Question

The work done in lifting a stone of mass 10 kg and specific gravity 3 from the bed of a lake to a height of 6 m inside the water is : (Take acceleration due to gravity as $10m/s^2$, and neglect the effect of viscous forces)

• A. $200J$
• B. $600J$
• C. $400J$
• D. $800J$

Answer 1

The correct option is C $400J$

Apparent weight $=mg-F_B$

where, $F_B$ is buoyant force and $mg$ is weight

We know that buoyant force $F_B$ will be:
$F_B=V{\rho }_{water}g=\frac{m}{{\rho }_{stone}}{\rho }_{water}g$

where $V$ is volume of stone =$\frac{m}{{\rho }_{stone}}$ and ${\rho }_{water}$ is density of water

As specific gravity is given as 3 it's density will be three times of water

$\frac{{\rho }_{stone}}{{\rho }_{water}}=3$

Hence, $F_B=\frac{mg}{3}$

Putting the values of $m,g$ and applying the formula for work done,

Work done $=$ Apparent weight $\times$ displacement

$=10\times g\times h-\frac{10\times g}{3}\times h$
$=\frac{2}{3}\times 10\times 10\times 6$
$=400J$