JEE Main 2017 — Circular Motion Question with Solution
From: JEE Main 2017 (Online) 9th April Morning Slot
Question
A conical pendulum of length 1 m makes an angle = 45o w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be: (Take g = 10 ms−2 )


This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: D
Correct answer
D2 m/s
Step-by-step explanation
FBD of pendulum is :
T sin =
T cos = mg
tan =
tan45o =
v2 = rg
v = = 2 m/s
T sin =
T cos = mg
tan =
tan45o =
v2 = rg
v = = 2 m/s
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Circular Motion chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2017, covering the Circular Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.