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Current Question (ID: 16524)

Question:

$\text{An electric field } \vec{E} = 10x\hat{i} \text{ exists in a certain region of space. Then the potential difference } V = V_0 - V_A, \text{ where } V_0 \text{ is the potential at the origin and } V_A \text{ is the potential at } x = 2 \text{ m is:}$

Options:

1. $10 \text{ V}$
2. $-20 \text{ V}$
3. $+20 \text{ V}$
4. $-10 \text{ V}$

Solution:

$\text{Hint: The relation between electric field and potential is given by, } E = -\frac{dV}{dx}. $ $\text{Step 1: Since } E = -\frac{dV}{dx} $ $\text{Step 2: Integrating both sides} $ $\Rightarrow \int_{V_0}^{V} dV = \int_{0}^{x} -E dx $ $\Rightarrow V - V_0 = -10 \int_{0}^{x} x dx $ $\Rightarrow V - V_0 = -5x^2 $ $\Rightarrow V_0 - V_A = 5x^2 $ $= 5 \times 4 = 20 \text{ V}$

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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
}