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Current Question (ID: 20223)

Question:

$\text{A battery of } 3.0 \text{ V is connected to a resistor dissipating } 0.5 \text{ W of power.}$ $\text{If the terminal voltage of the battery is } 2.5 \text{ V, the power}$ $\text{dissipated within the internal resistance is:}$

Options:

1. $0.10 \text{ W}$
2. $0.072 \text{ W}$
3. $0.125 \text{ W}$
4. $0.50 \text{ W}$

Solution:

$\text{Hint: } P = i^2 R$ $PR = 0.5 \text{ W}$ $\Rightarrow i^2 R = 0.5 \text{ W}$ $\text{Also, } V = E - ir$ $2.5 = 3 - ir$ $\Rightarrow ir = 0.5$ $\text{Power dissipated across 'r' } P_r = i^2 r$ $\text{Now } iR = 2.5$ $ir = 0.5$ $\text{On dividing: } \frac{R}{r} = 5$ $\text{Now } \frac{P_R}{P_r} = \frac{i^2 R}{i^2 r} \Rightarrow \frac{P_R}{P_r} = \frac{R}{r} \Rightarrow \frac{P_R}{P_r} = 5$ $\Rightarrow P_r = \frac{P_R}{5}$ $\Rightarrow P_r = \frac{0.50}{5} \Rightarrow 0.10 \text{ W}$

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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
}