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A uniform solid sphere of copper is rotating about a diameter with an angular speed ‘ $ω$ ’. Its temperature is increased by $80^{\circ} C$ . $α_{c u}$ being the coefficient of linear expansion of copper, new angular speed of the sphere will be

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$ω (1 + 180 α_{c u})$

$\frac{ω}{1 + 80 α_{c u}}$

$\frac{ω}{1 + 160 α_{c u}}$

$\frac{ω}{1 + 180 α_{c u}}$

answer is C.

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Detailed Solution

Using conservation of angular momentum

$I_{1} ω_{1} = I_{2} ω_{2} m k_{1}^{2} ω = m k_{2}^{2} ω_{2} k_{1}^{2} ω_{1} = k_{1}^{2} {(1 + α t)}^{2} ω_{2} ω_{1} = (1 + 2 α Δ t) ω_{2} ω_{2} = \frac{ω}{1 + 160 α_{c u}}$

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