System of Particles and Rotational Motion - Result Question 7
7. If the linear density (mass per unit length) of a rod of length $3 m$ is proportional to $x$, where $x$ is the distance from one end of the rod, the distance of the centre of gravity of the rod from this end is
(a) $2.5 m$
(b) $1 m$
(c) $1.5 m$
(d) $2 m$
[2002]
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Answer:
Correct Answer: 7. (d)
Solution:
- (d) Consider an element of length $d x$ at a distance $x$ from end $A$.
Here, mass per unit length $\lambda$ of rod
$\lambda \propto x \Rightarrow \lambda=k x$
$\therefore d m=\lambda d x=k x d x$
Position of centre of gravity of rod from end $A$.
$ \therefore x _{C G}=\frac{\int_0^{3} x(k x d x)}{\int_0^{3} k x d x}=\frac{[\frac{x^{3}}{3}]_0^{3}}{[\frac{x^{2}}{2}]_0^{3}}=\frac{\frac{(3)^{3}}{3}}{\frac{(3)^{3}}{2}}=2 m $
