A) \[l\]
B) \[2l\]
C) \[\frac{1}{2}l\]
D) \[4l\]
Correct Answer: A
Solution :
Young's modulus of elasticity of material of wire \[Y=\frac{FL}{\pi {{r}^{2}}\Delta l}\] \[\therefore \] Increase in length \[\Delta l=\frac{FL}{\pi {{r}^{2}}Y}\] \[\frac{\Delta {{l}_{1}}}{\Delta {{l}_{2}}}=\frac{{{F}_{1}}}{{{F}_{2}}}\times \frac{{{L}_{1}}}{{{L}_{2}}}\times {{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}\] \[\frac{l}{\Delta {{l}_{2}}}=\frac{F}{2F}\times \frac{L}{2L}\times {{\left( \frac{2r}{r} \right)}^{2}}\] So, \[\Delta {{l}_{2}}=l\]You need to login to perform this action.
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