Question

A small solid ball is dropped from a height above the free surface of a liquid. It strikes the surface of the liquid at $t=0$. The density of the material of the ball is $500kg/m^3$ and that of liquid is $1000kg/m^3$. If the ball comes momentarily at rest at $t=2\sec$ then initial height of the ball from surface of liquid was (neglect viscosity).

• A. $20m$
• B. $10m$
• C. $15m$
• D. $25m$

Answer 1

The correct option is A $20m$

When ball strikes the watch (liquid)

surface it will stent experiences the

upthrust force

This force will retrod its motion & after some time its velocity

will become zero momentarily

because the upthrust force beings more

than the gravity will cause a Net

acceleration in upward direction

upthrust force $=pvg;P=1000kg/m$

gravity $=mg=\sigma vg;\sigma =500kg/m$

v = volume of ball.

$\Rightarrow$ upward force $=pvg-\sigma vg$

$\Rightarrow$ retardation $a=\frac{f}{mass}=\frac{pvg-\sigma vg}{m}$

or $a=\frac{pvg-\sigma vg}{\sigma \times v}=\frac{p}{\sigma }g-g$

or $a=g(\frac{1000}{500}-1)=9=10m/s^2$

suppose velocity diving strike was v

then $v-gt=0$ rest momentarily

$\Rightarrow v=gt=10\times 2=20m/s$

but $v=\sqrt{2gh}\Rightarrow h=\frac{v^2}{2g}=\frac{400}{2\times 10}=20m$