VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    A motor of power \[{{P}_{0}}\] is used to deliver 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 top. The ratio of \[{{P}_{1}}\], to \[{{P}_{0}}\] is

    A)  \[n:1~\]

    B)  \[{{n}^{2}}:1\]

    C)  \[{{n}^{3}}:1\]

    D)  \[{{n}^{4}}:1\]  

    Correct Answer: A

    Solution :

    Power of motor initially  \[={{P}_{0}}\] Let, rate of flow of motor \[=(x)\] Since, power, \[{{P}_{0}}=\frac{work}{time}=\frac{mgy}{t}\] \[=mg\left( \frac{y}{t} \right),\] \[\frac{y}{t}=x=\]rate of flow of water \[=mgx\] ...(i) If rate of flow of water is increased by n times, ie, (nx). Increased power \[{{P}_{1}}=\frac{mgy}{t}\] \[=mg\left( \frac{y}{t} \right)\] \[=mgn.x\] \[=nmgx\] ?..(ii) The ratio of power \[\frac{{{P}_{1}}}{{{P}_{0}}}=\frac{nmgx}{mgx}\] \[\frac{{{P}_{1}}}{{{p}_{0}}}=\frac{n}{1}\] \[\Rightarrow \] \[{{P}_{1}}:{{P}_{0}}=n:1\]


You need to login to perform this action.
You will be redirected in 3 sec spinner