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The masses of $200 \mathrm{~g}$ and $300 \mathrm{~g}$ are attached to the $20 \mathrm{~cm}$ and $70 \mathrm{~cm}$ marks of a light meter rod, respectively. The moment of inertia of the system about an axis passing through $50 \mathrm{~cm}$ mark is
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Verified Answer
The correct answer is:
$0.036 \mathrm{~kg} \mathrm{~m}^{2}$
Given, $1 m_{1}=200 \mathrm{~g}=0.2 \mathrm{~kg}$
$$
\begin{gathered}
m_{2}=300 \mathrm{~g}=0.3 \mathrm{~kg} \\
l_{1}=50-20=30 \mathrm{~cm} \\
\Rightarrow \quad l_{1}=0.3 \mathrm{~m} \\
l_{2}=70-50=20 \mathrm{~cm}=0.8 \mathrm{~m}
\end{gathered}
$$
$\therefore$ Moment of inertia of the system about an axis passing through $50 \mathrm{~cm}$ mark (i.e. mid-point)
$$
\begin{aligned}
&\longleftrightarrow 50 \mathrm{~cm} \longrightarrow+50 \mathrm{~cm} \longrightarrow \\
&+20 \mathrm{~cm}+\frac{14}{\longleftrightarrow}+1-l_{2} \rightarrow \\
&l=m_{1} l_{1}^{2}+m_{2} l_{2}^{2} \\
&=0.2 \times(0.3)^{2}+0.3 \times(0.2)^{2} \\
&=0.036 \mathrm{~kg} \mathrm{~m}^{2}
\end{aligned}
$$
$$
\begin{gathered}
m_{2}=300 \mathrm{~g}=0.3 \mathrm{~kg} \\
l_{1}=50-20=30 \mathrm{~cm} \\
\Rightarrow \quad l_{1}=0.3 \mathrm{~m} \\
l_{2}=70-50=20 \mathrm{~cm}=0.8 \mathrm{~m}
\end{gathered}
$$
$\therefore$ Moment of inertia of the system about an axis passing through $50 \mathrm{~cm}$ mark (i.e. mid-point)
$$
\begin{aligned}
&\longleftrightarrow 50 \mathrm{~cm} \longrightarrow+50 \mathrm{~cm} \longrightarrow \\
&+20 \mathrm{~cm}+\frac{14}{\longleftrightarrow}+1-l_{2} \rightarrow \\
&l=m_{1} l_{1}^{2}+m_{2} l_{2}^{2} \\
&=0.2 \times(0.3)^{2}+0.3 \times(0.2)^{2} \\
&=0.036 \mathrm{~kg} \mathrm{~m}^{2}
\end{aligned}
$$
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