Question

A wire of length \(100\text{ cm}\) is made of a material of Young’s modulus \(1.6\times10^{11}\text{ Nm}^{-2}\). If work done in stretching this wire by \(0.1\text{ cm}\) is \(2\text{ J}\), then the area of cross-section of the wire (in \(10^{-5}\text{ m}^2\)) is

• A. \(5.0\)
• B. \(1.25\)
• C. \(1.5\)
• D. \(2.5\)

Hint:

Memorize elastic potential energy formula: \[ W=\frac12\frac{YA(\Delta L)^2}{L} \] This directly solves most stretching problems.

Answer 1

The Correct Option is D

Step 1: Work done stretching a wire.
The elastic energy stored is $W=\dfrac12 F\,\Delta L$, and from Young's modulus $Y=\dfrac{FL}{A\,\Delta L}$ so $F=\dfrac{YA\,\Delta L}{L}$.
Step 2: Combine into one formula.
Substituting $F$, $W=\dfrac12\dfrac{YA(\Delta L)^2}{L}$.
Step 3: Convert the given data.
$L=100$ cm $=1$ m, $\Delta L=0.1$ cm $=10^{-3}$ m, $Y=1.6\times10^{11}$ N/m$^2$, $W=2$ J.
Step 4: Plug in.
$2=\dfrac12\cdot\dfrac{(1.6\times10^{11})A(10^{-3})^2}{1}=\dfrac12(1.6\times10^{11})(10^{-6})A$.
Step 5: Simplify.
$2=\dfrac12(1.6\times10^{5})A=0.8\times10^{5}A$.
Step 6: Solve for the area.
$A=\dfrac{2}{0.8\times10^{5}}=2.5\times10^{-5}$ m$^2$, that is $2.5$ in units of $10^{-5}$ m$^2$.
\[ \boxed{2.5} \]