Question

Two wires $A$ and $B$ have equal lengths and are made of the same material, but the diameter of wire $A$ is twice that of wire $B$. Then, for a given load:

• A. the extension of $B$ will be four times that of $A$
• B. the extensions of $A$ and $B$ will be equal
• C. the strain in $B$ is four times that in $A$
• D. the strains in $A$ and $B$ will be equal

Answer 1

Answer: (A), (C)

The correct options are
A the extension of $B$ will be four times that of $A$
C the strain in $B$ is four times that in $A$

$Y=\frac{F/\pi {\left(\frac{d}{2}\right)}^2}{\Delta l/l}$

$Y=\frac{F/\pi {\left(\frac{d_A}{2}\right)}^2}{\Delta l_A/l}$-----(1)

$Y=\frac{F/\pi {\left(\frac{d_B}{2}\right)}^2}{\Delta l_B/l}$-----(2)

Dividing (1) by (2)

$1=\frac{\pi {\left(\frac{d_B}{2}\right)}^2}{\pi {\left(\frac{d_A}{2}\right)}^2}\times \frac{\Delta l_B}{\Delta l_A}$

$1={\left(\frac{d_B}{d_A}\right)}^2\times \frac{\Delta l_B}{\Delta l_A}$

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

$\frac{\Delta l_A}{\Delta l_B}=\frac{1}{4}$

strain$=\frac{\Delta l}{l}$

So, extension in B will be four times of A and strain in B will be four times of A since ${\Delta }_B=4{\Delta }_A$ and length of A and B are same.