Question

A sphere of mass m and radius $r$ is projected in a gravity free space with speed $v$. If coefficient of viscosity of the medium in which, it moves is $\frac{1}{6\pi }$, then the distance travelled by the body before it stops, is:

• A. $\frac{mv}{2r}$
• B. $\frac{2mv}{r}$
• C. $\frac{mv}{r}$
• D. $\frac{mv}{4r}$

Answer 1

Answer: (C)

$\frac{mv}{r}$

Since the sphere is projected in gravity free space, the only force acting on the sphere viscous force.

Hence, using Newton's 2nd law

$m\frac{dv}{dt}=-6\pi \eta rv$

$\therefore m\frac{dv}{dx}\frac{dx}{dt}=-6\pi \eta rv$

$\therefore mvdv=-6\pi \eta rvdx$

Integrating both sides:

$\therefore m{\int }_v^{0}dv=-6\pi \eta r{\int }_0^{x}dx$

$mv=6\pi \eta rx=6\pi \left(\frac{1}{6\pi }\right)rx$

$\Rightarrow x=\frac{mv}{r}$