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Current Question (ID: 20045)

Question:

$\text{Two electric dipoles of dipole moments } 1.2 \times 10^{-30} \text{ C-m and } 2.4 \times 10^{-30} \text{ C-m are placed in two different uniform electric fields of strengths } 5 \times 10^4 \text{ NC}^{-1} \text{ and } 15 \times 10^4 \text{ NC}^{-1} \text{ respectively. The ratio of maximum torque experienced by the electric dipoles is } \frac{1}{x}. \text{ The value of } x \text{ is:}$

Options:

1. $2$
2. $4$
3. $6$
4. $8$

Solution:

$\text{Hint: The torque is maximum when the angle between the dipole and the electric field is } 90^\circ.$ $\text{Step 1: Find the ratio of maximum torque experienced by the electric dipoles.}$ $\text{The maximum torque acting on the dipole is given by:}$ $\tau = pE \sin \theta$ $\Rightarrow \tau_{\text{max}} = pE$ $\text{The ratio of maximum torque experienced by the electric dipoles is given by:}$ $\frac{\tau_1}{\tau_2} = \frac{p_1E_1}{p_2E_2}$ $\Rightarrow \frac{\tau_1}{\tau_2} = \frac{1.2 \times 10^{-30} \times 5 \times 10^4}{2.4 \times 10^{-30} \times 15 \times 10^4}$ $\Rightarrow \frac{\tau_1}{\tau_2} = \frac{1}{6}$ $\text{Step 2: Find the value of } x.$ $\text{The ratio of maximum torques experienced by electric dipoles is given by:}$ $\frac{\tau_1}{\tau_2} = \frac{1}{x} = \frac{1}{6}$ $\Rightarrow x = 6$ $\text{Hence, option (3) is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}