JEE Main202529 Jan 2025Morning ShiftPhysicsWork, Power and EnergyActual
A body of mass ' (m ) ' connected to a massless and unstretchable string goes in verticle circle of radius ' (R ) 'under gravity (g ). The other end of the string is fixed at the center of circle. If velocity at top of circular path is (n g R ), where, (n 1 ), then ratio of kinetic energy of the body at bottom to that at top of the circle is
Options
- A( n^2 n^2+4 )
- B( n^2+4 n^2 )
- C( n+4 n )
- D( n n+4 )
Correct answer
B. ( n^2+4 n^2 )
Step-by-step solution
aligned & v₀= v^2+2 g(2 R) & v₀= n^2 g R+4 g R & k_ bottom k_ top = v₀^2 v^2 = n^2+4 n^2 aligned