Question

Two elastic wire $A\&B$ having length ${\ell }_A=2m$ and ${\ell }_B=1.5m$ and the ratio of young's modules $Y_A:Y_B$ is $7:4$. If radius of wire $B\left(r_B\right)$ is $2mm$ then choose the correct value of radius of wire $A$. Given that due to application of the same force charge in length in both $A\&B$ is same

• A. $1.7mm$
• B. $1.9mm$
• C. $2.7mm$
• D. $2mm$

Answer 1

Answer: (A)

$1.7mm$

$Y=\frac{F\ell }{\pi r^2\Delta \ell }$
$\frac{Y_A}{Y_B}=\frac{F{\ell }_A}{\pi r_A^2\Delta \ell }\times \frac{\pi r_B^2\Delta \ell }{F{\ell }_B}$
$\frac{7}{4}=\frac{2}{1.5}\times \frac{2^2}{r_A^2}\Rightarrow r_A^2=\frac{4\times 2\times 2^2}{1.5\times 7}$
$r_A=1.7mm$