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NEET PHYSICSRotational motionMedium

Question

An energy of 484 J484 \text{ J} is spent in increasing the speed of a flywheel from 60 rpm60 \text{ rpm} to 360 rpm360 \text{ rpm}. The moment of inertia of the flywheel is:

A

0.7 kg-m20.7 \text{ kg-m}^2

B

3.22 kg-m23.22 \text{ kg-m}^2

C

30.8 kg-m230.8 \text{ kg-m}^2

D

0.07 kg-m20.07 \text{ kg-m}^2

Step-by-Step Solution

According to the work-energy theorem, the work done on the flywheel is equal to the change in its rotational kinetic energy. W=ΔK=12Iωf212Iωi2=12I(ωf2ωi2)W = \Delta K = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 = \frac{1}{2} I (\omega_f^2 - \omega_i^2) Given: Work done, W=484 JW = 484 \text{ J} Initial angular speed, ωi=60 rpm=60×2π60 rad/s=2π rad/s\omega_i = 60 \text{ rpm} = \frac{60 \times 2\pi}{60} \text{ rad/s} = 2\pi \text{ rad/s} Final angular speed, ωf=360 rpm=360×2π60 rad/s=12π rad/s\omega_f = 360 \text{ rpm} = \frac{360 \times 2\pi}{60} \text{ rad/s} = 12\pi \text{ rad/s} Substitute the values in the equation: 484=12I((12π)2(2π)2)484 = \frac{1}{2} I ((12\pi)^2 - (2\pi)^2) 484=12I(144π24π2)484 = \frac{1}{2} I (144\pi^2 - 4\pi^2) 484=12I(140π2)=70Iπ2484 = \frac{1}{2} I (140\pi^2) = 70 I \pi^2 Taking π2(227)2=48449\pi^2 \approx \left(\frac{22}{7}\right)^2 = \frac{484}{49}: 484=70×I×48449484 = 70 \times I \times \frac{484}{49} 1=I×7049=I×1071 = I \times \frac{70}{49} = I \times \frac{10}{7} I=710=0.7 kg-m2I = \frac{7}{10} = 0.7 \text{ kg-m}^2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionenergyincreasingflywheelmomentinertia

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