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Current Question (ID: 20208)

Question:

$\text{Two batteries with emf } 12 \text{ V and } 13 \text{ V are connected in parallel across a load resistor of } 10 \ \Omega. \text{ The internal resistances of the two batteries are } 1 \ \Omega \text{ and } 2 \ \Omega \text{ respectively. The voltage across the load lies between:}$ $1. \ 11.6 \ \text{V and } 11.7 \ \text{V}$ $2. \ 11.5 \ \text{V and } 11.6 \ \text{V}$ $3. \ 11.4 \ \text{V and } 11.5 \ \text{V}$ $4. \ 11.7 \ \text{V and } 11.8 \ \text{V}$

Options:

1. $11.6 \ \text{V and } 11.7 \ \text{V}$
2. $11.5 \ \text{V and } 11.6 \ \text{V}$
3. $11.4 \ \text{V and } 11.5 \ \text{V}$
4. $11.7 \ \text{V and } 11.8 \ \text{V}$

Solution:

$\text{Hint: } E_{eq} = \left[ \frac{E_1}{r_1} + \frac{E_2}{r_2} \right] r_{eq}$ $V_{\text{Load}} = \frac{\frac{12}{1} + \frac{13}{2}}{\frac{1}{1} + \frac{1}{2} + \frac{1}{10}}$ $= \frac{\frac{37}{2}}{\frac{10}{10+5+1}} = \frac{37}{2} \times \frac{10}{16}$ $\approx 11.56 \ \text{volt}$

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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
}