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 $10 m/s^2$)
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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.
The body has a mass \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) and moves in a circular path of radius \( r = 2 \, \text{m} \). At point A, the velocity is \( v_A = 10 \, \text{m/s} \).
At point A, the total energy is the sum of kinetic energy \( K_A \) and potential energy \( U_A \). The kinetic energy is calculated as:
\[
K_A = \frac{1}{2} m v_A^2 = \frac{1}{2} (0.1) (10)^2 = 5 \, \text{J}.
\]
Assuming the reference for potential energy is at the lowest point (O), the potential energy at point A is \( U_A = 0 \, \text{J} \) since the height is zero.
Mechanical energy is conserved at points B and C, meaning:
\[
K_A + U_A = K_B + U_B = K_C + U_C.
\]
The heights at points B and C are \( h_B = r \) and \( h_C = r \sin 30^\circ \), respectively. The potential energies at these points are:
\[
U_B = m g h_B = (0.1)(10)(2) = 2 \, \text{J},
\]
\[
U_C = m g h_C = (0.1)(10)(2 \sin 30^\circ) = 1 \, \text{J}.
\]
Using conservation of mechanical energy, the kinetic energies at points B and C are:
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 \( \boxed{\frac{2 + \sqrt{2}}{3}} \).
