The density of a non-uniform rod of length 1m is given by (x)= (1… - Vidyadip

MCQ

The density of a non-uniform rod of length 1m is given by $\rho(\text{x})=\alpha(1+\text{b}\text{x}^2)$ where a and b are constants and $0\leq\text{x}\leq1$ The centre of mass of the rod will be at:

✓
$\frac{3(2+\text{b})}{4(3+\text{b})}$
B
$\frac{4(2+\text{b})}{3(3+\text{b})}$
C
$\frac{3(3+\text{b})}{4(2+\text{b})}$
D
$\frac{4(3+\text{b})}{3(2+\text{b})}$

✓

Answer

Correct option: A.

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

According to the problem, density is given as $\rho(\text{x})=\alpha(1+\text{b}\text{x}^2)$ where a and b are constants and $0\leq\text{x}\leq1$

Let us first consider a small element of the rod at a distance x from one end of length dx.
So, mass of this element is
$\text{dm}=\rho(\text{dx})=\text{a}\big(1+\text{bx}^2\big)\text{dx}$
The centre of mass of the rod is,
$\text{X}_\text{cm}=\frac{\int\limits^1_0\text{x dm}}{\int\limits^1_0\text{dm}}=\frac{\int\limits^1_0\text{xa}(1+\text{bx}^2)}{\int\limits^1_0(1+\text{bx}^2)\text{dx}}$
$=\frac{\int\limits^1_0{(\text{x}+\text{bx}^3)\text{dx}}}{\int\limits^1_0\text{a}(1+\text{bx}^2)\text{dx}}=\frac{\Big[\frac{\text{x}^2}{2}+\frac{\text{bx}^4}{4}\Big]^1_0}{\Big[\text{x}+\frac{\text{bx}^3}{3}\Big]^1_0}$
$=\frac{\big[\frac{1}{2}+\frac{\text{b}}{4}\big]}{\big[1+\frac{\text{b}}{3}\big]}=\frac{3(2+\text{b})}{4(3+\text{b})}$

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