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

Density is given as `rho (x)= a(1+ bx^(2))` Where a and b are constants and `0 le x le 1`
Let `b rarr 0`, in this case `rho(x)= a=` constant Hence, centre of mass will be at x = 0.5 m (middle of the rod) Putting, b = 0 in all the options, only (a) gives 0.5