A rod of length L has non-uniform linear mass density given by $ρ (x) = a + b {(\frac{c}{L})}^{2}$ , where a and b are constants and $0 \leq x \leq L$ . The value of x for the center of mass of the rod is at:

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given, Linear mass density, $ρ (x) = a + b {(\frac{x}{L})}^{2}$

$X_{CM} = \frac{\int xdm}{\int dm}$

$\int dm = \int_{0}^{L} ρ (x) dx = \int_{0}^{L} [a + b {(\frac{x}{L})}^{2}] dx$

$= aL + \frac{bL}{3}$

$\int_{0}^{L} xdm = \int_{0}^{L} (ax + \frac{{bx}^{3}}{L^{2}}) dx = (\frac{{aL}^{2}}{2} + \frac{{bL}^{2}}{4})$

$∴ X_{CM} = \frac{(\frac{{aL}^{2}}{2} + \frac{{bL}^{2}}{4})}{aL + \frac{bL}{3}}$

$\Rightarrow X_{CM} = \frac{3 L}{4} (\frac{2 a + b}{3 a + b})$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE