Question:
A bead P sliding on a frictionless semi-circular string... bead Q ejected... relation between $t_P$ and $t_Q$ is
A bead P sliding on a frictionless semi-circular string... bead Q ejected... relation between $t_P$ and $t_Q$ is
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Brachistochrone principle: Curved path under gravity is faster.
Updated On: Mar 27, 2026
- $t_P>t_Q$
- $t_P>1.25 t_Q$
- $t_P = t_Q$
- $t_P<t_Q$
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The Correct Option is D
Solution and Explanation
Bead Q moves at constant speed $v$.
Bead P accelerates due to gravity on the curved path, so average speed $> v$.
Despite longer path, the increased speed results in shorter time.
Bead P accelerates due to gravity on the curved path, so average speed $> v$.
Despite longer path, the increased speed results in shorter time.
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