Question

In the arrangement as shown, $m_B=3m$, 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) ${\displaystyle \frac{7}{4}}m$
(B) $2m$
(C) ${\displaystyle \frac{9}{2}}m$
(D) ${\displaystyle \frac{9}{4}}m$

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, $m_B=3m$.
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, $m_B=3m$.
So, the volume of the block $B$ is, $V={\displaystyle \frac{3m}{2\rho }}$
The buoyancy force is given by,
$\Rightarrow V\times \rho \times g$
By substituting the volume of the block $B$, then
$\Rightarrow {\displaystyle \frac{3m}{2\rho }}\times \rho \times g$
On further simplification, then
$\Rightarrow {\displaystyle \frac{3mg}{2}}$
Now, the equation of motion when the block $B$ is inside the liquid and moving up, then
$Mg+{\displaystyle \frac{3}{2}}mg-3mg=\left(3m+M\right)a..............\left(2\right)$
The buoyancy force pushes the block $B$ upwards, so it is added.
By equation the equation (1) and equation (2), then
$$3mg-Mg=Mg+{\displaystyle \frac{3}{2}}mg-3mg$$
By arranging the above equation, then
$3mg+3mg-{\displaystyle \frac{3}{2}}mg=Mg+Mg$
By adding the above equation, then
$6mg-{\displaystyle \frac{3}{2}}mg=2Mg$
On cross multiplying the terms in LHS, then
${\displaystyle \frac{9}{2}}mg=2Mg$
By rearranging the terms, then
$Mg={\displaystyle \frac{9}{4}}mg$
By cancelling the terms on both sides, then
$M={\displaystyle \frac{9}{4}}m$
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 $m$.