NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
Two small balls of the same size, having masses m 1 and m 2 ( m 1 > m 2 ) are tied by a thin weightless thread and dropped from a certain height. Taking the force of buoyancy of air into account, the tension T in the thread during the flight, after the motion of the ball becomes uniform, will be
Options
- A( m 1 - m 2 ) g
- B( m 1 - m 2 ) g 2
- C( m 1 + m 2 ) g
- D( m 1 + m 2 ) g 2
Correct answer
B. ( m 1 - m 2 ) g 2
Step-by-step solution
As both the ball are of same size, force of buoyancy on each is same. Therefore, in equilibrium F + F = m 1 g + m 2 g or F = m 1 + m 2 g 2 Considering the equilibrium of lower ball, T + F = m 1 g T = m 1 g - F T = m 1 g - m 1 + m 2 g 2 T = m 1 - m 2 g 2