JEE Main 2026 · 22 January · morning · Single correct · +4 / −1
A simple pendulum has a bob with mass m and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗ is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ____ .
- (a) mg-qE
- (b) √(m²g²+q²E²)
- (c) m g+qE
- (d) √(m²g²-q²E²)
Answer and solution
Answer (b) √(m²g²+q²E²)
Solution
Physics sir · 3 steps
Gravity pulls down, the field pushes sideways, and the string must balance both at once.
- Forces on the bob. Gravity gives mg vertically down. The horizontal field gives qE sideways. The string tension T acts along the string.
- Equilibrium. At rest the string tilts until T points exactly opposite to the resultant of mg and qE. These two forces are perpendicular, so their resultant comes from Pythagoras.
T = √((mg)² + (qE)²) = √(m²g² + q²E²)
- So. The tension is √(m²g² + q²E²), which is option (b). The string settles at angle θ from the vertical with tan θ = qE/mg.
The idea to keep. When forces act at right angles, never add them directly. Balance components, or take the resultant with Pythagoras, and the string tension equals that resultant.
Answer from the NTA final answer key. Solution by PrepWiser.