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Current Question (ID: 20209)

Question:
$\text{On interchanging the resistances, the balance point of a meter bridge shifts to the left by } 10 \text{ cm.}$ $\text{The resistance of their series combination is } 1 \text{ k}\Omega. \text{ How much was the resistance on the left slot before interchanging the resistances?}$
Options:
  • 1. $990 \ \Omega$
  • 2. $505 \ \Omega$
  • 3. $550 \ \Omega$
  • 4. $910 \ \Omega$
Solution:
$\text{Hint: } \frac{R_1}{R_2} = \frac{l_1}{l_2}$ $\text{As on interchanging resistors balance point move towards left by } 10 \text{ cm.}$ $\frac{P}{Q} = \frac{55}{45}$ $\frac{P}{Q} = \frac{11}{9} \Rightarrow 9P = 11Q$ $P + Q = 1000$ $\frac{11Q}{9} + Q = 1000$ $\frac{20Q}{9} = 1000$ $Q = 50 \times 9 = 450 \ \Omega$ $\therefore P = 550 \ \Omega$

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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
}