Question

In the arrangement as shown, ${m_B} = 3\,m$, density of liquid is $\rho $ and density of block $B$ is $2\rho $. The system is released from rest so that the block $B$ moves up when in liquid and moves down when out of liquid with the same acceleration. Find the mass of block $A$:



(A) $\dfrac{7}{4}\,m$
(B) $2\,m$
(C) $\dfrac{9}{2}\,m$
(D) $\dfrac{9}{4}\,m$

Answer 1

Hint: The mass of the block $A$ is determined by using the equation of the motion of the block $B$ and the buoyancy force act on the block $B$. By equating both the equation of the block $B$, the mass of the block $A$ can be determined.

Useful formula:
The buoyancy force is given by,
$F = V \times \rho \times g$
Where, $F$ is the buoyancy force, $V$ volume of the block, $\rho $ is the density of the liquid and $g$ is the acceleration due to gravity.

Complete step by step solution:
Given that,
The mass of the block $B$ is, ${m_B} = 3\,m$.
The density of the liquid is, $\rho $.
The density of the block $B$ is, $2\rho $.
Now,
The block $B$ moves down as it be out of liquid, so the equation of the motion is given by,
$3mg - Mg = \left( {3m + M} \right)a\,.............\left( 1 \right)$
The above equation is defined as, the force of the block $B$ and the force of the block $A$ are moved in opposite directions, so both are subtracted, is equal to the total force due to block $A$ and block $B$.
The density of the block $B$ is, $2\rho $.
The mass of the block $B$ is, ${m_B} = 3\,m$.
So, the volume of the block $B$ is, $V = \dfrac{{3m}}{{2\rho }}$
The buoyancy force is given by,
$ \Rightarrow V \times \rho \times g$
By substituting the volume of the block $B$, then
$ \Rightarrow \dfrac{{3m}}{{2\rho }} \times \rho \times g$
On further simplification, then
$ \Rightarrow \dfrac{{3mg}}{2}$
Now, the equation of motion when the block $B$ is inside the liquid and moving up, then
$Mg + \dfrac{3}{2}mg - 3mg = \left( {3m + M} \right)a\,..............\left( 2 \right)$
The buoyancy force pushes the block $B$ upwards, so it is added.
By equation the equation (1) and equation (2), then
\[3mg - Mg = Mg + \dfrac{3}{2}mg - 3mg\]
By arranging the above equation, then
$3mg + 3mg - \dfrac{3}{2}mg = Mg + Mg$
By adding the above equation, then
$6mg - \dfrac{3}{2}mg = 2Mg$
On cross multiplying the terms in LHS, then
$\dfrac{9}{2}mg = 2Mg$
By rearranging the terms, then
$Mg = \dfrac{9}{4}mg$
By cancelling the terms on both sides, then
$M = \dfrac{9}{4}m$
Thus, the above equation shows the mass of the block $A$.

Hence, the option (D) is correct.

Note: The block $B$ moves in two conditions, so the two equations of the motion of the block $B$ is written and these two equations of the motion of the block $B$ is equated, then the mass of the block $A$ is determined in terms of $m$.