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  4. A wire, suspended vertically from one of its ends, is stretched by attaching a weight of 200N to the lower end. The weight stretches the wire by 1mm . If the elastic energy (in J ) stored in the wire is U , What is the value of (1/U) ?

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Q. A wire, suspended vertically from one of its ends, is stretched by attaching a weight of $200N$ to the lower end. The weight stretches the wire by $1mm$ . If the elastic energy (in $J$ ) stored in the wire is $U$ , What is the value of $\frac{1}{U}$ ?

NTA AbhyasNTA Abhyas 2022

Solution:

Elastic potential energy, $U=\frac{1}{2}\times F\times l=\frac{1}{2}\times 200\times 10^{- 3}=\text{0.1}J$
Therefore, $\frac{1}{U}=10$ .