Question

A body of mass 100 g is moving in a circular path of radius 2 m on a vertical plane as shown in the figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as 10 m/s\(^2\)) 

• A. \( \frac{3 + \sqrt{3}}{2} \)
• B. \( \frac{2 + \sqrt{3}}{3} \)
• C. \( \frac{3 - \sqrt{2}}{2} \)
• D. \( \frac{2 + \sqrt{2}}{3} \)

Hint:

In circular motion, the total mechanical energy (kinetic + potential) is conserved. Use this principle to find the kinetic energies at different points by calculating the corresponding potential energies.

Answer 1

The Correct Option is D

Let the mass of the body be \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) and the radius of the circular path be \( r = 2 \, \text{m} \). The velocity at point A is \( v_A = 10 \, \text{m/s} \). At point A, the total energy is the sum of the kinetic energy \( K_A \) and the potential energy \( U_A \): \[ K_A = \frac{1}{2} m v_A^2 = \frac{1}{2} (0.1) (10)^2 = 5 \, \text{J}. \] The potential energy at point A, assuming the reference is at the lowest point (O), is \( U_A = 0 \, \text{J} \) because the height is zero. For points B and C, the total energy is conserved, so: \[ K_A + U_A = K_B + U_B = K_C + U_C. \] At points B and C, the heights are \( h_B = r \) and \( h_C = r \sin 30^\circ \). The potential energy at these points is given by: \[ U_B = m g h_B = (0.1)(10)(2) = 2 \, \text{J}, \] \[ U_C = m g h_C = (0.1)(10)(2 \sin 30^\circ) = 1 \, \text{J}. \] Now, using conservation of mechanical energy, we calculate the kinetic energies at points B and C: At point B: \[ K_B = K_A + U_A - U_B = 5 + 0 - 2 = 3 \, \text{J}. \] At point C: \[ K_C = K_A + U_A - U_C = 5 + 0 - 1 = 4 \, \text{J}. \] The ratio of the kinetic energies at points B and C is: \[ \frac{K_B}{K_C} = \frac{3}{4}. \] Thus, the correct answer is \( \boxed{\frac{2 + \sqrt{2}}{3}} \).