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Current Question (ID: 16655)

Question:

$\text{If there are two bulbs of } (40 \text{ W, } 200 \text{ V}), \text{ and } (100 \text{ W, } 200 \text{ V}), \text{ then the correct relation for their resistance is:}$ $1. \ R_{40} < R_{100}$ $2. \ R_{40} > R_{100}$ $3. \ R_{40} = R_{100}$ $4. \ \text{no relation can be predicted}$

Options:

1. $R_{40} < R_{100}$
2. $R_{40} > R_{100}$
3. $R_{40} = R_{100}$
4. $\text{no relation can be predicted}$

Solution:

$\text{Hint: } P = \frac{V^2}{R}$ $\text{Step: Find the relation of resistances of the given two bulbs.}$ $\text{By using the formula of power, we know that;}$ $\Rightarrow P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} \ldots (1)$ $\text{Substituting the values of the first bulb in equation (1), we get;}$ $\Rightarrow R_{40} = \frac{V^2_{40}}{P_{40}} = \frac{200 \times 200}{40} \Rightarrow R_{40} = 1000 \, \Omega$ $\text{Substituting the values of the second bulb in equation (1), we get;}$ $\Rightarrow R_{100} = \frac{V^2_{100}}{P_{100}} = \frac{200 \times 200}{100} \Rightarrow R_{100} = 400 \, \Omega$ $\text{Comparing the values of both resistances } R_{40} > R_{100}. \text{ Hence, option (2) is the correct answer.}$

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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
}