Question

Two bars $A$ and $B$ of circular section and of the same volume and made of the same material are subjected to tension. If the diameter of $A$ is half that of $B$ and if the force supplied to both the rods is the same and is within the elastic limit, the ratio of the extension of $A$ to that of $B$ will be

• A. $16$
• B. $8$
• C. $4$
• D. $2$

Answer 1

The correct option is A $16$

Using
$Y=\frac{F/A}{△L/L}$
$Y=\frac{F/{△}_1}{△L_1/L_1}-\left(1\right)\text{For bar A}$
$Y=\frac{F/{△}_2}{△L_2/L_2}-\left(2\right)\text{For bar B}$
Volume is same for A and B
$So,R(\frac{d_1}{2}{)}^2{L}_1=R(\frac{d_2}{2}{)}^2{L}_2$
$d_1^2{L}_1=4d_1^2{L}_2(given{d}_1=\frac{1}{2}{d}_2)$
$\frac{L_1}{L_2}=4$
Dividing 1 and 2
$1=\frac{L_1△L_2{A}_2}{L_2△L_1{A}_1}$
$\frac{△L_1}{△L_2}=4\times 4(\because \frac{A_2}{A_1}=(\frac{d_2}{d_1}{)}^2=4)$
$\frac{△L_1}{△L_2}=16$