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Current Question (ID: 16631)

Question:

$\text{A circuit consists of five identical conductors as shown in the figure. The two similar conductors are added as indicated by the dotted lines. The ratio of resistances before and after the addition will be:}$

Options:

1. $\frac{7}{5}$
2. $\frac{3}{5}$
3. $\frac{5}{3}$
4. $\frac{6}{5}$

Solution:

$\text{Hint: Identify the Wheatstone Bridge.}$ $\text{Step 1: Find the initial effective resistance between the given two points.}$ $R_i = 5 \times 1 \Rightarrow 5 \, \Omega$ $\text{Step 2: Find the final effective resistance between the given two points.}$ $\text{After addition,}$ $\text{By Wheatstone condition, the circuit reduces to as drawn.}$ $\text{The equivalent resistance is given by;}$ $\Rightarrow R_f = 1 + \frac{2 \times 2}{2 + 2} + 1$ $\Rightarrow R_f = 1 + 1 + 1 = 3 \, \Omega$ $\text{The ratio of resistances before and after addition is given by;}$ $\Rightarrow \frac{R_i}{R_f} = \frac{5}{3}$ $\text{Hence, option (3) is the correct answer.}$

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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
}