Question

# The linear density of a rod of length 3m varies as $$\lambda=2 + x$$, then the position of the centre of mass of the rod is

A
712m
B
513m
C
127m
D
144m

Solution

## The correct option is C $$\dfrac{12}{7}m$$Linear density of rod varies with distance$$\dfrac{dm}{dx}=\lambda$$$$x_{cm}=\dfrac{\int dm \times x}{\int dm}$$$$=\dfrac {\int_{0}^{3}(\lambda\, dx)x}{\int_{0}^{3}\lambda \, dx}=\dfrac {\int_{0}^{3}(2+x)xdx)}{\int_{0}^{3}(2+x)dx}$$$$=\dfrac {\left [ x^2+\dfrac{x^3}{3} \right ]^3_0}{\left [ 2x+\dfrac{x^3}{3} \right ]^3_0}$$$$=\dfrac {9+9}{6+\dfrac{9}{2}}=\dfrac{36}{21}=\dfrac{12}{7}m$$Physics

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