JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A string is wrapped around the rim of a wheel of moment of inertia \(0.40 \mathrm{kgm}^2\) and radius \(10 \mathrm{~cm}\). The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of \(40 \mathrm{~N}\). The angular velocity of the wheel after \(10\) \(\mathrm{s}\) is \(\mathrm{x} \mathrm{rad} / \mathrm{s}\), where \(\mathrm{x}\) is _______.
- A \(100\)
- B \(199\)
- C \(198\)
- D \(99\)
Answer & Solution
Correct Answer
(A) \(100\)
Step-by-step Solution
Detailed explanation
\(\tau=\mathrm{FR}=\mathrm{I} \alpha \Rightarrow 40 \times 0.1=0.4 \alpha\) \(\alpha=10 \mathrm{rad} / \mathrm{s}^2\) \(\mathrm{~W}_{\mathrm{f}}=10 \times 10=100 \mathrm{rad} / \mathrm{s}\)
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