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NEET PHYSICSMedium

A body of mass 0.4 kg is whirled in a vertical circle making 2 rev/sec. If the radius of the circle is 2 m, then tension in the string when the body is at the top of the circle is:

A

41.56 N

B

89.86 N

C

109.86 N

D

122.4 N

Step-by-Step Solution

  1. Identify Given Values:
  • Mass (mm) = 0.4 kg
  • Frequency (ff) = 2 rev/sec
  • Radius (rr) = 2 m
  • Acceleration due to gravity (gg) 9.8 m/s2\approx 9.8 \text{ m/s}^2
  1. Calculate Angular Velocity (ω\omega): ω=2πf=2π(2)=4π rad/s\omega = 2\pi f = 2\pi(2) = 4\pi \text{ rad/s} [Source 43]

  2. Forces at the Top: At the highest point of a vertical circle, both the Tension (TT) and the Weight (mgmg) act downwards towards the center. The sum of these forces provides the necessary centripetal force (FcF_c). T+mg=Fc=mω2rT + mg = F_c = m\omega^2 r [Source 76, 82]

  3. Calculate Centripetal Force: Fc=0.4×(4π)2×2F_c = 0.4 \times (4\pi)^2 \times 2 Fc=0.8×16π2=12.8π2F_c = 0.8 \times 16\pi^2 = 12.8\pi^2 Using π29.87\pi^2 \approx 9.87: Fc12.8×9.87126.34 NF_c \approx 12.8 \times 9.87 \approx 126.34 \text{ N}

  4. Calculate Tension: Rearranging the force equation for TT: T=mω2rmgT = m\omega^2 r - mg T=126.34(0.4×9.8)T = 126.34 - (0.4 \times 9.8) T=126.343.92T = 126.34 - 3.92 T=122.42 NT = 122.42 \text{ N} This matches Option 4.

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