Physics - W.E.P. Question with Solution | TestHub

Output power$=\eta \times$input power

$=0.75\times 400=300 \mathrm{W}$

Now, from$P=\frac{dw}{dt}=\frac{d\left(mgh\right)}{dt}$

$\frac{dm}{dt}=\frac{P}{gh}=\frac{300}{10\times 40}=\frac{3}{4} \mathrm{kg}$

$=\frac{3}{4}\times 60 \mathrm{kg} {\min }^{-1}$

$=45 \mathrm{kg}{\min }^{-1}$

Hence, water drawn in$10 \min$

$=45\times 10=450\mathrm{kg}{\mathrm{m}}^{-3}$

Since, density of water$=1000\mathrm{kg}{\mathrm{m}}^{-3}$

Therefore, from$m=\rho V$

$V=\frac{m}{\rho }=\frac{450}{1000}=0.45 {\mathrm{m}}^3$