Question

The density of a non-uniform rod of length 1 m is given by ρ(x) = a(1 + bx²) where, a and b are constants and 0 ≤ x ≤ 1. The centre of mass of the rod will be at
(a) $\frac{3(2+b)}{4(3+b)}$
(b) $\frac{4(2+b)}{3(3+b)}$
(c) $\frac{3(3+b)}{4(2+b)}$
(d) $\frac{4(3+b)}{3(2+b)}$

Answer 1

Answer


c is correct answer .......

Answer 2

Answer:

a) 3(2+b)/4(3+b)

Explanation:

Length of the rod = 1m (Given)

Density of the rod = p(x) = a ( 1+bx²) (Given)

Constants = a and b (Given)

where a and b are constants and 0<x<1

Let b → 1 in this case and p(x) = a constant.

Thus, the centre of the mass will be = 0.5m ( middle of the rod)

Hence, when b→0, the density becomes uniform and thus the centre of mass is at x = 0.5. Putting b = 0 in all the options, Only option (a) tends to 0.5 as b→0.

Thus, the centre of mass of the rod will be at  3(2+b)/4(3+b).