When a $8 \mathrm{~m}$ long wire is stretched by a load of $10 \mathrm{~kg}-\mathrm{wt}$, it is elongated by…
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 $\left(g=10 \mathrm{~ms}^{-2}\right)$
$7.5 \mathrm{~J}$
$0.05 \mathrm{~J}$
$5 \mathrm{~J}$
$0.075 \mathrm{~J}$
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}
$