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A uniform solid brass cylinder of mass M=0.5 Kg and radius R=0.03m is placed in frictionless bearings and set to rotate about its geometrical axis with an angular velocity of 60 rad/s. After the cylinder has reached the specified state of rotation, it is heated (without any mechanical contact) from room temperature 20°C to 100°C. The fractional change in angular velocity of the cylinder is
$(α = 2 \times 10^{- 5} /^{\circ} C)$

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$- 2.3 \times 10^{- 3}$

$3.2 \times 10^{- 3}$

$- 3 .2 \times 10^{- 3}$

$2.3 \times 10^{- 3}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{- ΔI}{I} = \frac{- 2 ΔR}{R} - 2 α Δt \\ [I \propto R^{2}, \frac{Δ}{I} = \frac{2 ΔR}{R}] \end{array}$

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