When it enters the lake (density $$\sigma$$) it is fully immersed. Two constant forces now act on it:

Its weight: $$W = \rho Vg,$$

The buoyant force: $$B = \sigma Vg.$$

Since $$\sigma > \rho$$, the buoyant force is greater than the weight, so after entry the net force acts upward. In fact, the net upward force is

$$F_{\text{net}} = B - W = (\sigma Vg - \rho Vg) = (\sigma - \rho)Vg.$$

As the body sinks into the water, it will decelerate due to this net upward force. Its initial kinetic energy is

$$K = \frac{1}{2}m v^2 = \frac{1}{2}(\rho V)(2gh) = \rho Vgh.$$

Let the additional distance the body sinks below the water surface be $$d$$ (that is, the maximum extra immersion beyond the water level). As it sinks this extra distance, two contributions affect the energy:

The weight does positive work (adding energy) equal to $$\rho Vg\,d,$$

The buoyant force does negative work (removing energy) equal to $$\sigma Vg\,d.$$

Thus, the net work done while moving a distance $$d$$ is

$$W_{\text{net}} = (\rho Vg - \sigma Vg)d = -(\sigma - \rho)Vg\,d.$$

Note that this net work exactly opposes the kinetic energy gained before entering the water. At the point of maximum immersion, the body comes momentarily to rest. Therefore, equate the kinetic energy with the work done against the net force:

$$\rho Vgh = (\sigma - \rho)Vg\,d.$$

Solving for $$d$$:

$ d = \frac{\rho Vgh}{(\sigma - \rho)Vg} = \frac{h\rho}{\sigma-\rho}. $

Thus, the maximum depth the body sinks into the lake is

$$\boxed{\frac{h\rho}{\sigma-\rho}}.$$

Comparing this with the given options, the correct one is Option A.

Fluid Mechanics

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