JEE Main 2025 Jan 28 Shift 1, Physics Q17: Work Energy Theorem And Its Applications

JEE Main2025Jan 28, Shift 1Physics

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
$\frac{2 q E ℓ}{m}$
B
$\frac{q E ℓ}{4 m}$
C
$\frac{q E ℓ}{m}$
D
$\frac{q E ℓ}{2 m}$

Solution

The particle moves on a circular arc (string keeps it at distance $ℓ$ from $O$ ). Initially it sits at angle $60°$ above the $+ x$ axis at position $(ℓ cos 60°, ℓ sin 60°) = (ℓ /2, ℓ 3 /2)$ .

When the particle crosses the x-axis, its position is $(ℓ, 0)$ .

The horizontal displacement is $Δ x = ℓ - ℓ /2 = ℓ /2$ .

The electric force $q E$ acts in $+ x$ , so the work done by the electric force between the two positions is

$W = q E \cdot Δ x = q E \cdot \frac{ℓ}{2}$ .

The string tension is always perpendicular to motion, so it does no work. By work-energy theorem:

$\frac{1}{2} m v^{2} = \frac{q E ℓ}{2} \Rightarrow v = \frac{q E ℓ}{m}$ .

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JEE Main 2025 Jan 28 Shift 1, Physics Q17: Work Energy Theorem And Its Applications

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