Q145 / 160 · TS EAMCET 2014 · Physics
Two wires of the same material and length but diameters in the ratio $1: 2$ are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio.
- Correct answerA$16: 1$
- B$1: 1$
- C$2: 1$
- D$4: 1$
Solution
We know that,
Elastic potential energy in a stretched wire
$$
\begin{aligned}
U & =\frac{1}{2} \text { Young's modulus } \times(\text { strain })^2 \\
U & =\frac{1}{2} \times \frac{F}{A} \times \frac{1}{I} \times \frac{F l}{Y \cdot A} \\
U & =\frac{1}{2} \frac{F^2}{A^2} \cdot \frac{1}{Y} \\
\Rightarrow U & \propto \frac{1}{r^4}
\end{aligned}
$$
So,
$$
\begin{aligned}
\frac{U_1}{U_2} & =\frac{1 / r_1^4}{1 / r_2^4}=\left(\frac{r_2}{r_1}\right)^4 \quad\left[\because \frac{d_1}{d_2}=\frac{r_1}{r_2}=\frac{1}{2}\right] \\
\frac{U_1}{U_2} & =\left(\frac{2}{1}\right)^2=\frac{16}{1} \\
U_1: U_2 & =16: 1
\end{aligned}
$$
Hence, the correct answer is option: 'A'.