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 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. $\frac{7}{4}m$
• B. 2m
• C. $\frac{9}{2}m$
• D. $\frac{9}{4}m$

Answer 1

The correct option is D $\frac{9}{4}m$

Let mas of block A $m_A$ = M
Magnitude of acceleration in both up and down $=a$
Block b moves down as it be out of liquid. so writing equation of motion.
$3mg-Mg=(3m+M)a------\left(1\right)$
Density of block $B=2\rho$
Mass of block $B=3\text{m}$
Volume of bock $B=\frac{3m}{2\rho }$
Buoyancy force acting up on block $B=\text{volume of B}\times \text{density of liquid}\times g=\frac{3m}{2\rho }\rho g=\frac{3mg}{2}$
Now writing equation of motion when bock B is inside the liquid and going up,
$Mg+\frac{3}{2}mg-3mg=(3m+M)a-------\left(2\right)$
Equating left side of equation (1) and (2),
$3mg-Mg=Mg+\frac{3mg}{2}-3mg$
$Mg=\frac{9}{4}mg$
$M=\frac{9}{4}\text{m}$