Q136 / 160 · AP EAMCET 2021 (24 Aug Shift 1) · Physics
When a $8 \mathrm{~m}$ long wire is stretched by a load of $10 \mathrm{~kg}-\mathrm{wt}$, it is elongated by $1.5 \mathrm{~mm}$. The energy stored in the wire in this process is $\left(g=10 \mathrm{~ms}^{-2}\right)$
- A$7.5 \mathrm{~J}$
- B$0.05 \mathrm{~J}$
- C$5 \mathrm{~J}$
- Correct answerD$0.075 \mathrm{~J}$
Solution
Given that, weight of load, $w=10 \mathrm{~kg}-\mathrm{wt}$
$$
F=w=10 \times 10 \mathrm{~N}=100 \mathrm{~N}
$$
Elongation in wire, $\Delta l=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m}$
Length of wire, $l=8 \mathrm{~m}$
We know that, elastic potential energy stored in wire,
$$
U=\frac{1}{2} F \Delta l=\frac{1}{2} \times 100 \times 1.5 \times 10^{-3}=0.075 \mathrm{~J}
$$
Hence, the correct answer is option: 'D'.