Question

A solid sphere is rolling on a frictionless surface with translational velocity 'V'. It climbs the inclined plane from 'A' to 'B' and then moves away from Bon the smooth horizontal surface. The value of 'V' should be ______.

पर्याय

• ${\displaystyle \ge {\left[\frac{10\text{gh}}{7}\right]}^{\frac{1}{2}}}$

• ${\displaystyle \sqrt{2gh}}$

• ${\displaystyle \frac{10gh}{7}}$

• ${\displaystyle \sqrt{gh}}$

Answer 1

A solid sphere is rolling on a frictionless surface with translational velocity 'V'. It climbs the inclined plane from 'A' to 'B' and then moves away from Bon the smooth horizontal surface. The value of 'V' should be ${\displaystyle \underline{\ge {\left[\frac{10\text{gh}}{7}\right]}^{\frac{1}{2}}}}$.

Explanation:

${\displaystyle \frac{1}{2}{\text{mv}}^2+\frac{1}{2}I{\omega }^2=\text{mgh}}$

${\displaystyle \frac{1}{2}{\text{mv}}^2+\frac{1}{2}\frac{2}{5}{\text{mr}}^2\times \frac{{\text{v}}^2}{{\text{r}}^2}=\text{mgh}}$

${\displaystyle \frac{{\text{v}}^2}{2}+\frac{2{\text{v}}^2}{10}=\text{gh}}$

${\displaystyle \frac{5{\text{v}}^2+2{\text{v}}^2}{10}=\text{gh}}$

${\displaystyle {\text{v}}^2=\frac{10}{7}\text{gh}}$

${\displaystyle \text{v}=\sqrt{\frac{10}{7}\text{gh}}}$