JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Work, Power & Energy

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) $\frac{2 qE ℓ}{m}$
(B) $\frac{qE ℓ}{4 m}$
(C) $\frac{qE ℓ}{m}$ ✓ Correct
(D) $\frac{qE ℓ}{2 m}$

Answer: (C)

Solution

The particle at A is attached by an inextensible string of length

ℓ

to the fixed point O, with the string initially making a 60° angle with the x-axis; the field

E

is directed along the x-axis. Released from rest, the particle accelerates under the constant force

q E

along the field direction.

From the initial geometry, the horizontal separation of A from O is

ℓ cos 6 0^{\circ} = ℓ /2

; the particle crosses the x-axis once it has moved this same horizontal distance in the field direction.

Applying the work-energy theorem for this horizontal displacement:

q E \cdot \frac{ℓ}{2} = \frac{1}{2} m v^{2} - 0

v = \frac{q E ℓ}{m}

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Work, Power & Energy

Solution

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