NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
A door 1 . 6 m wide requires a minimum force of 1 N to be applied at the free end to open or close it. The minimum force that is required at a point 0 . 4 m away from the hinges for opening or closing the door is
Options
- A1 . 2 N
- B2 . 4   N
- C3 . 6   N
- D4 N
Correct answer
D. 4 N
Step-by-step solution
It is given in the question, torque τ = 1.6 × 1 = 1.6 N m So, when d = 0 .4 m , F = τ d = 1.6 0.4 = 4 N .