Conservation Of Energy

$v_1=\frac{30}{\sqrt{\frac{2h}{g}}},v_2=\frac{120}{\sqrt{\frac{2h}{g}}}$
$(0.01)u=(0.2)\frac{30\sqrt{g}}{\sqrt{2h}}+(0.01)\frac{120\sqrt{g}}{\sqrt{2h}}$
$u=300+60=360m{s}^{-1}$

The correct option is(C): 3

To find the value of , we utilize the principles of conservation of mechanical energy and circular motion dynamics.

First, let's consider the energy of the body. Since the plane and the loop are smooth, there is no work done by friction. The total mechanical energy is conserved throughout the motion.
Let the reference level for potential energy ( ) be the lowest point of the loop.
At the initial point on the inclined plane, the body is released from height at rest ( ).
Initial Mechanical Energy,

At the highest point of the vertical circle, the height is (the diameter of the circle). Let the velocity of the body at this point be .
Final Mechanical Energy,
By the Law of Conservation of Energy, :

$ mg N 3mg N = 3mg N = 3mg v^2 g h = \alpha R \alpha = 4$. This corresponds to option 2.