Physics Elasticity Question 123
Question: A steel ring of radius r and cross-section area ?A? is fitted on to a wooden disc of radius $ R(R>r) $ . If Young’s modulus be E, then the force with which the steel ring is expanded is
[EAMCET 1986]
Options:
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} $
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Answer:
Correct Answer: B
Solution:
Initial length (circumference) of the ring = 2pr Final length (circumference) of the ring = 2pR Change in length = 2pR ? 2pr. $ r=\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) $
BETA
