Question

A body of cubic shape of side $2cm$ is falling in the water. Density of body is $=3g{m}^{-3}$. Find out the value of acceleration of cube.
(Density of water, ${\rho }_w=1g{m}^{-3},g=10000cm{s}^{-2}$).

• A. $\frac{500}{9}gcm-3$
• B. $\frac{2000}{3}gcm-3$
• C. $\frac{4000}{3}gcm-3$
• D. $\frac{1000}{3}gcm-3$

Answer 1

The correct option is B $\frac{2000}{3}gcm-3$

Given:
Side of cube $=2cm$
Volume of mass,
$V$$=sid{e}^3=2^3=8c{m}^3,$
$\rho =3gc{m}^{-3}$
Mass of body,$m=\rho V=24g$

Buoyant force ($F_B$) is eaual to the weight of displaced water by the body.
$F_B=V\times {\rho }_w\times g$
$F_B=8\times 1\times 1000=8000dyne$

Force of gravity on body, $F_g=m\times g=\rho \times V\times g$
$F_g=3\times 8\times 1000=24000dyne$

Resultant force on the body,
$F=F_g-F_B=24000-8000$
$F=16000dyne$

Value of acceleration,
$a=\frac{F}{m}$
$a=\frac{16000}{24}=\frac{2000}{3}gcm-3$