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Current Question (ID: 20159)

Question:

$\text{Assume that an electric field } \vec{E} = 30x^2 \hat{i} \text{ exists in space. Then the potential difference } V_A - V_0, \text{ where } V_0 \text{ is the potential at the original and } V_A \text{ the potential at } x = 2 \text{ m is:}$ $1.\ -120 \text{ V}$ $2.\ -80 \text{ V}$ $3.\ 80 \text{ V}$ $4.\ 120 \text{ V}$

Options:

1. $-120 \text{ V}$
2. $-80 \text{ V}$
3. $80 \text{ V}$
4. $120 \text{ V}$

Solution:

$\text{Hint: } \Delta V = - \int \vec{E} \cdot d\vec{r}$ $V_A - V_0 = - \int_{0}^{r = 2} \vec{E} \cdot d\vec{r}$ $= - \int_{x=0}^{r=2} 30x^2 dx = -30 \left[ \frac{x^3}{3} \right]_{0}^{2}$ $= -10 \left[ 2^3 - 0^3 \right] = -80 \text{ V}$ $\text{[Note: Unit of potential is given wrong in this question. Its unit should be volt not the joule.]}$

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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
}