JEE Main 2025 · 28 January · morning · Single correct · +4 / −1
A particle of mass ' m ' and charge ' q ' is fastened to one end ' A ' of a massless string having equilibrium length l, whose other end is fixed at point ' O '. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is
- (a) √((qE l)/(2 m))
- (b) √((qE l)/(4 m))
- (c) √((qE l)/m)
- (d) √((2 qE l)/m)
Answer and solution
Answer (c) √((qE l)/m)
Solution
Physics sir · 3 steps
The string only steers the particle; the field does all the work, and it only counts the x-displacement.
- What does work. The plane is frictionless and horizontal, so gravity and the normal force do no work. The string tension is always along the radius, perpendicular to the circular path, so it does no work either. Only the electric force qE, along +x, does work.
- Find the x-shift. The particle moves on a circle of radius l about O. At A its x-coordinate is l cos 60° = l/2, and on the x-axis it is l. So it moves l/2 along the field.
1/2mv² = qE(l - l cos 60°) = qEl/2
- So. v² = qEl/m, so v = √(qEl/m), which is option (c).
The idea to keep. A constant force does work equal to the force times the displacement along it, whatever curved path is taken. Tension in a string always acts at right angles to circular motion, so it does no work.
Answer from the NTA final answer key. Solution by PrepWiser.