Correct Option: (a)

The problem states that the density of the cylinder is equal to that of water ($\rho_c = \rho_w$). This implies that when the cylinder is fully submerged, the buoyant force $B$ exactly equals the gravitational force $mg$.

$$ B_{max} = \rho V g = mg $$

Phase 1: Lifting inside water

While the cylinder is completely submerged (moving from depth $d$ until the top surface touches the water level), the tension $T$ in the cord is:

$$ T = mg – B_{max} = mg – mg = 0 $$

Since the tension is zero, the work done by the pulling agency during this phase is zero.

Phase 2: Emerging from water

As the cylinder is pulled out, let $y$ be the length of the cylinder outside the water. The submerged length is $(l – y)$.

The buoyant force becomes: $B(y) = \rho A (l-y) g$.

The tension required to lift it slowly (equilibrium) is:

$$ T(y) = mg – B(y) = \rho A l g – \rho A (l-y) g = \rho A g y $$

Here, $\rho A g$ is a constant. This shows that Tension increases linearly from $0$ (at $y=0$) to $mg$ (at $y=l$).

Calculation of Work:

The work done is the integral of tension over the displacement $l$ (or the area under the T-y graph shown above).

$$ W = \int_0^l T(y) dy = \int_0^l (\rho A g y) dy $$ $$ W = \rho A g \left[ \frac{y^2}{2} \right]_0^l = \frac{1}{2} \rho A g l^2 $$

Substituting $m = \rho A l$:

$$ W = \frac{1}{2} (m) g l = 0.5 mgl $$