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Current Question (ID: 16744)

Question:

$\text{The metre bridge shown is in a balanced position with } \frac{P}{Q} = \frac{l_1}{l_2}. \text{ If we now interchange the position of the galvanometer and the cell, will the bridge work? If yes, what will be the balanced condition?}$

Options:

1. $\text{Yes, } \frac{P}{Q} = \frac{l_1 - l_2}{l_1 + l_2}$
2. $\text{No, no null point}$
3. $\text{Yes, } \frac{P}{Q} = \frac{1}{l_1}$
4. $\text{Yes, } \frac{P}{Q} = \frac{1}{l_2}$

Solution:

$\text{Hint: For a balanced Wheatstone bridge, } \frac{P}{Q} = \frac{R}{S} \text{ or } \frac{P}{R} = \frac{Q}{S}.$ $\text{Step 1: Find the initial balanced condition.}$ $\text{Let resistance for segments of meterbridge wire be,}$ $R_1 = kl_1 \text{ and } R_2 = kl_2$ $\text{Initially,}$ $\frac{P}{Q} = \frac{R_1}{R_2} = \frac{l_1}{l_2}$ $\text{Step 2: Interchange the cell and the galvanometer.}$ $\text{When cell and galvanometer are interchanged,}$ $\text{Resistance of branch ABD, } R_B = P + R_1$ $= \frac{Ql_1}{l_2} + kl_1 \quad \text{(1)}$ $\text{Resistance of branch ACD, } R_C = Q + R_2$ $= Q + kl_2 \quad \text{(2)}$ $\text{From (1) and (2) . . . .}$ $R_B = \frac{l_1}{l_2} R_C$ $\text{Step 3: Find the potential drop across the galvanometer.}$ $\text{Current i by battery distributes in branches ABD and ACD:}$ $i_1 = \frac{R_C i}{R_B + R_C} = \frac{i_2}{l_1 + l_2} \text{ and } i_2 = \frac{R_B i}{R_B + R_C} = \frac{i_1}{l_1 + l_2}$ $\frac{V_{AB}}{V_{AC}} = \frac{i_1 P}{i_2 Q} \Rightarrow \frac{l_2}{l_1} \times \frac{l_1}{l_2} = 1$ $\Rightarrow V_{AB} = V_{AC}$ $\text{So, } V_B = V_C \Rightarrow \text{zero deflection in galvanometer}$

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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
}