A body slides without friction from a height H = 60 cm and then loops the loop of radius R = 20 cm at the bottom of incline. The ratio of forces exerted on the body by the track at A, B and C are (g = 10 ms^–2)

From the data,

H=3R

velocity at A,B and C

$V_A=\sqrt{2gH}=\sqrt{2g\left(3R\right)}=\sqrt{6gR}$

$V_B=\sqrt{4gR}$

$V_C=\sqrt{2gR}$

now the reaction force at A, B and C are

$$\begin{gathered} R_1=\frac{m{V_A}^2}{R}+mg\cos 0^{°} \\ =\frac{m\times 6gR}{R}+mg \\ =7mg \end{gathered}$$

$$\begin{gathered} R_2=\frac{m{V_B}^2}{R}+mg\cos {90}^{°} \\ =\frac{m\times 4gR}{R}+0 \\ =4mg \end{gathered}$$

$$\begin{gathered} R_3=\frac{m{V_A}^2}{R}+mg\cos {180}^{°} \\ =\frac{m\times 2gR}{R}-mg \\ =mg \end{gathered}$$

Hence the correct answer is 7 : 4 : 1.