Question

Figure shows the graph of elastic potential energy $\left(U\right)$ stored versus extension, for a steel wire $(Y=2\times {10}^{11}\text{Pa})$ of volume $200\text{cc}$. If area of cross-section $\left(A\right)$ and original length $\left(L\right)$, then

• A. $A={10}^{-4}{\text{m}}^2$
• B. $A={10}^{-3}{\text{m}}^2$
• C. $L=1.5\text{m}$
• D. $L=2\text{m}$

Answer 1

Answer: (A), (D)

$L=2\text{m}$

We can assume, elastic potential energy $\left(U\right)$ equal to $\frac{1}{2}k{x}^2$ as curve is a parabola symmetric about $\text{Y-axis}$ and open upward.

So, $U=\frac{1}{2}k{x}^2......\left(1\right)$

Where, $k=\frac{YA}{L}$, from analogue of spring and solid.

$x=$ Extention in the body.

Now, from the graph at $x=0.2\text{mm}$, $U=0.2\text{J}$

Using above point and equation $\left(1\right)$, we get,

$0.2=\frac{1}{2}\times k(0.2\times {10}^{-3}{)}^2$

$\Rightarrow k={10}^7\text{N/m}$

Also, from analogue of spring and solid,

$k=\frac{YA}{L}$

$\Rightarrow {10}^7=\frac{2\times {10}^{11}\times A}{L}$

$\Rightarrow A=5\times {10}^{-5}L......\left(2\right)$

Further, we know that,

$\text{volume= Area}\times \text{Length}$

Putting the values in this equation,
$\Rightarrow 200\times {10}^{-6}=A\times L...\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$,
$\Rightarrow 2\times {10}^{-4}=5\times {10}^{-5}\times L^2$

$\Rightarrow L^2=4$

$\therefore L=2\text{m}$

Now, from $\left(2\right)$,

$A=5\times {10}^{-5}\times 2$

$\therefore A={10}^{-4}{\text{m}}^2$

Hence, option $\left(a\right)$ and $\left(d\right)$ are correct answers.