A) \[AE\frac{R}{r}\]
B) \[AE\left( \frac{R-r}{r} \right)\]
C) \[\frac{E}{A}\left( \frac{R-r}{A} \right)\]
D) \[\frac{Er}{AR}\]
Correct Answer: B
Solution :
Initial length (circumference) of the ring = 2pr Final length (circumference) of the ring = 2pR Change in length = 2pR ? 2pr. \[\text{strain}=\frac{\text{change in length}}{\text{original length}}\]\[=\frac{2\pi (R-r)}{2\pi r}\]\[=\frac{R-r}{r}\] Now Young's modulus \[E=\frac{F/A}{l/L}=\frac{F/A}{(R-r)/r}\] \\[F=AE\left( \frac{R-r}{r} \right)\]You need to login to perform this action.
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