Question

A wooden cube floating in water supports a mass m = 0.2 kg on its stop. When the mass is removed the cube rises by 2 cm. The side of the cube is - (density of water ${10}^{3}kg/m^3$)

• A. 6 cm
• B. 12 cm
• C. 8 cm
• D. 10 cm

Answer 1

The correct option is D 10 cm

Cube is floating, therefore

Weight of liquid displaced $=$ Weight of cube

let the volume of liquid displaced $=V$

Density of liquid $=\rho$

Mass of cube $=M$

Then,

$(M+m)g=\rho vg$

$⟹M+m=\rho v$ ............(1)

When mass is removed, cube reise by $l=2cm$

Therefore,

$Mg=(V-lA)\rho g$

where, $AS$ is the are of face of cube

$⟹M=\rho (V-lA)$ .................(2)

Eliminating $M$ by subtracting (2) from (1)

$m=\rho lA$

$⟹A=\frac{m}{\rho l}$

Let side of cube $=a$

Then,

$a^2=A=\frac{m}{\rho l}$

$⟹a=\sqrt{\frac{m}{\rho l}}$

If liquid is water, then $\rho =1000kg/m^3$

$I=0.02m$ (given)

$m=0.2kg$ (given)

Putting these value, we get $a=0.1m=10cm$