}{2}M{V}^2+\frac{1}{2}\frac{1}{2}M{R}^2\frac{V^2}{R^2}=Mgh$

$\frac{1}{2}M{V}^2+\frac{1}{4}M{V}^2=Mgh$

$\frac{3}{4}M{V}^2=Mgh$

$h=\frac{3V^2}{4g}$

Hence, option $\left(B\right)$ is correct answer.