Physics Circle home Student Login

Home›Chapters›Circular Motion PYQs›Conical Pendulum & Funnel›Question 12

JEE Main 2022 (24 Jun, Shift 1) · Conical Pendulum & Funnel

Conical Pendulum & Funnel question from JEE Main 2022 (24 Jun, Shift 1)

Q12JEE Main 2022 · 24 JUN · SHIFT 1Medium

Conical Pendulum & Funnel

A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is Kπ\frac{K}{\pi} rev./min. The value of K is:

(Assume the string is massless and unstretchable)

Show answer and solution

Answer: (c)

  1. Tension gives the centripetal force: T=mω2LT = m\omega^2 L.
  2. 80=0.1×ω2×2⇒ω2=40080 = 0.1 \times \omega^2 \times 2 \Rightarrow \omega^2 = 400, ω=20\omega = 20 rad/s
  3. In rev/min: n=202π×60=600πn = \frac{20}{2\pi} \times 60 = \frac{600}{\pi} rev/min
  4. So K=600K = 600, the answer is (c).
Report a mistake

More questions on Conical Pendulum & Funnel