AP EAMCET · PHYSICS · Mechanical Properties of Solids
The length of a metal wire is \(l_1\) when the tension in it is \(F_1\) and \(l_2\) when the tension is \(F_2\). Then, original length of the wire is
- A \(\frac{l_1 F_1+l_2 F_2}{F_1+F_2}\)
- B \(\frac{l_2-l_1}{F_2-F_1}\)
- C \(\frac{l_1 F_2-l_2 F_1}{F_2-F_1}\)
- D \(\frac{l_1 F_1-l_2 F_2}{F_2-F_1}\)
Answer & Solution
Correct Answer
(C) \(\frac{l_1 F_2-l_2 F_1}{F_2-F_1}\)
Step-by-step Solution
Detailed explanation
We have, \(F_1 \propto\) \(\left(l_1-l\right)\), where \(\ell=\) original length Similarly, \(F_2 \propto \quad\left(l_2-l\right)\), The ratio, \(\frac{F_1}{F_2}=\frac{l_1-l}{l_2-l}\) Solving, we get \(\Rightarrow \quad l=\frac{F_2 l_1-F_1 l_2}{F_2-F_1}\)
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