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VITEEE 2009 Paper

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Question : 4 of 120

Marks: +1, -0

A motor of power

P_{0}

is used to deliever water at a certain rate through a given horizontal pipe. To increase the rate of flow of water through the same pipe n times, the power of the motor is increased to

P_{1}.

P_{1}\to P_{0}

is

n:1
$n^{2}:1$
$n^{3}:1$
$n^{4}:1$

Solution:

Power of motor initially =

P_{0}

Let, rate of flow of motor = (x)

Since, power

P_{0} = \frac{\text{work}}{\text{time}} = \frac{mgy}{t} = mg\left(\frac{y}{t}\right)

\frac{y}{t}

= x = rate of flow of water = mgx...(i)

If rate of flow of water is increased by n times, i.e., (nx).

Increased power,

P_{1} = \frac{mgy'}{t} = mg\left(\frac{y'}{t}\right) = nmgx

...(ii)

The ratio of power,

\frac{P_{1}}{P_{0}} = \frac{nmgx}{mgx} ; P_{1}:P_{0} = n:1

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