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Current Question (ID: 16464)

Question:
$\text{The electric flux from a cube of edge } l \text{ is } \phi. \text{ What will be its value if the edge of the cube is made } 2l \text{ and the charge enclosed is halved?}$
Options:
  • 1. $\frac{1}{2} \phi$
  • 2. $2 \phi$
  • 3. $4 \phi$
  • 4. $\phi$
Solution:
$\text{Hint: Recall the concept of electric flux through a surface due to a charge.}$ $\text{Step 1: Write the equation of electric flux.}$ $\phi = \frac{q_{\text{in}}}{\varepsilon_0}$ $\text{Step 2: Write the new electric flux after increasing the length of the cube and decreasing charge enclosed by it.}$ $\phi_1 = \frac{q_{\text{in}}}{2 \varepsilon_0}$ $\text{Step 3: Now, relate } \phi \text{ and } \phi_1.$ $\frac{\phi}{\phi_1} = \frac{q_{\text{in}} / \varepsilon_0}{q_{\text{in}} / 2 \varepsilon_0} = 2$ $\Rightarrow \phi_1 = \frac{\phi}{2}$

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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
}