AP EAMCET · PHYSICS · Motion In Two Dimensions
The kinetic energy of a particle moving along a circle of radius ' \(R\) ' depends on the distance ' \(\mathrm{s}\) ' as \(\mathrm{K}=\mathrm{as}^2\) where ' \(\mathrm{a}\) ' is a constant. Then the force acting on the particle is
- A \(\frac{2 \mathrm{as}^2}{\mathrm{R}}\)
- B \(2 \mathrm{as}\left[1+\frac{\mathrm{s}^2}{\mathrm{R}^2}\right]^{1 / 2}\)
- C \(2 \mathrm{as}\)
- D \(2 \mathrm{a}\left(\frac{\mathrm{R}}{\mathrm{s}}\right)^{1 / 2}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{as}\left[1+\frac{\mathrm{s}^2}{\mathrm{R}^2}\right]^{1 / 2}\)
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