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Current Question (ID: 16849)

Question:

$\text{A galvanometer has a coil resistance of } 100 \, \Omega \text{ and gives a full-scale deflection for } 30 \, \text{mA of current. If it is to work as a voltmeter in the } 30 \, \text{V range, how much resistance does it require to be added?}$

Options:

1. $900 \, \Omega$
2. $1800 \, \Omega$
3. $500 \, \Omega$
4. $1000 \, \Omega$

Solution:

$\text{(1) Hint: The potential drop across the galvanometer and the resistance will be equal to } 30 \, \text{V.}$ $\text{Step 1: Find the net potential drop in the circuit.}$ $\text{Let the resistance added in series } = R$ $\text{The potential drop in the circuit } = (100 + R) \times 30 \times 10^{-3} \, \text{V}$ $\text{Step 2: Find the value of } R.$ $\text{The net potential drop in the circuit } = 30 \, \text{V}$ $\Rightarrow (100 + R) \times 30 \times 10^{-3} = 30$ $R = \frac{30}{30 \times 10^{-3}} - 100 = 900 \, \Omega$

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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
}