Question

If the work done is stretching a wire by $1mm$ is $W$. Then, the work required to stretch another wire of same material but with half the radius of cross-section and double the length by $2mm$ is.

• A. $4W$
• B. $W$
• C. $\frac{1}{4}W$
• D. $\frac{1}{2}W$

Answer 1

The correct option is D $\frac{1}{2}W$

Given,

$W_1=W△l_1=1mm={10}^{-3}m{r}_2=\frac{r_1}{2}{l}_2=2l_1△l_2=2mm=2\times {10}^{-3}mm$

We know that

$W=\frac{1}{2}\times F\times △l$

Here we have substitute $F$

$F=\frac{YA△l}{l}$

We can write

$W\propto \frac{A△l\times △l}{l}$

$\frac{W_1}{W_2}=\frac{A_1{\left(△l_1\right)}^2}{l_1}\times \frac{l_2}{A_2{\left(△l_2\right)}^2}=\frac{{\left(r_1\right)}^2{\left(△l_1\right)}^2}{l_1}\times \frac{l_2}{{\left(r_2\right)}^2{\left(△l_2\right)}^2}$

Substitute all value in above equation

$\frac{W_1}{W_2}={\left(\frac{r_1}{r_2}\right)}^2\times {\left(\frac{△l_1}{△l_2}\right)}^2\times \left(\frac{l_2}{l_1}\right)=2^2\times {\left(\frac{1}{2}\right)}^2\times 2=4\times \frac{1}{4}\times 2W_2=\frac{1}{2}W$

Correct option is D