Question

A solid ball of density half that of water falls freely under gravity from a height of $19.6\text{m}$ and then enters a stream of water. Upto what depth will the ball go? [Take $g=9.8{\text{m/s}}^2]$

• A. $29.4\text{m}$
• B. $4.9\text{m}$
• C. $9.8\text{m}$
• D. $19.6\text{m}$

Answer 1

The correct option is D $19.6\text{m}$


From law of conservation of mechanical energy, velocity at the surface of water

$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 19.6}=19.6\text{m/s}$

Net retardation inside the water,

$a=\frac{\text{Upthrust - weight}}{\text{mass}}$

$\Rightarrow a=\frac{V\left(2\rho \right)g-V\left(\rho \right)g}{V\left(\rho \right)}=9.8{\text{m/s}}^2$ [upward]

[$\rho$ is the density of ball and 2$\rho$ is the density of water]

So, using $v^2-u^2=2as$

$\Rightarrow 0^2-{19.6}^2=2\times (-9.8)\times s$

Since, at maximum depth ball will be at instantaneous rest,

$\Rightarrow s=19.6\text{m}$

$\begin{array}{c}\text{Why this question}?\\ \text{Concept -As acceleration is in upward direction, ball will}\\ \text{first retard and come to the maximum depth where its speed}\\ \text{is zero and then accelerate upwards and come to the water}\\ \text{surface.}\end{array}$