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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Rotational Motion

BITSAT 2013 · Physics · Rotational Motion

Two rings of radius R and nR made of same material have the ratio...

Two rings of radius R\mathrm{R} and nR\mathrm{nR} made of same material have the ratio of moment of inertia about an axis passing through centre is 1: 8. The value of n\mathrm{n} is

Options

  1. A2
  2. B222 \sqrt{2}
  3. C4
  4. D1/21 / 2

Correct answer

A. 2

Step-by-step solution

Ratio of moment of inertia of the rings \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right)\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{2}=(λl1λl2)(R1R2)2=(2πR2πnR)(RnR)2(λ= linear density of  wire =constant)⇒I1I2=1n3=18 (given) ∴n3=8⇒n=2\begin{aligned}=&\left(\frac{\lambda l_{1}}{\lambda l_{2}}\right)\left(\frac{R_{1}}{R_{2}}\right)^{2}=\left(\frac{2 \pi \mathrm{R}}{2 \pi \mathrm{n} \mathrm{R}}\right)\left(\frac{\mathrm{R}}{\mathrm{n} \mathrm{R}}\right)^{2} \\ &(\lambda=\text{ linear density\text{ of } wire }=\mathrm{constant}) \\ \Rightarrow \quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{1}{\mathrm{n}^{3}}=\frac{1}{8} \quad \text{ (given) } \\ \therefore \quad \mathrm{n}^{3}=8 \Rightarrow \mathrm{n}=2 \end{aligned}$$
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