A rod of length $L$ has non-uniform linear mass density given by $\rho(x)=a+b\left(\frac{x}{L}\right)^{2}$, 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:

(C) $\frac{3}{4}\left(\frac{2 a+b}{3 a+b}\right) \mathrm{L}$

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

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

(D) $\frac{3}{2}\left(\frac{2 a+b}{3 a+b}\right) \mathrm{L}$

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Conservation of Momentum

A rod of length

L

has non-uniform linear mass density given by

ρ (x) = a + b (\frac{x}{L})^{2}

, 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:

Answer: (C)

The centre of mass is

x_{c m} = \frac{\int _{0}^{L} x ρ ( x ) d x}{\int _{0}^{L} ρ ( x ) d x}

with

ρ (x) = a + b \frac{x ^{2}}{L ^{2}}

. The numerator is

\int_{0}^{L} (a x + b \frac{x ^{3}}{L ^{2}}) d x = \frac{a L ^{2}}{2} + \frac{b L ^{2}}{4} = \frac{L ^{2} ( 2 a + b )}{4}

, and the denominator is

\int_{0}^{L} (a + b \frac{x ^{2}}{L ^{2}}) d x = a L + \frac{b L}{3} = \frac{L ( 3 a + b )}{3}

. Dividing one by the other gives

x_{c m} = \frac{3 L}{4} (\frac{2 a + b}{3 a + b})

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option