Question

An engine pumps a liquid of density d continuously through a pipe of area of cross section A. If the speed of the liquid through the pipe is v, then the power of the engine is

• A. $Ad{v}^3/2$
• B. $Adv$
• C. $Ad{v}^2/2$
• D. $Ad{v}^2$

Answer 1

The correct option is A $Ad{v}^3/2$

Power $=\frac{W.D}{t}=\frac{\Delta (K.E)}{t}$ $=\frac{\frac{1}{2}m{v}^2}{t}$

If $m$ be the mass pumped in time $t$ then: $m=dAvt$
$\Rightarrow power=\frac{\frac{1}{2}\times dAvt\times v^2}{t}$ $=\frac{dA{v}^3}{2}$