Question

If power is increased by $50\% $ and work done is constant, then time taken is reduced by

Answer 1

Hint
The power is given by the work done per unit time. So in the first case we use the formula of power and in the second case we use the same formula to find the new power for different time and constant work. On comparing the two cases we can find the time taken in the second case.
Formula Used: In this solution, we will be using the following formula
$P = \dfrac{W}{t}$
where $P$ is the power
 $W$ is the work done and
$t$ is the time taken to perform the work.

Complete step by step answer
The power of a system in physics refers to the rate of work done by the system. So it can be calculated by dividing the work done by a system by the time taken to do the work. Therefore we can write the formula as,
$P = \dfrac{W}{t}$
Now the work done is taken constant in both the cases, so we take the work done for both the cases as $W$. We consider the power in the first case as ${P_1}$ and the power in the second case as ${P_2}$. And let the time taken in the first case be ${t_1}$ and in the second case be ${t_2}$. Now according to the formula,
${P_1} = \dfrac{W}{{{t_1}}}$ is the first case and ${P_2} = \dfrac{W}{{{t_2}}}$ is the second case.
Now taking the ratio of the powers in the first and the second case, we get
$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{W \mathord{\left/
 {\vphantom {W {{t_1}}}} \right.} {{t_1}}}}}{{{W \mathord{\left/
 {\vphantom {W {{t_2}}}} \right.} {{t_2}}}}}$
Now by cancelling the work $W$ in both the cases, we get
$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{t_2}}}{{{t_1}}}$
Now in the question it is given that the power is increased by $50\% $. So we get the power in the second case as,
${P_2} = {P_1} + \dfrac{{50}}{{100}}{P_1}$
This is equal to,
${P_2} = \dfrac{{15}}{{10}}{P_1}$
So on substituting this value of ${P_2}$ is the above equation we get,
$\dfrac{{{P_1}}}{{\dfrac{{15}}{{10}}{P_1}}} = \dfrac{{{t_2}}}{{{t_1}}}$
On cancelling ${P_1}$ from both the numerator and denominator, we get
$\dfrac{{10}}{{15}} = \dfrac{{{t_2}}}{{{t_1}}}$
The change is the time is given by,
$t\% = \dfrac{{{t_1} - {t_2}}}{{{t_1}}} \times 100\% = \left( {1 - \dfrac{{{t_2}}}{{{t_1}}}} \right) \times 100\% $
Now substituting the calculated value of $\dfrac{{{t_2}}}{{{t_1}}}$ we get
$t\% = \left( {1 - \dfrac{{10}}{{15}}} \right) \times 100\% $
On doing the calculation in the bracket we get,
$t\% = \left( {\dfrac{{15 - 10}}{{15}}} \right) \times 100\% = \dfrac{5}{{15}} \times 100\% $
On doing the above calculation, we get the percentage reduction in time as,
$t\% = 33.33\% $
Hence when the power is increased by $50\% $ then time gets reduced by $33.33\% $.

Note
Here we have taken the change in time as ${t_1} - {t_2}$ but not ${t_2} - {t_1}$ since the time is reduced and the time in the second case will be less than the time in the first case. The SI unit of power is given by watt which is joule per second.