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Current Question (ID: 16736)

Question:

$\text{Figure (a) below shows a Wheatstone bridge in which } P, Q, R, S \text{ are fixed resistances, } G \text{ is a galvanometer, and } B \text{ is a battery.}$ $\text{For this particular case, the galvanometer shows zero deflection.}$ $\text{Now, only the positions of } B \text{ and } G \text{ are interchanged, as shown in the figure (b).}$ $\text{The new deflection of the galvanometer:}$

Options:

1. $\text{is to the left}$
2. $\text{is to the right}$
3. $\text{is zero}$
4. $\text{depends on the values of } P, Q, R, S$

Solution:

$\text{Hint: } \frac{P}{Q} = \frac{S}{R}$ $\text{Step 1: Identify the condition of zero deflection in case (a).}$ $\text{For null deflection } V_B = V_D$ $\Rightarrow \frac{P}{Q} = \frac{S}{R} \text{ or } \frac{P}{S} = \frac{Q}{R}$ $\text{Step 2: Check the balance condition when the positions of } B \text{ and } G \text{ are interchanged.}$ $\text{When battery } B \text{ and galvanometer } G \text{ are interchanged, the position of the galvanometer is as shown below.}$ $\text{Now, the ratio of resistances across the galvanometer is given by;}$ $\Rightarrow \frac{P}{Q} = \frac{S}{R} \text{ still valid}$ $\text{Hence, the galvanometer still shows zero deflection because Wheatstone's bridge is balanced.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}