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The moment of inertia of a meter scale of mass $0.6 \mathrm{~kg}$ about an axis perpendicular to the scale and located at the $20 \mathrm{~cm}$ position on the scale in $\mathrm{kg}-\mathrm{m}^2$ is :
(Breadth of the scale is negligible)
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(Breadth of the scale is negligible)
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Verified Answer
The correct answer is:
0.078
$m=0.6 \mathrm{~kg}$
Mass per unit length $=\frac{0.6}{100} \mathrm{~kg} / \mathrm{cm}$
Mass of part $A B, m_1=\frac{0.6}{100} \times 20=\frac{0.6}{5} \mathrm{~kg}$
Mass of part $B C, m_2=\frac{0.6}{100} \times 80=\frac{0.6 \times 4}{5}=\frac{2.4}{5} \mathrm{~kg}$
Moment of inertia,
$I=m_1\left(\frac{A B}{2}\right)^2+m_2 \frac{(B C)^2}{2}$
$=\frac{0.6}{5} \times\left(\frac{20}{2} \times 10^{-2}\right)^2+\frac{2.4}{5} \times\left(\frac{80}{2} \times 10^{-2}\right)^2$
$=\frac{0.6}{5} \times 10^{-2}+\frac{2.4}{5} \times\left(4 \times 10^{-1}\right)^2$
$=\frac{0.6}{5} \times 10^{-2}+\frac{2.4}{5} \times 16 \times 10^{-2}$
$=\left(\frac{0.6+38.4}{5}\right) \times 10^{-2}$
$=7.8 \times 10^{-2}$
$=0.078 \mathrm{~kg}-\mathrm{m}^2$
Mass per unit length $=\frac{0.6}{100} \mathrm{~kg} / \mathrm{cm}$
Mass of part $A B, m_1=\frac{0.6}{100} \times 20=\frac{0.6}{5} \mathrm{~kg}$
Mass of part $B C, m_2=\frac{0.6}{100} \times 80=\frac{0.6 \times 4}{5}=\frac{2.4}{5} \mathrm{~kg}$
Moment of inertia,
$I=m_1\left(\frac{A B}{2}\right)^2+m_2 \frac{(B C)^2}{2}$
$=\frac{0.6}{5} \times\left(\frac{20}{2} \times 10^{-2}\right)^2+\frac{2.4}{5} \times\left(\frac{80}{2} \times 10^{-2}\right)^2$
$=\frac{0.6}{5} \times 10^{-2}+\frac{2.4}{5} \times\left(4 \times 10^{-1}\right)^2$
$=\frac{0.6}{5} \times 10^{-2}+\frac{2.4}{5} \times 16 \times 10^{-2}$
$=\left(\frac{0.6+38.4}{5}\right) \times 10^{-2}$
$=7.8 \times 10^{-2}$
$=0.078 \mathrm{~kg}-\mathrm{m}^2$
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