Question

From a uniform wire, two circular loops are made. One loop P of radius r and the other Q of radius nr. If the moment of inertia of Q about an axis passing through its center and perpendicular to its plane is 8 times that of P about a similar axis, the value of n is (diameter of the wire is very much smaller than r or nr):
$
  {\text{A}}{\text{. 8}} \\
  {\text{B}}{\text{. 6}} \\
  {\text{C}}{\text{. 4}} \\
  {\text{D}}{\text{. 2}} \\
 $

Answer 1

Hint: Find the moment of inertia of the two loops in terms of their masses and radius. As P and Q will have different masses, so try to relate their masses in terms of ‘n’. Then compare them using the condition given in question, to get n.

Formula used:
$MOI = M{R^2}$, of a ring about a perpendicular axis passing through its centre.

Complete Step-by-Step solution:
Let mass of P is M which is equal to $2\pi r.dx$
Then as the radius of Q is nr, so mass of Q will be
 $
   = 2\pi (nr).dx \\
   = n.M \\
$
Now MOI of P about an axis passing through centre and perpendicular to its plane will be: ${I_p} = M{r^2}$ (similar that of a ring)
MOI of Q about a similar axis:${I_Q} = (nM).{(nr)^2} = {n^3}.M{r^2}$
As,
 $
  {I_Q} = 8.{I_p} \\
  {n^3}.M{r^2} = 8.M{r^2} \\
  {n^3} = 8 \\
  n = 2 \\
 $
The correct option is (D).

Note: To do these types of questions, the MOI of different objects along the perpendicular axes through their centre should be known. As for MOI about other axes, parallel and perpendicular axes theorem can be used.