Q137 / 160 · AP EAMCET 2006 · Physics

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 $\left.=11 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)$

  • A
    0.0055
  • B
    0.011
  • Correct answer
    C
    0.022
  • D
    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. $\frac{I^{\prime}-I}{I}=\frac{(0.500055)^2-(0.5)^2}{(0.5)^2}$ $=0.00022$ $=0.022 \%$

Hence, the correct answer is option: 'C'.

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