Q168 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
Two electric dipoles of dipole moments $3.9 \times 10^{-30} \mathrm{Cm}$ and $5.2 \times 10^{-30} \mathrm{Cm}$ are placed in two different uniform electric fields of strengths $16 \times 10^4 N C^{-1}$ and $4 \times 10^4 N C^{-1}$ respectively. What is the ratio of maximum torque experienced by the electric dipoles?
- A$12: 1$
- Correct answerB$3: 1$
- C$9: 1$
- D$1: 9$
Solution
The torque $\tau$ experienced by an electric dipole of moment $p$ in a uniform electric field $E$ is given by $\tau = pE \sin \theta$.
The maximum torque $\tau_{max}$ occurs when $\theta = 90^{\circ}$, so $\tau_{max} = pE$.
For the first dipole, $p_1 = 3.9 \times 10^{-30} \text{ Cm}$ and $E_1 = 16 \times 10^4 \text{ N C}^{-1}$.
$\tau_{max,1} = (3.9 \times 10^{-30}) \times (16 \times 10^4) = 62.4 \times 10^{-26} \text{ Nm}$.
For the second dipole, $p_2 = 5.2 \times 10^{-30} \text{ Cm}$ and $E_2 = 4 \times 10^4 \text{ N C}^{-1}$.
$\tau_{max,2} = (5.2 \times 10^{-30}) \times (4 \times 10^4) = 20.8 \times 10^{-26} \text{ Nm}$.
The ratio of the maximum torques is $\dfrac{\tau_{max,1}}{\tau_{max,2}} = \dfrac{62.4 \times 10^{-26}}{20.8 \times 10^{-26}} = \dfrac{62.4}{20.8} = 3$.
Therefore, the ratio is $3:1$.
Answer: $3: 1$
The maximum torque $\tau_{max}$ occurs when $\theta = 90^{\circ}$, so $\tau_{max} = pE$.
For the first dipole, $p_1 = 3.9 \times 10^{-30} \text{ Cm}$ and $E_1 = 16 \times 10^4 \text{ N C}^{-1}$.
$\tau_{max,1} = (3.9 \times 10^{-30}) \times (16 \times 10^4) = 62.4 \times 10^{-26} \text{ Nm}$.
For the second dipole, $p_2 = 5.2 \times 10^{-30} \text{ Cm}$ and $E_2 = 4 \times 10^4 \text{ N C}^{-1}$.
$\tau_{max,2} = (5.2 \times 10^{-30}) \times (4 \times 10^4) = 20.8 \times 10^{-26} \text{ Nm}$.
The ratio of the maximum torques is $\dfrac{\tau_{max,1}}{\tau_{max,2}} = \dfrac{62.4 \times 10^{-26}}{20.8 \times 10^{-26}} = \dfrac{62.4}{20.8} = 3$.
Therefore, the ratio is $3:1$.
Answer: $3: 1$
Hence, the correct answer is option: 'B'.