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Current Question (ID: 20348)

Question:

$\text{A galvanometer of resistance } 100 \, \Omega \text{ has } 50 \text{ divisions on its scale}$ $\text{and has sensitivity of } 20 \, \mu\text{A/division. It is to be converted to a}$ $\text{voltmeter with three ranges of } 0-2 \, \text{V}, \, 0-10 \, \text{V and } 0-20 \, \text{V. The}$ $\text{appropriate circuit to do so is:}$

Options:

1. $\begin{array}{l} R_1 = 1900 \, \Omega \\ R_2 = 8000 \, \Omega \\ R_3 = 10000 \, \Omega \end{array}$
2. $\begin{array}{l} R_1 = 19900 \, \Omega \\ R_2 = 9900 \, \Omega \\ R_3 = 1900 \, \Omega \end{array}$
3. $\begin{array}{l} R_1 = 1900 \, \Omega \\ R_2 = 9900 \, \Omega \\ R_3 = 19900 \, \Omega \end{array}$
4. $\begin{array}{l} R_1 = 2000 \, \Omega \\ R_2 = 8000 \, \Omega \\ R_3 = 10000 \, \Omega \end{array}$

Solution:

$\text{Hint: } V = i_g(G + R)$ $\text{sensitivity} = 20 \, \mu\text{A/div}$ $\text{Total division} = 50$ $\text{maximum current through galvanometer can be}$ $I_{\text{max}} = (50)(20 \, \mu\text{A}) = 10^{-3} \, \text{A}$ $I_{\text{max}} = \frac{2}{100 + R_1} = 10^{-3}$ $\frac{2}{10^3} = 100 + R_1$ $R_1 = 2000 - 100$ $R_1 = 1900 \, \Omega$ $R_G = 100 \, \Omega$ $I_{\text{max}} = \frac{10}{R_G + R_1 + R_2} = 10^{-3}$ $R_2 + 2000 = \frac{10}{10^3}$ $R_2 = 1000 - 2000 = 8000 \, \Omega$ $I_{\text{max}} = \frac{20}{100 + 1900 + 8000 + R_3}$ $10000 + R_3 = 20000$ $R_3 = 10000 \, \Omega$

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{
  "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
}