Question

A small block of wood of relative density $0.5$ is submerged in water at a depth of $5\,m$. When the block is released, it starts moving upwards, the acceleration of the block is $\left( {g = 10\,m/{s^2}} \right)$
A. $5\,m/{s^2}$
B. $10\,m/{s^2}$
C. $7.5\,m/{s^2}$
D. $15\,m/{s^2}$

Answer 1

Hint-We need to find acceleration when the block of wood is released.
Foe that we need to consider all the forces acting on the Block. The forces acting on the block are force of gravity acting downward, force of buoyancy acting upward
Force is the product of mass and acceleration.
$F = ma$
The net force can be written as
$ma = {F_b} - W$
Where, ${F_b}$ is the force of buoyancy, W is the weight.
Force of buoyancy is given as
${F_b} = V \times \rho \times g$
Where, $V$ is the volume of liquid displaced, $\rho $ is the density of liquid displaced and $g$ is the acceleration due to gravity.
Using these we can find the answer to this question.

Step by step solution:
Given
${\text{depth}} = 5cm$
Relative density $ = 0.5$
Acceleration due to gravity, $g = 10\,m/{s^2}$
Relative density is the ratio of density of the substance to density of water
$\rho = \dfrac{{{\rho _s}}}{{{\rho _w}}}$
Where, ${\rho _s}$ is the density of substance and ${\rho _w}$ is the density of water.
Therefore,
$\dfrac{{{\rho _{wood}}}}{{{\rho _{water}}}} = 0.5$
We need to find acceleration when it is released.
Let us consider all the forces acting on the Block.
The forces acting on the block are force of gravity acting downward, force of buoyancy acting upward
Force is the product of mass and acceleration.
$F = ma$
The net force can be written as
$ma = {F_b} - W$ ……………….(1)
Where, ${F_b}$ is the force of buoyancy, W is the weight.
Force of buoyancy is given as
${F_b} = V \times \rho \times g$
Where, $V$ is the volume of liquid displaced, $\rho $ is the density of liquid displaced and $g$ is the acceleration due to gravity.
Therefore,
${F_b} = V \times {\rho _{water}} \times g$
Weight,
$W = mg = V \times {\rho _{wood}} \times g$
Since, $\rho = \dfrac{m}{V}$
Net force,
$ma = V \times {\rho _{wood}} \times a$
Substituting all these values in equation (1), we get
$V \times {\rho _{wood}} \times a = V \times {\rho _{water}} \times g - V \times {\rho _{wood}} \times g$
$ \Rightarrow a = \dfrac{{{\rho _{water}}}}{{{\rho _{wood}}}} \times g - g = \dfrac{1}{{0.5}}g - g$
$\therefore a = 10\,m/{s^2}$

So, the correct answer is option B.

Note:Remember that the relative density of wood is given. Relative density is the ratio of density of the substance to density of water
$\rho = \dfrac{{{\rho _s}}}{{{\rho _w}}}$
Where, ${\rho _s}$ is the density of substance and ${\rho _w}$ is the density of water.
Hence don’t substitute the given value of relative density for the density of wood. Relative density of wood is the value obtained by dividing density of wood with density of water.