JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point \(A\) is \(10 \mathrm{~m} / \mathrm{s}\). The ratio of its kinetic energies at point \(B\) and \(C\) is :
(Take acceleration due to gravity as \(10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(\frac{2+\sqrt{2}}{3}\)
- B \(\frac{2+\sqrt{3}}{3}\)
- C \(\frac{3+\sqrt{3}}{2}\)
- D \(\frac{3-\sqrt{2}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3+\sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{1}{2} \mathrm{~m} \times 100+0=\frac{1}{2} \mathrm{mV}_{\mathrm{B}}^2+\mathrm{mg}\left(\mathrm{R}-\frac{\mathrm{R} \sqrt{3}}{2}\right) \\ & 100=\mathrm{V}_{\mathrm{B}}^2+2 \mathrm{gR}\left(1-\frac{\sqrt{3}}{2}\right] \\ &…
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