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 $10\,m{s^{ - 2}}$, and neglect the effect of viscous forces).
(A) $200\,J$
(B) $600\,J$
(C) $400\,J$
(D) $800\,J$

Answer 1

Hint: The work done to lift the stone inside the water is equal to the product of the apparent weight of the stone and the displacement of the stone inside the water. The apparent weight is equal to the total weight of the stone subtracted from the buoyant force on the stone by the water.

Useful formula
The apparent weight of the stone inside the water is given by,
$AW = mg - {F_B}$
Where, $AW$ is the apparent weight, $m$ is the mass of the stone, $g$ is the acceleration due to gravity and ${F_B}$ is the buoyant force.
The buoyant force of the object inside the water is given by,
${F_B} = V{\rho _w}g$
Where, ${F_B}$ is the buoyant force, $V$ is the volume of the object, ${\rho _w}$ is the density of the water and $g$ is the acceleration due to gravity.
The work done is given by,
$W = AW \times h$
Where, $W$ is the work done, $AW$ is the apparent weight and $h$ is the displacement.

Complete step by step solution
Given that,
The mass of the stone is, $m = 10\,kg$,
The specific gravity is, $\dfrac{{{\rho _s}}}{{{\rho _w}}} = 3$,
The displacement of the stone, $h = 6\,m$,
The acceleration due to gravity is, $g = 10\,m{s^{ - 2}}$
Now,
The apparent weight of the stone inside the water is given by,
$AW = mg - {F_B}\,....................\left( 1 \right)$
Now,
The buoyant force of the object inside the water is given by,
${F_B} = V{\rho _w}g\,................\left( 2 \right)$
Already we know that the volume is equal to the $\dfrac{m}{\rho }$, here the volume is the volume of the object, so substitute the value of mass of the stone and density of the stone, then
${F_B} = \dfrac{m}{{{\rho _s}}}{\rho _w}g$
By rearranging the terms, then
${F_B} = m\dfrac{{{\rho _w}}}{{{\rho _s}}}g$
By the given specific gravity value, then the above equation is written as,
${F_B} = \dfrac{{mg}}{3}$
By substituting the above equation in the equation (1), then the equation (1) is written as,
$AW = mg - \dfrac{{mg}}{3}\,.....................\left( 3 \right)$
Now,
The work done is given by,
$W = AW \times h\,................\left( 4 \right)$
By substituting the equation (3) in the equation (4), then
$W = \left( {mg - \dfrac{{mg}}{3}} \right)\, \times h$
By taking the terms common, then the above equation is written as,
$W = \left( {1 - \dfrac{1}{3}} \right)\, \times mgh$
By cross multiplying the terms, then the above equation is written as,
$W = \left( {\dfrac{{3 - 1}}{3}} \right)\, \times mgh$
On further simplification, then the above equation is written as,
$W = \left( {\dfrac{2}{3}} \right)\, \times mgh$
By substituting the mass, acceleration due to gravity and displacement in the above equation, then
$W = \left( {\dfrac{2}{3}} \right)\, \times 10 \times 10 \times 6$
On multiplying the terms, then
$W = \dfrac{{1200}}{3}$
On dividing the terms, then
$W = 400\,J$

Hence, the option (C) is the correct answer.

Note: In apparent weight the weight of the object is subtracted from the buoyant force because the weight of the force is acting downwards, but the buoyant force is pushing the object upwards, so both are in different directions, so both are subtracted.