A wire of length 6 m is stretched such that its radius is reduced by 20%. Which one of the following is the value of increase in its length?
[amp_mcq option1=”50%” option2=”56.25%” option3=”62.25%” option4=”75%” correct=”option2″]
This question was previously asked in
UPSC CAPF – 2020
The wire is stretched such that its radius is reduced by 20%. The new radius is $R_2 = R_1 – 0.20R_1 = 0.80R_1$.
Let the new length be $L_2$. The new volume is $V_2 = \pi R_2^2 L_2 = \pi (0.80R_1)^2 L_2 = \pi (0.64R_1^2) L_2$.
Since the volume remains constant, $V_1 = V_2$:
$6\pi R_1^2 = \pi (0.64R_1^2) L_2$
Assuming $R_1 > 0$, we can cancel $\pi R_1^2$:
$6 = 0.64 L_2$
$L_2 = \frac{6}{0.64} = \frac{600}{64} = \frac{75}{8} = 9.375$ m.
The increase in length is $L_2 – L_1 = 9.375 – 6 = 3.375$ m.
The percentage increase in length is $\frac{\text{Increase in length}}{\text{Original length}} \times 100\% = \frac{3.375}{6} \times 100\%$.
$\frac{3.375}{6} = \frac{3375}{6000} = \frac{675}{1200} = \frac{135}{240} = \frac{27}{48} = \frac{9}{16}$.
Percentage increase $= \frac{9}{16} \times 100\% = 0.5625 \times 100\% = 56.25\%$.