Q64 / 90 · JEE Main 2022 (24 Jun Shift 2) · Physics

Two identical charged particles each having a mass 10 g and charge 2.0×10-7C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g=10 ms-2 ]

  • Correct answer
    A
    12 cm
  • B
    10 cm
  • C
    8 cm
  • D
    5 cm

Solution

Normal reaction applied by table on object will be, N=mg.

Both charges are same, hence a repulsive force will act between them.

Value of force acting will be, kQ2L2.

Now the maximum value of friction force which table can provide is,

=μN=μmg

For equilibrium, kQ2L2=μmg.

L=kQ2μmg=k0.25mgQ=2kmgQ

=29×10910×10-3×102×10-7=0.12 m=12 cm

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

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