A) \[\frac{3mgl}{\pi {{r}^{2}}Y}\]
B) \[\frac{2mgl}{3\pi {{r}^{2}}Y}\]
C) \[\frac{3mgl}{2\pi {{r}^{2}}Y}\]
D) \[\frac{3mgl}{4\pi {{r}^{2}}Y}\]
Correct Answer: C
Solution :
Youngs modulus,\[Y=\frac{mgl}{{{a}_{1}}{{l}_{1}}}\] \[\therefore \] \[{{l}_{1}}=\frac{mgl}{Y\pi {{r}^{2}}}\] ... (i) and \[Y=\frac{mg(2l)}{{{a}_{2}}{{l}_{2}}}=\frac{mg(2l)}{\pi {{(2r)}^{2}}{{l}_{2}}}\] or \[{{l}_{2}}=\frac{mgl}{2Y\pi {{r}^{2}}}\] ? (ii) From Eqs. (i) and (ii), we have \[\therefore \] \[{{l}_{1}}+{{l}_{2}}=\frac{mgl}{Y\pi {{r}^{2}}}+\frac{mgl}{2Y\pi {{r}^{2}}}+\frac{3mgl}{2Y\pi {{r}^{2}}}\]
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