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:


The metal ring of radius = r

The cross-sectional area = A

Iinitial = 2π r

Ifinal = 2π R

Δ l = change in length = 2 π (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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