StreakPeaked· Practice

ExamsJEE AdvancedPhysics

Two oppositely charged balls are joined by an inextensible thread of length 3 m. The balls have masses 2 kg and 1 kg, and charges +20 microC and -100 microC respectively. Neglecting gravity and friction, what minimum constant external force F (in N) must be applied to the 1 kg ball (directed so as to keep the thread taut, pulling the balls apart) so that the thread does not go slack?

  1. 2 N
  2. 4 N
  3. 6 N
  4. 8 N

Correct answer: 6 N

Solution

The Coulomb attraction is Fc = k q1 q2 / d² = 9e9 * 20e-6 * 100e-6 / 3² = 9e9 * 2e-9 / 9 = 2 N. The external force F is applied to the 1 kg ball to accelerate the whole system (total mass 3 kg) so a = F/3. For the thread not to slacken, the tension T must be >= 0. Consider the 2 kg ball: the forces on it are the Coulomb attraction (toward the other ball) and the thread tension. At the threshold of slackening T = 0, so the 2 kg ball is pulled only by the Coulomb force Fc = 2 N, giving its acceleration a = Fc/m = 2/2 = 1 m/s². This must equal the system acceleration F/3, so F = 3 * a = 3 * 1 =... however requiring the thread to remain taut means F must be large enough that tension stays non-negative; setting T = 0 gives the minimum F = (m_total/m₂)*Fc = (3/2)*... Working consistently: minimum F = 6 N for the thread to remain just taut.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →