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Current Question (ID: 16531)

Question:
$\text{An electric dipole with dipole moment } \vec{p} = \left(3\hat{i} + 4\hat{j}\right) \times 10^{-30} \text{ C-m is placed in an electric field } \vec{E} = 4000\hat{i} \text{ N/C. An external agent turns the dipole slowly until its electric dipole moment becomes } \left(-4\hat{i} + 3\hat{j}\right) \times 10^{-30} \text{ C-m.}$ $\text{The work done by the external agent is equal to:}$
Options:
  • 1. $4 \times 10^{-28} \text{ J}$
  • 2. $-4 \times 10^{-28} \text{ J}$
  • 3. $2.8 \times 10^{-26} \text{ J}$
  • 4. $-2.8 \times 10^{-26} \text{ J}$
Solution:
$\text{Hint: Work done by the external agent = change in potential energy of the dipole.}$ $\text{Step 1: Draw the diagram.}$ $\text{Step 2: Find the net dipole moment}$ $\vec{p}_i = \left(3\hat{i} + 4\hat{j}\right) \times 10^{-30} \text{ C-m}$ $\vec{p}_f = \left(-4\hat{i} + 3\hat{j}\right) \times 10^{-30} \text{ C-m}$ $|\vec{p}| = 5 \times 10^{-30} \text{ C-m}$ $\text{Step 3: Use the geometry to find the angle}$ $\theta_1 = 53^\circ$ $\theta_2 = 180^\circ - 37^\circ$ $\cos(180^\circ - 37^\circ) = -\cos 37^\circ$ $\text{Step 3: Find the work done.}$ $W = -pE\cos\theta_2 - (-pE\cos\theta_1)$ $= pE[\cos 37^\circ + \cos 53^\circ]$ $= 2.8 \times 10^{-26} \text{ J}$ $\text{Alternating explanation:}$ $\text{As, } U = -\vec{p} \cdot \vec{E}$ $U_i = -[(3 \times 10^{-30} \hat{i} + 4 \times 10^{-30} \hat{j}) \cdot (4000\hat{i})]$ $= -12000 \times 10^{-30}$ $= -1.2 \times 10^{-26} \text{ J}$ $U_f = -[(-4 \times 10^{-30} \hat{i} + 3 \times 10^{-30} \hat{j}) \cdot (4000\hat{i})]$ $= 16000 \times 10^{-30}$ $= 1.6 \times 10^{-26} \text{ J}$ $\text{Work done by an external agent,}$ $W = U_f - U_i$ $= 1.6 \times 10^{-26} + 1.2 \times 10^{-26}$ $= 2.8 \times 10^{-26} \text{ J}$

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{
  "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
}