JEE Advanced 2026 — Paper 2 — PHY-II solution

JEE Advanced2026Paper 2PHY-II

Consider an electric dipole comprising two charges $+ q$ and $- q$ each with mass $m$ , separated by a fixed distance $d$ and initially at rest with its dipole moment pointing along $\hat{i}$ . A uniform electric field $E \hat{j}$ is turned on at time $t = 0$ and it is turned off at $t = t_{f}$ , when the dipole moment makes an angle $θ_{f}$ with $\hat{i}$ . Neglecting any sources of energy loss, correct option(s) is/are:

A
The center of mass of the dipole is deflected towards $\hat{j}$ in the presence of the field.
B
If the magnitude of the final angular velocity $ω_{f} = \frac{2 q E}{m d}$ then $θ_{f} = \frac{π}{6}$
C
If $θ_{f} = π /4$ , then the change in kinetic energy of the dipole is given by $23 q E d$ .
D
For $θ_{f} = π /4$ , the dipole rotates around its center of mass with a constant angular velocity after $t > t_{f}$ .

Solution

Forces and torque. The two equal and opposite charges experience equal and opposite forces in a uniform field, so the net force on the centre of mass is zero. Option (A) is incorrect.

Rotational dynamics. Moment of inertia about the centre of mass: $I = 2 m (d /2)^{2} = \frac{m d ^{2}}{2}$ . The work done by the torque from $0$ to $θ_{f}$ equals $Δ K E$ :

$Δ K E = q E d sin θ_{f} = \frac{1}{2} I ω_{f}^{2} = \frac{m d ^{2}}{4} ω_{f}^{2}$ .

(B). Setting $ω_{f} = \frac{2 q E}{m d}$ gives $\frac{m d ^{2}}{4} \cdot \frac{2 q E}{m d} = \frac{q E d}{2} = q E d sin θ_{f} \Rightarrow sin θ_{f} = \frac{1}{2} \Rightarrow θ_{f} = \frac{π}{6}$ . Correct.

(C). For $θ_{f} = π /4$ , $Δ K E = q E d sin (π /4) = \frac{q E d}{2}$ , not $23 q E d$ . Incorrect.

(D). After $t > t_{f}$ the field is off, so no torque acts on the dipole. By conservation of angular momentum it rotates with constant angular velocity about its centre of mass. Correct.

$∴$ Correct options are (B) and (D).

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