TS EAMCET · Physics · Motion In Two Dimensions
Assume proton is rotating along a circular path of radius \(1 \mathrm{~m}\) under a centrifugal force of \(4 \times 10^{-12} \mathrm{~N}\). If the mass of proton is \(1.6 \times 10^{-27} \mathrm{~kg}\), then its angular velocity of rotation is
- A \(5 \times 10^7 \mathrm{rad} / \mathrm{s}\)
- B \(10^{15} \mathrm{rad} / \mathrm{s}\)
- C \(2.5 \times 10^7 \mathrm{rad} / \mathrm{s}\)
- D \(5 \times 10^{14} \mathrm{rad} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(5 \times 10^7 \mathrm{rad} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Given, \(r=1 \mathrm{~m}\) \(\begin{aligned} & F_{\text {centrifugal }}=4 \times 10^{-12} \mathrm{~N} \\ & m=1.6 \times 10^{-27} \mathrm{~kg} \\ & \omega=? \end{aligned}\)…
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