Question

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I₁. The same rod is bent into a ring and its moment of inertia about diameter is I₂. If `I_1/I_2` is `(xpi^2)/3`, then the value of x will be ______.

पर्याय

• 8

• 7

• 6

• 5

Answer 1

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I₁. The same rod is bent into a ring and its moment of inertia about diameter is I₂. If `I_1/I_2` is `(xpi^2)/3`, then the value of x will be 8.

Explanation:

Let 'l' be the rod's length and 'r' be the ring's radius.

`I_1 = (ml^2)/3`

`l_2 = (mr^2)/2` & 2πr = 1

Hence, `l_2 = (m(l/(2pi))^2)/(2)`

So, `I_1/I_2 = ((ml^2)/3)/((ml^2)/(4pi^2 xx 2))`

`I_1/I_2 = (ml^2)/(3) xx (4pi^2 xx 2)/(ml^2)`

`l_1/l_2 = 8/3 pi^2`

Hence, x = 8