Question

A solid cube floats in water half immersed and has small vertical oscillations of time period $\frac{\pi }{5}\text{s}$. Its mass $\left(\text{in kg}\right)$ is:
$(\text{Take}g=10{\text{ms}}^{-2})$

• A. $4$
• B. $2$
• C. $1$
• D. $0.5$

Answer 1

The correct option is A $4$

Given,
Time period of oscillation $\left(T\right)=\frac{\pi }{5}s$
Density of liquid $\left(\rho \right)=1000{\text{kg/m}}^3$
Since, cube is half immersed i.e $h=\frac{l}{2}$ is the height of cube immersed in water, where $l$ is the side of the cube.
Applying the formula for floatation,
$mg=\rho V^{\prime }g$
$\sigma V=\rho V^{\prime }$
$\sigma Al=\rho Ah$
$\frac{h}{l}=\frac{\sigma }{\rho }$
where, $l\to$ edge length of cubical block
$\sigma \to$ Density of block
$\rho \to$ density of liquid
$m\to$mass of cube
$V\to$volume of cube
$V^{\prime }\to$volume of immersed part
$A\to$base area of cube

$\Rightarrow \frac{l}{2\times l}=\frac{\sigma }{\rho }$
$\Rightarrow \sigma =\frac{\rho }{2}$
i.e density of cube is half the density of liquid
$\Rightarrow \sigma =500{\text{kg/m}}^3$
Time period of oscillations is given as,
$\Rightarrow T=2\pi \sqrt{\frac{m}{\rho Ag}}....\left(i\right)$
Where $m=\sigma l^3=$mass of cubical block
$A=l^2=$cross-sectional area of cubical block
Eq.$\left(i\right)$ can be expressed as,
$\Rightarrow T=2\pi \sqrt{\frac{\sigma l}{\rho g}}$
Substituting $\frac{\sigma }{\rho }=\frac{1}{2}$
$\Rightarrow \frac{\pi }{5}=2\pi \sqrt{\frac{l}{2g}}$
$\Rightarrow \frac{l}{20}=\frac{1}{100}$
$\left(\text{or}\right)l=0.2\text{m}$
Now mass of cube is,
$m=V\times \sigma =l^3\sigma =(0.2{)}^3\times 500$
$\Rightarrow m=4\text{kg}$