The density of a non-uniform rod of length 1m is given by ρ(x)=a(1+bx2) where, a, and b are constants and 0x1. The centre of mass of the rod will be at

1. \(\frac{3(2+b)}{4(3+b)}\)
2. \(\frac{4(2+b)}{3(3+b)}\)
3. \(\frac{3(3+b)}{4(2+b)}\)
4. \(\frac{4(3+b)}{3(2+b)}\)

(a) Hint: Apply the concept of centre of mass.

Step 1: Find the position of centre of mass.

 Density is given as ρ(x)=a(1+bx2) where a and b are constants and 0x1 Let b0 , in this case ρ(x)=a= constant 

Hence. center of mass will be at x 0.5m. (middle of the rod)
Putting, b = Oin all the options, only (a) gives 0.5.