Question

A motor of power $P$ is employed to deliver water at a certain rate from a given pipe. In order to obtain thrice as much water from the same pipe in the same time, another motor of power $P^{\prime }$ is required. Given : $g=9.3489m/s^2$. Now, the ratio $P^{\prime }/P$ equals :

• A. $27$
• B. $81$
• C. $561$
• D. $9.3489$

Answer 1

Answer: (A)

$27$

Mass of water flowing through volume V per second = $M=\frac{(Volume\times Density)}{Time}$

$M=\frac{(Area\times Length\times Density)}{Time}$

$M=Area\times Velocity\times Density$

$M=A\times V\times D$

Now the $K.E.=\frac{1}{2}M{V}^2$

Now putting the value of M in the equation

$K.E.=\frac{1}{2}A\times V\times D\times V^2$

$K.E.=\frac{1}{2}A\times D\times V^3$

Let assume new mass is $M_1$

Now comparing both:

$\frac{M_1}{M}=\frac{A\times V_1\times D}{A\times V\times D}=\frac{V_1}{V}$

Now the Power comparison:

$\frac{P_1}{P}=\frac{\frac{1}{2}A\times D\times {V_1}^3}{\frac{1}{2}A\times D\times V^3}$

$\frac{P_1}{P}={\left(\frac{M_1}{M}\right)}^3$

$M_1=3M$

$P_1=27P$