# A Metal Ring Of Initial Radius R And Cross-Sectional Area A Is Fitted Onto A Wooden Disc Of Radius R > R. If Young's Modulus Of Metal Is Y Then Tension In The Ring Is:

Given:

The metal ring of radius = r

The cross-sectional area = A

Iinitial = 2Ï€ r

Ifinal = 2Ï€ R

Î”Â l = change in length = 2 Ï€ (R – r)

Thermal strain = $$\frac{\Delta l}{l_{i}} = \frac{R – \gamma}{r}$$

Therefore, F = yA $$(\frac{\Delta l}{l_{i}}) = yA (\frac{R-r}{r})$$

Therefore, the tension in the ring is yA $$(\frac{R-\gamma}{r})$$

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