Question

44. Two loops P and Q are made from a uniform wire.
The radii of P and Q are R1, and R2, respectively
and their moment of inertia about axes normally
through centre are I2/ I1=4 .
then find R2/R1​

Answer 1

Answer:

4^1/3

Explanation:

I2/I1=4

hope it helps u

Answer 2

Answer:

$4^{\dfrac{1}{3}}=\dfrac{R2}{R1}$

Explanation:

Loop P -   Radius = R1

Loop Q -   Radius = R2

Moment of inertia

 I = M R²

Mass = density x volume

M= L . A . ρ

M= 2 π R x A x ρ

So

I =( 2 π R x A x ρ) R²

Density and cross sectional area is same for both the wires

I ∝ R³

$\dfrac{I2}{I1}=\left(\dfrac{R2}{R1}\right)^3$

Given that

I2/I1= 4

$4=\left(\dfrac{R2}{R1}\right)^3$

$4^{\dfrac{1}{3}}=\dfrac{R2}{R1}$