TS EAMCET · Physics · Mechanical Properties of Solids
The length of a metal wire is found to be \(L_1\) and \(L_2\) when the tension of \(T_1\) and \(T_2\) are applied to it respectively. The natural length of the wire is
- A \(\frac{L_1 T_1+L_2 T_2}{T_2+T_1}\)
- B \(\frac{L_1+L_2}{2}\)
- C \(\frac{L_1 T_2+L_2 T_1}{T_2+T_1}\)
- D \(\frac{L_1 T_2-L_2 T_1}{T_2-T_1}\)
Answer & Solution
Correct Answer
(D) \(\frac{L_1 T_2-L_2 T_1}{T_2-T_1}\)
Step-by-step Solution
Detailed explanation
As \(Y=\) Young's modulus of material \(=\frac{F / A}{\Delta L / L_0}\) \(\Rightarrow \quad \Delta L=\frac{L_0 F}{Y A} \Rightarrow L-L_0=\frac{L_0 F}{Y A}\) \(\Rightarrow \quad L=L_0+\frac{L_0 F}{Y A}\) Here, \(\quad L_1=L_0+\frac{L_0 T_1}{Y A}\) \(\ldots\) (i) and…
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