JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2022
- Hard
- MCQ
A wire of length \(L\) is hanging from a fixed support. The length changes to \(L _{1}\) and \(L _{2}\) when masses \(1 \,kg\) and \(2 \,kg\) are suspended respectively from its free end. Then the value of \(L\) is equal to ..................
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Answer & Solution
Correct Answer
(C) \(2 L _{1}- L _{2}\)
Step-by-step Solution
Detailed explanation
By Hooke's Law
so \(F \alpha \Delta L\)
\(\frac{ F _{1}}{ F _{2}}=\frac{\Delta L _{1}}{\Delta L _{2}}\)
\(\frac{10}{20}=\frac{\left(L_{1}-L\right)}{\left(L_{2}-L\right)}\)
\(L =2 L _{1}- L _{2}\)
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