A rod of length has non – uniform linear mass density given by λ(x)=a+bx3L where a and b are constants and 0≤x≤L. The value of x for the centre of mass of the rod is at

xc=∫dmx∫dm=∫0L(a+bx3L)dx x∫0L(a+bx3L)dx

A rod of length has non – uniform linear mass density given by λ(x)=a+bx3L where a and b are constants and 0≤x≤L. The value of x for the centre of mass of the rod is at

xc=∫dmx∫dm=∫0L(a+bx3L)dx x∫0L(a+bx3L)dx

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A rod of length has non – uniform linear mass density given by $λ (x) = a + \frac{b x^{3}}{L}$ where a and b are constants and $0 \leq x \leq L .$ The value of $x$ for the centre of mass of the rod is at

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$\frac{3}{2} (\frac{a + b}{2 a + b}) L$

$\frac{3}{4} (\frac{2 a + b}{3 a + b}) L$

$\frac{2}{5} (\frac{5 a + 2 b L^{2}}{4 a + b L^{2}}) L$

$\frac{3}{2} (\frac{2 a + b}{3 a + b}) L$

answer is C.

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Detailed Solution

$x_{c} = \frac{\int d m x}{\int d m} = \frac{\int_{0}^{L} (a + \frac{b x^{3}}{L}) d x x}{\int_{0}^{L} (a + \frac{b x^{3}}{L}) d x}$

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A rod of length has non – uniform linear mass density given by λ(x)=a+bx3L where a and b are constants and 0≤x≤L. The value of x for the centre of mass of the rod is at

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A rod of length has non – uniform linear mass density given by $λ (x) = a + \frac{b x^{3}}{L}$ where a and b are constants and $0 \leq x \leq L .$ The value of $x$ for the centre of mass of the rod is at