Question

A small body slides down a smooth uneven surface from a height $H,$ which eventually emerges into a circular loop of radius $R\left(< H\right)$ what should be the value of $H,$ so that the force on the body at A is $\sqrt{2}$ times its weight?

• A. $H=\frac{3R}{2}$
• B. $H=5R$
• C. $H=\frac{5R}{2}$
• D. None of these

Answer 1

Answer: (A)

Force on body at $A=\sqrt{2} mg$
$\Rightarrow F=\sqrt{F_{g}^{2}+F_{c}^{2}}$
$\Rightarrow F_{c}=m g=m v^{2} / R$
$\Rightarrow v^{2}=g R$
Using energy conservation
$m g H=m v^{2} / 2+m g R$
$=3 m g R / 2$
$H=3 R / 2$