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A metal ring of inner diameter $D_{1}$ and cross-sectional area 'A' is fitted onto a wooden disc of diameter $D_{2} . (D_{2} > D_{1})$ . If 'Y' is the young's modulus of the metal then the tension in the ring is

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$\frac{{AYD}_{2}}{D_{1}}$

$\frac{{YD}_{1}}{{AD}_{2}}$

$\frac{{Y(D}_{2} - D_{1})}{{AD}_{1}}$

$\frac{{AY(D}_{2} - D_{1})}{D_{1}}$

answer is B.

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Detailed Solution

$T = \frac{y A e}{L}$

$T = \frac{y A \frac{(D_{2} - D_{1})}{2}}{\frac{D_{1}}{2}}$

$T = \frac{y A (D_{2} - D_{1})}{D_{1}}$

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