Question

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

• A. 66.66 %
• B. 33.33 %
• C. 50 %
• D. remains constant

Answer 1

The correct option is B

33.33 %

Let the initial power be $P_1$ and the final power be $P_2$.
Let the initial and final time be $t_1$ and $t_2$ respectively.

$\Rightarrow P_2=P_1+50\%\times P_1=P_1+\frac{P_1}{2}=\frac{3P_1}{2}$
$\Rightarrow P_2=\frac{3P_1}{2}$

We know that, Power $=\frac{\text{Work done}}{Time}\Rightarrow P=\frac{W}{t}$
$\therefore W=P\times t$
If work done is constant, $P_1{t}_1=P_2{t}_2$
$\Rightarrow \frac{t_2}{t_1}=\frac{P_1}{P_2}=\frac{2}{3}$
$\Rightarrow t_2=\frac{2}{3}{t}_1$

Decrease in time $=t_1-t_2=t_1-\frac{2}{3}{t}_1=\frac{t_1}{3}$
% Decrease in time $=\frac{\text{Decrease in time}}{Initialtime}\times 100$
$=\frac{\frac{t_1}{3}}{t_1}\times 100=33.33\%$