Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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 g 2 (M 3 /6m 2 ).... plz solve


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


Grade:12

1 Answers

Sandeep Pathak
askIITians Faculty 25 Points
7 years ago
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 = ∫0T(M-m t)g gt dt = g2[Mt2/2-mt3/3]0T= g2[MT2/2-mT3/3]
where, T = M/m
This gives
W = g2 (M3/6m2)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free