Question

A hollow sphere of radius r rolls down a rough inclined plane of inclination ${30}^{\circ }$ with the horizontal. If it starts from rest at the top and reaches the bottom of the plane 2.5 m long, the velocity of the centre of the sphere is ( Take g = 10 m $s^2$ )

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Acceleration of rolling object on inclined plane is given by a = $\frac{gsin\theta }{1+\frac{k^2}{R^2}}$

For hollow sphere I = $\frac{2}{3}$ M $R^2$ = M $K^2$ $\Rightarrow$ $\frac{K^2}{R^2}$

$\therefore$ a = $\frac{\left(10\right){30}^{\circ }}{1+\frac{2}{3}}$ = 3 m/ $s^2$

$\therefore$ $v^2$ - $u^2$ = 2as $\Rightarrow$ = $\sqrt{2as}$ = $\sqrt{2\ast 3\ast \frac{5}{2}}$ = $\sqrt{15}$m /s