1. Use conservation of mechanical energy

The body moves in a vertical circle of radius $R=2\,\text{m}$. Its mass is $m=100\,\text{g}=0.1\,\text{kg}$ and speed at point $A$ is $v_A=10\,\text{m/s}.$

So kinetic energy at $A$ is $K_A=\frac12 m v_A^2=\frac12(0.1)(10)^2=5\,\text{J}.$

2. Interpret the geometry from the standard figure

In the usual vertical-circle figure for this question:

• $A$ is the lowest point,
• $B$ and $C$ are points on the circle at heights determined by radii making angles $45^\circ$ and $60^\circ$ from the downward vertical respectively (equivalently their heights above $A$ are used below).

For a point at angle $\theta$ from the downward vertical, height above the lowest point is $h=R(1-\cos\theta).$

Thus:

• For point $B$ at $\theta=45^\circ$, $h_B=2\left(1-\frac{1}{\sqrt2}\right)=2-\sqrt2.$

• For point $C$ at $\theta=60^\circ$, $h_C=2\left(1-\frac12\right)=1.$

3. Find kinetic energies at $B$ and $C$

Loss in kinetic energy equals gain in potential energy: $K = K_A - mgh.$

Since $mg=(0.1)(10)=1,$ we get directly:

For $B$: $K_B=5-1\cdot(2-\sqrt2)=3+\sqrt2.$

For $C$: $K_C=5-1\cdot 1=4.$

4. Compute the ratio

$\frac{K_B}{K_C}=\frac{3+\sqrt2}{4}.$

5. Match with options

Given options are:

• A: $\dfrac{3-\sqrt2}{2}$
• B: $\dfrac{2+\sqrt3}{3}$
• C: $\dfrac{2+\sqrt2}{3}$
• D: $\dfrac{3+\sqrt3}{2}$

But $\frac{3+\sqrt2}{4}$ is not present among the options.

So either the figure/option set in the prompt is incomplete or the stored answer is incorrect.