JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2026
- Hard
- MCQ
A bead \(P\) sliding on a frictionless semi-circular string \((A C B)\) and it is at point \(S\) at \(t =0\) and at this instant the horizontal component of its velocity is \(v\). Another bead \(Q\) of the same mass as \(P\) is ejected from point \(A\) at \(t =0\) along the horizontal string \(A B\), with the speed \(v\), friction between the beads and the respective strings may be neglected in both cases. Let \(t_p\) and \(t_Q\) be the respective times taken by beads \(P\) and \(Q\) to reach the point \(B\), then the relation between \(t_p\) and \(t_Q\) is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(t_{p} < t_{Q}\)
Step-by-step Solution
Detailed explanation
Horizontal displacement of Q is more than P. \(X_{Q}>X_{p}\)
Horizontal component of velocity is same
So \(t _{ p }=\frac{ x _{ p }}{ v }\)
\(t_Q=\frac{x_Q}{v}\)
\(t_{p} < t_{Q}\)
Horizontal component of velocity is same
So \(t _{ p }=\frac{ x _{ p }}{ v }\)
\(t_Q=\frac{x_Q}{v}\)
\(t_{p} < t_{Q}\)
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