Question

A block of mass $1\text{kg}$ is kept over a fixed smooth wedge. Block is attached to a sphere of same mass through fixed massless pulleys. Sphere is dipped inside the water as shown. If specific gravity of material of sphere is $2$. Acceleration of sphere while it is in water (in ${\text{m/s}}^2)$ is

Answer 1

Given, mass of block=mass of sphere, $m=1\text{kg}$
specific gravity of material of sphere, $s=2$.

FBD of the given system:


Force of buoyancy $\left(F_B\right)$ on sphere acts in upward direction as shown in figure.

Analyzing force for block,

$mg\sin {30}^{\circ }-T=ma$

Substituting the values,

$\Rightarrow 1\times 10\times \frac{1}{2}-T=1\times a$

$\Rightarrow 5-T=a....\left(1\right)$

Balancing force for sphere,

$T+F_B-mg=ma...\left(2\right)$

Now, $F_B=\rho gv$
where, density of liquid displaced $\rho =1000\text{kg}{\text{m}}^{-3}$
volume of liquid displaced=volume of sphere,$v=\frac{m}{{\rho }_{sphere}}=\frac{1}{1000\times 2}$

Putting the value of $\rho$ and $v$ in the expression of $F_B$,
$F_B=(1000\times 10\times \frac{1}{1000\times 2})$

$=\left(1000\right)\times 10\times \frac{1}{2\times 1000}$

$\therefore F_B=5\text{N}$

From equation $\left(2\right)$,
$\Rightarrow T+5-1\times 10=1\times a$
$\Rightarrow T+5-10=a$
$\Rightarrow T-5=a...\left(3\right)$

From $\left(1\right)$ and $\left(3\right)$,
$5-T+T-5=a+a$
$\therefore a=0{\text{m/s}}^2$

Hence acceleration of the sphere is $a=0{\text{m/s}}^2$.

Answer accepted: $0,0.0$