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Current Question (ID: 20265)

Question:

$\text{Circuit I is converted to voltmeter after adding resistance } R \text{ and current in circuit I is } 1 \text{ mA.}$ $\text{Circuit II is converted to ammeter after adding resistance } r \text{ as shown.}$ $\text{If current through battery in circuit II is } 10 \text{ mA and current through galvanometer is } 1 \text{ mA,}$ $\text{then find } R \text{ and } r \text{ if resistance of galvanometer is } 54 \, \Omega.$

Options:

1. $R = 49946 \, \Omega, \; r = 54 \, \Omega$
2. $R = 6 \, \Omega, \; r = 49946 \, \Omega$
3. $R = 49946 \, \Omega, \; r = 49946 \, \Omega$
4. $R = 49946 \, \Omega, \; r = 6 \, \Omega$

Solution:

$\text{Hint: } I = \frac{V_g}{R+G}$ $1 \times 10^{-2} = \frac{50}{54+R}$ $54 + 50000 = R$ $R = 49946 \, \Omega$ $\text{Now, } i_g = 1 \text{ mA and current through } r = 9 \text{ mA}$ $\text{So, } 54 \times 1 = r \times 9$ $r = \left( \frac{54}{9} \right) = 6 \, \Omega$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}