A solid sphere rolls on a smooth horizontal surface at `10 m//s` and then rolls up a smooth inclined plane of inclination `30^@` with horizontal. The mass of the sphere is `2 kg`. Find the height attained by the sphere before it stops (in `m`).

Options

A. 700 cm
B. 701 cm
C. 7.1 m
D. None of these

Answer

C

Explanation

If a body rolls on a horizontal surface, it possesses both translational and rotational kinetic energies. The net kinetic energy is given by
`K_("net")=(1)/(2)mv^(2)(1+(K^(2))/(R^(2)))`,
where K is the radius of gyration.
So from law of conservation of energy,
`(1)/(2)mv^(2)(1+(K^(2))/(R^(2)))=mgh`,
where h is the height attained by the sphere.
i.e., `(1)/(2)xx2xx(10)^(2)(1+(2)/(5))=2xx9.8xxh`
or `h=(700)/(98)=7.1m`