CBSE Class 12-science Answered

Figure shows the forces acting on pith ball when it is in equilibrium.

Tension force T acting on the rope is resolved into Tcosθ and Tsinθ along

vertical and horizontal directions respectively.

Where θ is the angle made by the string with vertical as shown in figure

Vertical component of tension force balances the weight of the ball .

T cosθ = m g

where m = 0.5 kg is mass of the ball and g is acceleration due to gravity .

Above expression is written as

T cosθ = 0.5 × 9.8  N = 4.9 N ............................(1)

Horizontal component of tension force balances the electrostatic force F_e between charged balls.

T sinθ = F_e = K × ( q² / d² )

where K = 1/(4πε_o ) = 9 × 10⁹ N m² C^-2 is Coulomb constant,  q = 5 μC is the charge on each pith ball

and d is distance between pith balls .

Hence above expression is written as

  ..........................(2)

From (1) and (2) , we get

.......................(3)

when θ is small , tanθ ≈ θ = (arc)/radius = (d/2) / 1 = d/2

Hence eqn.(3) becomes

d³ = 0.092 m³

hence , we get d = 0.451 m