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Q.

A thin circular metal disc of radius 500mm is set rotating about a central axis normal to its plane upon raising its temperature gradually, the radius increases to 507.5 mm, the percen-tage change in the rotational KE will be–––

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a

-3%

b

-1.5%

c

1.5%

d

3%

answer is D.

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Detailed Solution

Kinetic energy, $\large K=\frac {L^2}{2I}$ For a disc $\large I=\frac 12mr^2$
$\large \therefore K=\frac {L^2}{2.\frac 12mr^2}=\frac {L^2}{mr^2}$
Since external torque is zero, angular momentum of the disc remains conserved.
$\large \therefore \frac {\Delta k}{k}=-2.\frac {\Delta r}{r}$
$\large \therefore \frac {\Delta k}{k}\times 100%=-2\times \left ( \frac {7.5}{500} \times 100\right )%$
$\large =-3%$

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