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A rod of length is $3 \mathrm{~m}$ and its mass acting per unit length is directly proportional to distance $x$ from one of its end, then its centre of gravity from that end will be at:
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Verified Answer
The correct answer is:
$2 \mathrm{~m}$
Here $r=k x$ where
$k=$ constant mass of small element of $d x$ length is
$\begin{aligned}
& d m=k x \cdot d x \\
& \mathrm{X}_{\mathrm{CM}}=\frac{\int x d m}{\int d m}=\frac{\int_0^3 x(x d x)}{\int_0^3 x d x} \\
&=\frac{\left[\frac{x^3}{3}\right]_0^3}{\left[\frac{x^2}{2}\right]_0^3}=\frac{\frac{27}{3}}{\frac{9}{2}}=2
\end{aligned}$
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