JEE Main 2014 · 11 Apr · Shift 1 · Physics Q6

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PhysicsSingle correct+4 / -1Question 6 of 90

A thin bar of length L has a mass per unit length $\lambda$ , that increases linearly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is $\lambda_o$ , then the distance of the centre of mass from the lighter end is :

\dfrac L2-\dfrac{\lambda_oL^2}{4M}

\dfrac L3+\dfrac{\lambda_oL^2}{8M}

\dfrac L3+\dfrac{\lambda_oL^2}{4M}

\dfrac{2L}3-\dfrac{\lambda_oL^2}{6M}

Answer and worked solution

Official answer: D

Let $\lambda=\lambda_o+kx$ . The total mass is $M=\lambda_oL+\frac12kL^2$ , so $k=\dfrac{2(M-\lambda_oL)}{L^2}$ . Then
$x_{cm}=\frac1M\int_0^Lx(\lambda_o+kx)\,dx=\frac{\lambda_oL^2/2+kL^3/3}{M}=\frac{2L}3-\frac{\lambda_oL^2}{6M}.$

Topics:Centre of mass

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 6