The temperature of a thin uniform circular disc, of one metre diameter is increased by $10^{\circ}…
The temperature of a thin uniform circular disc, of one metre diameter is increased by $10^{\circ} \mathrm{C}$. The percentage increase in moment of inertia of the disc about an axis passing through its centre and perpendicular to the circular face : (linear coefficient of expansion $\left.=11 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)$
0.0055
0.011
0.022
0.044
Solution
Increase in area of disc
$\Delta A=A(2 \alpha) \Delta t$
$=\pi(0.5)^2\left(2 \times 11 \times 10^{-6}\right) \times 10$
$=0.000055 \pi$
New area of the disc,
$A^{\prime}=A+\Delta A$
$\therefore \quad A^{\prime}=\pi(0.5)^2+0.000055 \pi$
or $\quad \pi r^{\prime 2}=0.250055 \pi \mathrm{m}^2$
or $\quad r^{\prime}=0.500055 \mathrm{~m}$
Increase in moment of inertia.
$\frac{I^{\prime}-I}{I}=\frac{(0.500055)^2-(0.5)^2}{(0.5)^2}$
$=0.00022$
$=0.022 \%$