Question

The radii and Young's moduli of two uniform wires $A$ and $B$ are in the ratio $2:1$ and $1:2$ respectively. Both wires are subjected to the same longitudinal force. If the increase in length of the wire $A$ is one percent, the percentage increase in length of the wire $B$ is:

• A. $1.0$
• B. $1.5$
• C. $2.0$
• D. $3.0$

Answer 1

Answer: (C)

$2.0$

Given,

$r_A:r_B=2:1$

$Y_A:Y_B=1:2$

$\frac{\Delta l_A}{l_A}\times 100=1$%

$\frac{\Delta l_B}{l_B}\times =?$

Young's modulus of the wire $A$ is given by

$Y_A=\frac{F/\pi r_A^2}{\frac{\Delta l_A}{l_A}}$. . . . .(1)

Young's modulus of the wire $B$ is given by

$Y_B=\frac{F/\pi r_B^2}{\frac{\Delta l_B}{l_B}}$. . . . . .(2)

Divide equation (1) by (2), we get

$\frac{Y_A}{Y_B}=\frac{\frac{F/\pi r_A^2}{\Delta l_A/l_A}}{\frac{F/\pi r_B^2}{\Delta l_B/l_B}}$

$\frac{Y_A}{Y_B}=(\frac{r_B}{r_A}{)}^2\times \frac{l_A}{\Delta l_A}\times \frac{\Delta l_B}{l_B}$

$\frac{1}{2}=(\frac{1}{2}{)}^2\times (\frac{l_A}{\Delta l_A})\times (\frac{\Delta l_B}{l_B})$

$\frac{\Delta l_B}{l_B}=2\frac{\Delta l_A}{l_A}$

$\frac{\Delta l_B}{l_B}=2\times 1$

$\frac{\Delta l_B}{l_B}=2$%

The correct option is C.