The density of a non-uniform rod of length `1m` is given by `rho (x) = a (1 + bx^(2))`
where a and b are constants and `0 le x le 1`.
The centre of mass of the rod will be at

Options

A. `(3 (2+ b))/(4(3+b))`
B. `(4 (2+ b))/(3(3+b))`
C. `(3 (3+ b))/(4(2+b))`
D. `(4 (3+ b))/(3(2+b))`

Answer

A

Explanation

Here, `rho (x) = a (1 + bx^(2))`
When `b rarr0, rho(x) = a = consatnt`
i.e., density of rod of length 1 m is constant. In that event, centre of mass of rod would lie at `0.5 m` (i.e. at centre of rod).
When we try `b rarr 0` in all four given options, we find choice (a) alone gives `x = (3(2 + b))/(4(3 + b)) = (6)/(12) = 0.5`.
Therefore, choice (a) is correct.