Q16 / 122 · NTA Abhyas (JEE Main) 2020 · Physics · Rotational Motion

A uniform solid sphere of mass M and radius R is lying on a rough horizontal plane. A constant force F=4 Mg acts vertically downwards at point P such that the line OP makes an angle of 60° with the horizontal as shown in the figure. The minimum value of the coefficient of friction μ so that sphere performs pure rolling, is

Figure

  • A
    37
  • B
    47
  • Correct answer
    C
    27
  • D
    25

Solution

Figure

Writing torque about the bottommost point,

4Mg×R2=75MR2aR

⇒a=10g7

f=μ5Mg=Ma

5μg=10g7⇒μ=27

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

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