Two small balls each of mass m are held by light strings of length ℓ as shown in the figure. The initial separation between the balls is r . At t = 0 , each ball carries a positive charge q 0 . Now,

Question

Two small balls each of mass m are held by light strings of length $ℓ$ as shown in the figure. The initial separation between the balls is r . At $t = 0$ , each ball carries a positive charge $q_{0}$ . Now, the charge of each ball slowly changes; and the variation is given by $q (t) = q_{0} (1 + bt)^{1.5}$ , where b is a positive constant. The balls are also released at $t = 0$ . Which of the following is correct? (Assume $q_{0}$ is very small)

TestHub · Accepted Answer

The balls are released from rest, so the speed initially increases as they start approaching.

The charge $q (t) = q_{0} (1 + b t)^{1.5}$ keeps growing, so the Coulomb repulsion grows until it stops the approach - the speed decreases back to zero.

Speed first increases, then decreases. Hence (C).

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JEE Main 2027 (Class 12+)