Assertion If a charged particle enters from outside in uniform magnetic field, then it will keep on rotating in a circular path.
Reson Magnetic force is always perpendicular to velocity vector.

Answer

(i) 8 (ii) 0.375

Explanation

The free body diagram of the ring is shown in the figure. The force acting on the ring are
(A) the weight mg acting vertically downwards,
(B) Normal force N by the wire track. Normal force on the ring could be either radially outwards or radially inwards depending on whether the ring presses against the inner surface or outer surface of the track. Ascertain whether normal force is inwards or outward assume that, to being with, it is inwards then from `sum vec(F) = m vec(a)` find the value of normal force, if it is +ve it is inward and if it is ve, it is outwards.
(C) Force of the spring kx. In the physical situation given the spring is extended, it will pull the ring. So the spring force kx is along the spring towards O.
Length of the spring in the position shown = R.
`(CP = CO = R, /_ COP = /_COP = /_OPC = 60^(@) , Delta COP` is equilateral)
Change in length of the spring `= R - ((4)/(3)) R = ((R)/(4)) kx = ((mg)/(R ))((R)/(4)) = (mg)/4)`
Now from `F_(t) = ma_(t) ((mg)/(4)) cos 30^(@) + mg cos 30^(@) = m a_(t) implies a_(t) = (5 sqrt(3))/(8) g`
Now consider in radial direction `n + kx cos 60^(@) = mg cos 60^(@)`
`implies N = m g cos 60^(@) - kx cos 60^(@)` `implies N = (mg)/(2) - [(mg)/(4)] (1)/(2) = (3mg)/(8)`