A uniform circular ring of radius ‘R’ density $ρ$ and young’s modulus ‘Y’ is placed on a smooth horizontal surface. The ring is rotated with a constant angular velocity $ω$ about is own axis is shown. The increase in the radius of the ring is (assume increment in radius is very small as compared to radius)

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$\frac{ρ ω^{2} R^{3}}{2 Y}$

$\frac{ρ ω^{2} R^{3}}{Y}$

$\frac{2 ρ ω^{2} R^{3}}{Y}$

$\frac{3 ρ ω^{2} R^{3}}{2 Y}$

answer is B.

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Detailed Solution

$F d θ = d m ω^{2} R$

$F d θ = (ρ A R d θ) ω^{2} R$

$∴$ stress, $F / A = ρ ω^{2} R^{2}$

Now, $\frac{F}{A} = Y \frac{Δ l}{l} = Y \frac{Δ R}{R}$

$[∵ l = 2 π R, Δ l = 2 π Δ R]$

$ρ ω^{2} R^{2} = Y \frac{Δ R}{R}$

$∴ Δ R = \frac{ρ ω^{2} R^{3}}{Y}$

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