JEE Main 2015 · 11 Apr · Shift 1 · Physics Q5

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PhysicsSingle correct+4 / -1Question 5 of 90

A uniform thin rod AB of length L has linear mass density $\mu(x)=a+\dfrac{bx}L$ , where $x$ is measured from A. If the CM of the rod lies at a distance of $\left(\dfrac7{12}L\right)$ from A, then a and b are related as :

a = b

a = 2b

2a = b

3a = 2b

Answer and worked solution

Official answer: C

$x_{cm}=\frac{\int_0^Lx\left(a+\frac{bx}L\right)dx}{\int_0^L\left(a+\frac{bx}L\right)dx}=L\,\frac{\frac a2+\frac b3}{a+\frac b2}=\frac{7L}{12}.$
So $6a+4b=7a+\frac72b$ , i.e. $b=2a$ .

Topics:Centre of mass

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 5