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NEET PHYSICSEasy

Two bodies of equal masses revolve in circular orbits of radii R₁ and R₂ with the same period. Their centripetal forces are in the ratio:

A

(R₂/R₁)²

B

R₁/R₂

C

(R₁/R₂)²

D

√R₁/R₂

Step-by-Step Solution

  1. Identify the Given Conditions:
  • Masses are equal (m1=m2=mm_1 = m_2 = m).
  • Time periods are equal (T1=T2=TT_1 = T_2 = T).
  • Radii are R1R_1 and R2R_2.
  1. Select the Relevant Formula: The centripetal force (FcF_c) required for a body of mass mm moving in a circle of radius RR is given by Fc=macF_c = m a_c. Since the time period (TT) is given, it is convenient to use the angular velocity form of centripetal acceleration, ac=ω2Ra_c = \omega^2 R, where ω=2πT\omega = \frac{2\pi}{T}. Fc=mω2R=m(2πT)2RF_c = m \omega^2 R = m \left(\frac{2\pi}{T}\right)^2 R [Source 47]

  2. Analyze the Proportionality: Since mm and TT (and thus ω\omega) are the same for both bodies, the term m(2π/T)2m(2\pi/T)^2 is a constant constant (kk). Fc=kRF_c = k R Therefore, the centripetal force is directly proportional to the radius (FcRF_c \propto R).

  3. Calculate the Ratio: F1F2=R1R2\frac{F_1}{F_2} = \frac{R_1}{R_2}

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