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Current Question (ID: 16683)

Question:

$\text{A set of } 'n' \text{ equal resistors, of value } 'R' \text{ each, are connected in series to a battery of emf } 'E' \text{ and internal resistance } 'R'. \text{ The current drawn is } I. \text{ Now, if } 'n' \text{ resistors are connected in parallel to the same battery, then the current drawn becomes } 10I. \text{ The value of } 'n' \text{ is:}$

Options:

1. $10$
2. $11$
3. $20$
4. $9$

Solution:

$\text{Hint: } I = \frac{E}{nR+R}$ $\text{Step: Find the value of } 'n' \text{ resistors are connected in parallel to the same battery.}$ $\text{The current in the circuit is given by:}$ $I = \frac{E}{nR+R} \quad \ldots (1)$ $10I = \frac{E}{\frac{R}{n}+R} = \frac{nE}{R+nR} \quad \ldots (2)$ $\text{From the equations } (1) \text{ and } (2) \text{ we get:}$ $\frac{nE}{R+nR} = 10 \frac{E}{nR+R}$ $\Rightarrow n = 10$ $\text{Hence, option } (1) \text{ is the correct answer.}$

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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
}