A ring of mass `m` can slide over a smooth vertical rod as shown in figure. The ring is connected to a spring of force constant `k = 4 mg//R`, where `2R` is the natural length of the spring . The other end of spring is fixed to the ground at a horizontal distance `2R` from base of the rod . If the mass is released at a height `1.5 J` then the velocity of the ring as it reaches the ground is

Options

A. `sqrt(gR)`
B. `2 sqrt(gR)`
C. `sqrt(2 gR)`
D. `sqrt(3 gR)`

Answer

B

Explanation

Initial length of spring `= sqrt((2 R)^(2) + (1.5R)^(2)) = 2.5 R`
So, `x = 2.5 R - 2 R = 0.5 R`
From constant of energy: `(1)/(2) mv^(2) = mgh + (1)/(2) kx^(2)`
`implies (1)/(2) mv^(2) = mg 1.5R+ (1)/(2) (4mg)/( R) (0.5R) ^(2)`
`implies v = 2 sqrt(gR)`