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QNEET-2016-Ph-1091: NEET 2016 Physics - Class 11 NEET-2016-Ph-1091 Solution: A particle of mass 10 g moves along a circle of radius 6.4 cm…

A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10−4 J8 \times 10^{-4}\ \text{J} by the end of the second revolution after the beginning of the motion? 0.18 m/s20.18\ \text{m/s}^2 0.2 m/s20.2\ \text{m/s}^2 0.1 m/s20.1\ \text{m/s}^2 0.15 m/s20.15\ \text{m/s}^2

Multiple Choice Options

  • A 0.18 m/s 2
  • B 0.2 m/s 2
  • C 0.1 m/s 2
  • D 0.15 m/s 2

Step-by-Step Verified Solution

Given: Mass of particle, M = 10g = 10 1000 k g radius of circle R = 6.4 cm Kinetic energy E of particle = 8 _ 10 �4 J acceleration a t = ? 1 2 m v 2 = E _ 1 2 ( 10 1000 ) v 2 = 8 _ 10 _ 4 _ v 2 = 16 _ 10 _ 2 _ v = 4 _ 10 _ 1 = 0.4 m / s Now, using v 2 = u 2 + 2a t s (s = 4 � R) ( 0.4 ) 2 = 0 2 + 2 a t ( 4 _ 22 7 _ 6.4 100 ) _ a t = ( 0.4 ) 2 _ 7 _ 100 8 _ 22 _ 6.4 = 0.1 m / s 2

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