Question

A small block of wood of specific gravity $0.5$ is submerged at a depth of $1.2 \,m$ in a vessel filled with the water. The vessel is accelerated upwards with an acceleration $a_{0}=g / 2$. Time taken by the block to reach the surface, if it is released with zero initial velocity is

• A. 0.6 s
• B. 0.4 s
• C. 1. 2 s
• D. 1 s

Answer 1

Answer: (B)

Let $m$ be the mass of block and $a$ its acceleration in upward direction. Then.

$a=\frac{\text { upward thrust }-\text { weight }}{m}=\frac{F-w}{m}$
$\Rightarrow a=\frac{\left(\frac{m}{0.5 \times 10^{3}}\right) \times\left(10^{3}\right) \times\left(g+a_{0}\right)-m g}{m}$
$\Rightarrow a=2\left(g+\frac{g}{2}\right)-g=2 g$
Acceleration of block relative to water
$ a_{r} =a-a_{0}=2 g-\frac{g}{2}=\frac{3 g}{2}$
$ \therefore h =\left(\frac{1}{2}\right) a_{r} t^{2} $
$\Rightarrow t =\sqrt{\frac{2 h}{a_{r}}}=\sqrt{\frac{2 h}{\frac{3}{2} g}}=2 \sqrt{\frac{h}{3 g}}=2 \sqrt{\frac{1.2}{3 \times 10}} m$
$\Rightarrow t =0.4\, s$