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Current Question (ID: 16735)

Question:
$\text{Three resistances } P, Q, \text{ and } R, \text{ each of } 2 \, \Omega \text{ and an unknown resistance } S \text{ form the four arms of a Wheatstone bridge circuit.}$ $\text{When the resistance of } 6 \, \Omega \text{ is connected in parallel to } S, \text{ the bridge gets balanced.}$ $\text{What is the value of } S?$
Options:
  • 1. $2 \, \Omega$
  • 2. $3 \, \Omega$
  • 3. $6 \, \Omega$
  • 4. $1 \, \Omega$
Solution:
$\text{Apply the condition of the Wheatstone bridge.}$ $\text{Step 1: Draw the equivalent circuit diagram and find the net resistance of } S \text{ and } 6 \, \Omega \text{ the resistor.}$ $\text{As resistances } S \text{ and } 6 \, \Omega \text{ are in parallel, their effective resistance} = \frac{6S}{6+S} \, \Omega$ $\text{Step 2: Find the value of the resistance } S.$ $\frac{P}{Q} = \frac{R}{\frac{6S}{6+S}}$ $\Rightarrow \frac{2}{2} = \frac{2(6+S)}{6S}$ $\Rightarrow 3S = 6 + S$ $\Rightarrow S = 3 \, \Omega$ $\text{Hence, option (2) is the correct answer.}$

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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
}