A drop with initial mass M falls under the action of the force of gravity and evaporates uniformly losing a mass m each second.What is the work performed by the force of gravity during the time interval from the start to the complete evaporation of the drop?(air resistances may be neglected).The answer is g2(M3/6m2)....plz solve

Let us first calculate infinitesimal work done, dW by the force of gravity on the drop in the interval t to t+dt. Let’s assume the mass of the drop at time t be M(t) and its velocity at this instant be v(t). Let dx be the displacement of drop in time dt. Then,

dW = M(t)g dx = M(t)g v(t) dt

where, 
M(t) = M – m t  (Since the mass evaporates at uniform rate)
v(t) = gt           (Assuming drop to be point mass and using equation of motion. velocity of a point                               mass is independent of its mass.)

Hence, work done W = ∫dW = ∫₀^T(M-m t)g gt dt = g²[Mt²/2-mt³/3]₀^T= g²[MT²/2-mT³/3]
where, T = M/m

This gives

W = g² (M³/6m²)