Question

The mass of a bucket containing water is $M_0$. The bucket is pulled steadily up from a well of depth d. Due to a hole in the bucket the water is pouring out at a uniform rate and as a result the mass of the bucket with water at the top of the well reduces to M. Then the amount of work done in pulling up the bucket is?

• A. $(M_0-M)gd$
• B. $\frac{1}{2}(M_0-M)gd$
• C. $(M_0+M)gd$
• D. $\frac{1}{2}(M_0+M)gd$

Answer 1

The correct option is D $\frac{1}{2}(M_0+M)gd$

change in mass of bucket per unit length is $\frac{M_0-M}{d}$

so at any height $x$ from bottom of well mass of bucket is given by $M_0-($$\frac{M_0-M}{d}).x$

force applied is weight of the bucket at any height $x$

and work done is $froce\left(F\right).displacement\left(ds\right)$

$dW=[$$M_0-($$\frac{M_0-M}{d}).x$$]g.dx$

after integration both side $0toW,0tod$

we get

$W=\frac{1}{2}(M_0+M)gd$

so correct answer is option D.