Two loops ' A ' and ' B ' of radii ' R 1 ' and ' R 2 ' are made from uniform wire. If moment of inertia of ' A ' is ' I A ' and that &

Question

Two loops ' A ' and ' B ' of radii ' R 1 ​ ' and ' R 2 ​ ' are made from uniform wire. If moment of inertia of ' A ' is ' I A ​ ' and that ' B ' is ' B B ​ ', then R 2 ​ / R 1 ​ is [ I B ​ I A ​ ​ = 27 ]

TestHub · Accepted Answer

Let the linear mass density of the uniform wire be \lambda.

The mass of loop $A$ is $M_{A} = λ (2 π R_{1})$ and the mass of loop $B$ is $M_{B} = λ (2 π R_{2})$ .

The moment of inertia of a circular loop about its central axis is given by $I = M R^{2}$ .

For loop $A$ : $I_{A} = M_{A} R_{1}^{2} = (2 πλ R_{1}) R_{1}^{2} = 2 πλ R_{1}^{3}$

For loop $B$ : $I_{B} = M_{B} R_{2}^{2} = (2 πλ R_{2}) R_{2}^{2} = 2 πλ R_{2}^{3}$

Taking the ratio of the moments of inertia:

$\frac{I _{A}}{I _{B}} = (\frac{R _{1}}{R _{2}})^{3}$

Given that $\frac{I _{A}}{I _{B}} = 27$ , we get:

$(\frac{R _{1}}{R _{2}})^{3} = 27$

$\frac{R _{1}}{R _{2}} = 3$

Therefore, $\frac{R _{2}}{R _{1}} = \frac{1}{3}$ , which is $1 : 3$ .

Answer: $1 : 3$

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