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NEET PHYSICSLAWS OF MOTIONEasy

Question

A bob is whirled in a horizontal plane by means of a string with an initial speed of ω\omega rpm. The tension in the string is TT. If speed becomes 2ω2\omega while keeping the same radius, the tension in the string becomes:

A

4T

B

T/4

C

\sqrt{2}T

D

T

Step-by-Step Solution

  1. Identify the Principle: For a bob whirled in a horizontal circle, the tension (TT) in the string provides the necessary centripetal force (FcF_c).
  2. Formula: The centripetal force is given by Fc=mω2rF_c = m \omega^2 r, where mm is the mass, ω\omega is the angular velocity, and rr is the radius [Source 51, 52]. Therefore, T=mω2rT = m \omega^2 r.
  3. Analyze the Change:
  • Initial Tension: T1=T=mω2rT_1 = T = m \omega^2 r.
  • Final Angular Velocity: ω=2ω\omega' = 2\omega.
  • Radius remains constant.
  • Final Tension: T2=m(ω)2r=m(2ω)2r=m(4ω2)r=4(mω2r)T_2 = m (\omega')^2 r = m (2\omega)^2 r = m (4\omega^2) r = 4 (m \omega^2 r).
  1. Conclusion: Since mω2r=Tm \omega^2 r = T, the new tension T2=4TT_2 = 4T.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONwhirledhorizontalstringinitialtension

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