MHT CET · Physics · Rotational Motion
Two circular loops \(\mathrm{A}\) and \(\mathrm{B}\) of radii \({ }^{\prime} \mathrm{R}^{\prime}\) and ' \(\mathrm{NR}^{\prime}\) respectively are made from a uniform wire. Moment of inertia of B about its axis is 3 times that of A about its axis. The value of \(\mathrm{N}\) is
- A \([5]^{\frac{1}{3}}\)
- B \([3]^{\frac{1}{3}}\)
- C \([4]^{\frac{1}{3}}\)
- D \([2]^{\frac{1}{3}}\)
Answer & Solution
Correct Answer
(B) \([3]^{\frac{1}{3}}\)
Step-by-step Solution
Detailed explanation
If \(m\) is the mass per unit length of the wire, then mass of loop \(\mathrm{A} =\mathrm{M}_{\mathrm{A}}=2 \pi \mathrm{Rm}\) mass of loop \(\mathrm{B} =\mathrm{M}_{\mathrm{B}}=2 \pi \mathrm{NRm}\) \(\therefore \frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}=\mathrm{N}\)
\(\mathrm{I}_{\mathrm{A}}=\mathrm{M}_{\mathrm{A}} \mathrm{R}^{2}, \mathrm{I}_{\mathrm{B}}=\mathrm{M}_{\mathrm{B}}(\mathrm{NR})^{2}=\mathrm{M}_{\mathrm{B}} \mathrm{N}^{2} \mathrm{R}^{2}\)
\(\therefore \frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}=\frac{\mathrm{M}_{\mathrm{B}} \mathrm{N}^{2}}{\mathrm{M}_{\mathrm{A}}}=\mathrm{N}^{3}\)
\(\therefore 3=\mathrm{N}^{3}\)
\(\therefore \mathrm{N}=(3)^{\frac{1}{3}}\)
\(\mathrm{I}_{\mathrm{A}}=\mathrm{M}_{\mathrm{A}} \mathrm{R}^{2}, \mathrm{I}_{\mathrm{B}}=\mathrm{M}_{\mathrm{B}}(\mathrm{NR})^{2}=\mathrm{M}_{\mathrm{B}} \mathrm{N}^{2} \mathrm{R}^{2}\)
\(\therefore \frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}=\frac{\mathrm{M}_{\mathrm{B}} \mathrm{N}^{2}}{\mathrm{M}_{\mathrm{A}}}=\mathrm{N}^{3}\)
\(\therefore 3=\mathrm{N}^{3}\)
\(\therefore \mathrm{N}=(3)^{\frac{1}{3}}\)
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