Question

A solid sphere rolls down from top of inclined plane, $7m$ high, without slipping. Its linear speed at the foot of plane is $(g=10m/s^2)$

• A. $\sqrt{70}m/s$
• B. $\sqrt{\frac{140}{3}}m/s$
• C. $\sqrt{\frac{280}{3}}m/s$
• D. $\sqrt{100}m/s$

Answer 1

Answer: (D)

Given, height of inclined plane, $h=7m$
$g=10m/s^2$
By conservation of energy, Potential energy lost by the solid sphere in rolling down the inclined plane = Kinetic energy gained by the sphere.
$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}m{v}^2+\frac{1}{2}\cdot \frac{2}{5}m{R}^2\cdot {\left(\frac{v}{R}\right)}^2$
$=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2$
$mgh=\frac{7}{10}m{v}^2$
$\therefore v^2=\frac{10}{7}gh$
$\Rightarrow v=\sqrt{\frac{10}{7}gh}=\sqrt{\frac{10}{7}\times 10\times 7}$
$=\sqrt{100}m/s$