JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
A body of \(\mathrm{m} \mathrm{kg}\) slides from rest along the curve of vertical circle from point \(A\) to \(B\) in friction less path. The velocity of the body at \(B\) is _______. (given, \(\mathrm{R}=14 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2\) and \(\sqrt{2}=1.4\) )
- A \(19.8 \mathrm{~m} / \mathrm{s}\)
- B \(21.9 \mathrm{~m} / \mathrm{s}\)
- C \(16.7 \mathrm{~m} / \mathrm{s}\)
- D \(10.6 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(21.9 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Apply \(W.E.T.\) from \(A\) to \(B\) \(\Rightarrow \mathrm{W}_{\mathrm{mg}}=\mathrm{K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{A}}\)…
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