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Current Question (ID: 20304)

Question:

$\text{Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length } l_1 \text{ is } 40 \text{ cm. Now an unknown resistance } x \text{ is connected in series with } P \text{ and the new balancing length is found to be } 80 \text{ cm measured from the same end. Then the value of } x \text{ will be:}$

Options:

1. $10 \, \Omega$
2. $20 \, \Omega$
3. $30 \, \Omega$
4. $40 \, \Omega$

Solution:

$\frac{P}{Q} = \frac{l_1}{l_2}$ $\text{Given: } P = 4 \, \Omega, \; l_1 = 40 \text{ cm}, \; l_2 = 80 \text{ cm}$ $\frac{4}{Q} = \frac{40}{80}$ $Q = 8 \, \Omega$ $\text{Now, } x + P = Q$ $x + 4 = 8$ $x = 4 \, \Omega$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}