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Current Question (ID: 16857)

Question:

$\text{On connecting a shunt of } 10 \, \Omega, \text{ the deflection in a moving coil galvanometer}$ $\text{falls from } 40 \text{ divisions to } 6 \text{ divisions. What is the resistance of the}$ $\text{galvanometer?}$

Options:

1. $\frac{120}{3} \, \Omega$
2. $\frac{30}{7} \, \Omega$
3. $\frac{170}{3} \, \Omega$
4. $\frac{150}{7} \, \Omega$

Solution:

$\text{Let } G \text{ be the resistance of the galvanometer.}$ $\text{Using the formula: } \frac{G}{G + S} = \frac{n_2}{n_1}$ $\text{where } S = 10 \, \Omega, n_1 = 40, n_2 = 6$ $\frac{G}{G + 10} = \frac{6}{40}$ $\Rightarrow 40G = 6(G + 10)$ $\Rightarrow 40G = 6G + 60$ $\Rightarrow 34G = 60$ $\Rightarrow G = \frac{60}{34} = \frac{30}{17} \, \Omega$

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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
}