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

0.6 s

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0.4 s

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1.2 s

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1 s

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Solution

The correct option is B $0.4 s$
Let $m$ be the mass of block and $a$ its acceleration in upward direction. Then,
$a = \frac{u p w a r d t h r u s t - w e i g h t}{m} = \frac{F - w}{m}$
$\Rightarrow a = \frac{(\frac{m}{0.5 \times 10^{3}}) \times (10^{3}) \times (g + a_{0}) - m g}{m}$
$\Rightarrow a = 2 (g + \frac{g}{2}) - g = 2 g$
Acceleration of block relative to water
$a_{r} = a - a_{0} = 2 g - \frac{g}{2} = \frac{3 g}{2}$
$∴ h = (\frac{1}{2}) a_{r} t^{2}$
$\Rightarrow t = \sqrt{\frac{2 h}{a_{r}}} = \sqrt{\frac{2 h}{\frac{3}{2} g}} = 2 \sqrt{\frac{h}{3 g}} = 2 \sqrt{\frac{1.2}{3 \times 10}} m$
$\Rightarrow t = 0.4 s$

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Q. A small block of wood of specific gravity

0.5

is submerged at a depth of

1.2 m

in a vessel filled with water. The vessel is accelerated upwards with an acceleration

a_{0} = \frac{g}{2}

. Time taken by the block to reach the surface, if it is released with zero initial velocity is

\frac{x}{10} s

. Then find the value of

x

Two block of wood of masses $m_{1}$ and $m_{2}$ and each of specific gravity 0.5 are submerged at a depth of $h_{1}$ and $h_{2}$ in
a vessel $(h_{2} > h_{1})$ filled with water, which is accelerated upwards with an acceleration $\frac{g}{2}$ . the difference in
time taken by the blocks to reach the surface, when released with zero velocity is :