A particle of mass m slides on a frictionaless surface ABCD, starting from rest as shown in figure. The part BCD is a circular arc. If it looses contact at point P, the maximum height attained by the particle from point C is

Options

A. `R (2 + (1)/( 2 sqrt2))`
B. `R (2- (1)/(2 sqrt2))`
C. 3 R
D. `(1 + (1)/( 2 sqrt2)) R`

Answer

D

Explanation

Applying law of conservation of energy
`2mg R - mg ""(R)/(sqrt2) = (1)/(2) m u ^(2)`
`implies u = sqrt ( 4 gR - sqrt2 gR)`
Maximum height attained is
`H _(max) = (R)/( sqrt2) + ( u ^(2) sin ^(2) 45 ^(@))/( 2g)`
`= (R)/(sqrt2) + ( 4 g R - sqrt2 g R)/( 2 gxx 2) `
`= (R)/( sqrt2) + R - ( R)/( 2 sqrt2)`
`= R ((2 + 2sqrt2 -1)/( 2 sqrt2))= (( 1 + 2 sqrt2)/( 2 sqrt2)) R = (1 + (1)/(2 sqrt2)) R`