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NEET PHYSICSLAWS OF MOTIONMedium

Question

A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed of 10 rpm. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is:

A

5 rpm

B

10 rpm

C

20 rpm

D

14 rpm

Step-by-Step Solution

  1. Identify the Force: For a ball whirled in a horizontal circle, the tension (TT) in the string provides the necessary centripetal force (FcF_c) to maintain the circular motion.
  2. Formula: The centripetal force is given by Fc=mω2RF_c = m \omega^2 R, where mm is mass, ω\omega is angular velocity, and RR is radius [Source 44, 64]. T=mω2RT = m \omega^2 R
  3. Proportionality: Since the mass (mm) and radius (RR) are kept constant, the tension is directly proportional to the square of the angular speed. Tω2T \propto \omega^2
  4. Calculation:
  • Initial Tension: T1ω12T_1 \propto \omega_1^2
  • Final Tension: T2ω22T_2 \propto \omega_2^2
  • Given: T2=14T1T_2 = \frac{1}{4} T_1 T2T1=ω22ω12\frac{T_2}{T_1} = \frac{\omega_2^2}{\omega_1^2} 14=(ω210)2\frac{1}{4} = \left( \frac{\omega_2}{10} \right)^2 Taking the square root of both sides: 12=ω210\frac{1}{2} = \frac{\omega_2}{10} ω2=5 rpm\omega_2 = 5 \text{ rpm}

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 MOTIONwhirledhorizontalcircleradiusstring

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