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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
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B
513m
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C
127m
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D
144m
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Solution

The correct option is C $$\dfrac{12}{7}m$$
Linear density of rod varies with distance
$$\dfrac{dm}{dx}=\lambda$$
$$x_{cm}=\dfrac{\int dm \times x}{\int dm}$$
$$=\dfrac {\int_{0}^{3}(\lambda\, dx)x}{\int_{0}^{3}\lambda \, dx}=\dfrac {\int_{0}^{3}(2+x)xdx)}{\int_{0}^{3}(2+x)dx}$$
$$=\dfrac {\left [ x^2+\dfrac{x^3}{3} \right ]^3_0}{\left [ 2x+\dfrac{x^3}{3} \right ]^3_0}$$
$$=\dfrac {9+9}{6+\dfrac{9}{2}}=\dfrac{36}{21}=\dfrac{12}{7}m$$
741201_681269_ans_2e20fd0d3a8e49649a77b6c93941048a.PNG

Physics

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