A rod of length L has non-uniform linear mass density given by ρ (x) = a + b(x/L)², where a and b are constants and 0 ≤ x ≤ L. The value of x for the centre of mass of the rod is at: (a) 3/2((a + b)/(2a + b))L (b) 4/3((a + b)/(2a + 3b))L (c) 3/4((2a + b)/(3a + b))L (d) 3/2((2a + b)/(3a + b))L

Answer: (c). M = ∫₀L (a + (bx²)/(L²))dx = aL + bL/3 ∫₀L xρ dx = aL²/2 + bL²/4 xcm = (L(2a + b)/4)/((3a + b)/3) = 3/4((2a + b)/(3a + b))L So the answer is (c).

A rod of length L has non-uniform linear mass density given by ρ (x)… | JEE Main 2020 (9 Jan, Shift 2)

Q6JEE Main 2020 · 9 JAN · SHIFT 2Medium

Continuous Bodies: Standard Results & Variable Density

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:

Show answer and solution

Answer: (c)

$M = \int_0^L \left(a + \frac{bx^2}{L^2}\right)dx = aL + \frac{bL}{3}$
$\int_0^L x\rho\,dx = \frac{aL^2}{2} + \frac{bL^2}{4}$
$x_{cm} = \frac{L(2a + b)/4}{(3a + b)/3} = \frac{3}{4}\left(\frac{2a + b}{3a + b}\right)L$
So the answer is (c).

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Continuous Bodies: Standard Results & Variable Density question from JEE Main 2020 (9 Jan, Shift 2)

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