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Current Question (ID: 16509)

Question:

$\text{Some equipotential surfaces are shown in the figure. The electric field at points } A, B \text{ and } C \text{ are respectively:}$

Options:

1. $1 \text{ V/cm}, \frac{1}{2} \text{ V/cm}, 2 \text{ V/cm (all along +ve X-axis)}$
2. $1 \text{ V/cm}, \frac{1}{2} \text{ V/cm}, 2 \text{ V/cm (all along -ve X-axis)}$
3. $\frac{1}{2} \text{ V/cm}, 1 \text{ V/cm}, 2 \text{ V/cm (all along +ve X-axis)}$
4. $\frac{1}{2} \text{ V/cm}, 1 \text{ V/cm}, 2 \text{ V/cm (all along -ve X-axis)}$

Solution:

$\text{Hint: } \mathbf{E} = -\frac{\Delta V}{\Delta x}$ $\text{Step: Find the electric field at the points } A, B \text{ and } C.$ $\text{The electric field (} \mathbf{E} \text{) is given by:}$ $E = -\frac{\Delta V}{\Delta x}$ $\text{The negative sign indicates that the electric field points in the direction of decreasing potential.}$ $\text{The electric field at } A \text{ between 40 V and 30 V:}$ $\Delta V = 40 - 30 = 10 \text{ V}, \quad \Delta x = 20 \text{ cm}$ $E_A = \frac{10}{20} = 0.5 \text{ V/cm (along +ve x-axis)}$ $\text{The electric field at } B \text{ between 30 V and 20 V:}$ $\Delta V = 30 - 20 = 10 \text{ V}, \quad \Delta x = 10 \text{ cm}$ $E_B = \frac{10}{10} = 1 \text{ V/cm (along +ve x-axis)}$ $\text{The electric field at } C \text{ between 20 V and 10 V:}$ $\Delta V = 20 - 10 = 10 \text{ V}, \quad \Delta x = 5 \text{ cm}$ $E_C = \frac{10}{5} = 2 \text{ V/cm (along +ve x-axis)}$ $\text{Hence, option (3) is the correct answer.}$

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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
}