First slide

Moment of intertia

Question

From a given sample of uniform wire, two circular loops P and Q are made, P of radius r and Q of radius nr. If the M.I. of Q about its axis is 4 times that of P about its axis (assuming wire diameter much smaller than either radius of the loops), the value of n is:

Moderate

A

${(4)}^{\frac{2}{3}}$
B

${(4)}^{\frac{1}{3}}$
C

${(4)}^{\frac{1}{2}}$
D

${(4)}^{\frac{1}{4}}$

Solution

MOMENT OF INERTIA OF A RING = $M R^{2} = (λ 2 π R) R^{2} = λ 2 π R^{3}$
$λ$ is mass per unit length of wire
THEREFORE $\frac{I_{Q}}{I_{P}} = \frac{{R_{Q}}^{3}}{{R_{P}}^{3}} = \frac{n^{3}}{1}$
$n^{3} = 4 \Rightarrow n = 4^{1 / 3}$

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