Import Question JSON

Current Question (ID: 16519)

Question:

$\text{The figure shows some of the equipotential surfaces. The magnitude and direction of the electric field are given by:}$

Options:

1. $200 \text{ V/m, making an angle } 120^\circ \text{ with the } x\text{-axis}$
2. $100 \text{ V/m, pointing towards the negative } x\text{-axis}$
3. $200 \text{ V/m, making an angle } 60^\circ \text{ with the } x\text{-axis}$
4. $100 \text{ V/m, making an angle } 30^\circ \text{ with the } x\text{-axis}$

Solution:

$\text{Hint: } dV = -\vec{E} \cdot d\vec{r}$ $\text{Step: Find the magnitude and direction of the electric field.}$ $\text{As potential decreases in the direction of } \vec{E}, \vec{E} \text{ would be in such a direction that } V \text{ decreases along its direction. Also } \vec{E} \text{ is perpendicular to equipotential surfaces.}$ $\text{Using } dV = -\vec{E} \cdot d\vec{r}$ $\Rightarrow \Delta V = -E \times \Delta r \cos \theta$ $\Rightarrow E = \frac{-\Delta V}{\Delta r \cos \theta}$ $\Rightarrow E = \frac{10 \times 10^{-2} \cos 120^\circ}{-10}$ $E = \frac{10 \times 10^{-2} (-\sin 30^\circ)}{-10}$ $\Rightarrow E = \frac{-10^2}{-1/2} = 200 \text{ V/m}$ $\text{The direction of } \vec{E} \text{ be perpendicular to the equipotential surface i.e., at } 120^\circ \text{ with } x\text{-axis.}$ $\text{Hence, option (1) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}