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 xle 1.`
Let `b to 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 the options , only (A) given 0.5
Note :- We should not check options by puting a = 0 , because ` rho =0` for `a =0`