JEE Mains · Physics · STD 11- 8. mechanical properties of solids
Two wires of same length and radius are joined end to end and loaded. The Young's modulii of the materials of the two wires are \(Y_{1}\) and \(Y_{2}\). The combination behaves as a single wire then its Young's modulus is:
- A \(y=\frac{Y_{1} Y_{2}}{Y_{1}+Y_{2}}\)
- B \(y=\frac{2 Y_{1} Y_{2}}{3\left(Y_{1}+Y_{2}\right)}\)
- C \(Y=\frac{2 Y_{1} Y_{2}}{Y_{1}+Y_{2}}\)
- D \({Y}=\frac{{Y}_{1} {Y}_{2}}{2\left({Y}_{1}+{Y}_{2}\right)}\)
Answer & Solution
Correct Answer
(C) \(Y=\frac{2 Y_{1} Y_{2}}{Y_{1}+Y_{2}}\)
Step-by-step Solution
Detailed explanation
In series combination \(\Delta l =l_{1}+l_{2}\) \(Y =\frac{ F / A }{\Delta l / l} \Rightarrow \Delta l=\frac{ F l}{ AY }\) \(\Rightarrow \Delta l \propto \frac{l}{ Y }\) Equivalent length of rod after joining is \(=2 l\) As, lengths are same and force is also same in series…
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