Q50 / 122 · NTA Abhyas (JEE Main) 2020 · Physics · Rotational Motion

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

  • A
    1.2 N
  • B
    2.4 N
  • C
    3.6 N
  • Correct answer
    D
    4 N

Solution

It is given in the question, torque τ=1.6×1=1.6 N m 

So, when d=0.4 m,

F=τd=1.60.4=4 N.

Hence, the correct answer is option: 'D'.

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