Question

A bucket weighing $1.5kg$ is pulled up from the bottom of a well $10m$ deep with $8kg$ of water in it. If there is a hole in the bucket so that water leaks out at a steady rate as it rises, leaving only $7kg$ when it reaches the top, what is the work needed?

Answer 1

R.E.F image

Solution $\to$ lets divide the weights

into 2 parts $\to$

1] 1.5 kg bucket +7kg water

2] 1 kg water which falls in the middle

for (1), work $=8.5gh=8.5\times 10\times g=85g$

now for (2), assume bucket as shown

let the mass of extra water be m

(ie excluding (1)st case)

$\therefore$ work for (2) $=\int mgdh$

now $\frac{dm}{dh}=c\Rightarrow m=ch$

now for $h=10$ $m=1\Rightarrow c=1/10$

$\therefore m=h/10$

$\therefore$ work for (2) $={\int }_0^{10}g\left(h/10\right)dh=\left(g/10\right)\frac{h^2}{z}{|}_0^{10}$

$=\left(\frac{g}{10}\right)50=5g$

$\therefore$ work = work for (1) + work for (2)

$=85g+5g$

$=90g\to$ ans