Back to Directory
NEET PHYSICSEasy

A particle moves in a circle of radius 5 cm with constant speed and time period 0.2\pi s. The acceleration of the particle is:

A

25 m/s²

B

36 m/s²

C

5 m/s²

D

15 m/s²

Step-by-Step Solution

  1. Identify Given Values:
  • Radius R=5 cm=5×10−2 mR = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}.
  • Time Period T=0.2π sT = 0.2\pi \text{ s}.
  1. Formula for Centripetal Acceleration: For an object in uniform circular motion, the acceleration is centripetal and given by ac=ω2Ra_c = \omega^2 R, where ω\omega is the angular velocity .
  2. Relate Angular Velocity to Time Period: The relationship is ω=2πT\omega = \frac{2\pi}{T} .
  3. Substitute and Calculate: ac=(2πT)2Ra_c = \left( \frac{2\pi}{T} \right)^2 R ac=(2π0.2π)2(5×10−2)a_c = \left( \frac{2\pi}{0.2\pi} \right)^2 (5 \times 10^{-2}) ac=(20.2)2(0.05)=(10)2(0.05)=100×0.05=5 m/s2a_c = \left( \frac{2}{0.2} \right)^2 (0.05) = (10)^2 (0.05) = 100 \times 0.05 = 5 \text{ m/s}^2
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: PHYSICS Question for NEET | Sushrut