A molecule of a substance has a permanent electric dipole moment of magnitude \( 10^{-31}\,\mathrm{C\,m} \). A mole of this substance is 100% polarised by applying a strong electrostatic field of magnitude \( 10^{8}\,\mathrm{V\,m^{-1}} \). The direction of field is suddenly changed by an angle of \( 60^\circ \). Find the heat released by substance in aligning its dipoles along the new direction of the field. [Take 1 mole = \( 6 \times 10^{23} \) molecules]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3 J
The potential energy of a dipole in a uniform field is \( U = -pE\cos\theta \), where \( \theta \) is the angle between the dipole and the field. When the dipoles turn to line up with the new field, their potential energy falls and the decrease in potential energy is released as heat.
For one mole, the total dipole moment is \( N p \).
\( \displaystyle N p E = 6\times10^{23} \times 10^{-31}\,\mathrm{C\,m} \times 10^{8}\,\mathrm{V\,m^{-1}} = 6\,\mathrm{J} \)
Just after the field turns, the dipoles are at \( 60^\circ \) to it; at the end they are along it.
\( \displaystyle U_i = -N p E\cos 60^\circ = -6 \times \frac{1}{2} = -3\,\mathrm{J} \)
\( \displaystyle U_f = -N p E\cos 0^\circ = -6\,\mathrm{J} \)
\( \displaystyle \text{Heat released} = U_i - U_f = -3\,\mathrm{J} - (-6\,\mathrm{J}) = 3\,\mathrm{J} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Potential Energy of a Dipole in an External Field – …
