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Current Question (ID: 20349)

Question:

$\text{A moving coil galvanometer, having a resistance } G, \text{ produces full scale deflection when a current } I_g \text{ flows through it.}$ $\text{This galvanometer can be converted into (i) an ammeter of range } 0 \text{ to } I_0 \ (I_0 > I_g) \text{ by connecting a shunt resistance } R_A \text{ to it and (ii) into a voltmeter of range } 0 \text{ to } V \ (V = GI_0) \text{ by connecting a series resistance } R_V \text{ to it. Then,}$

Options:

1. $R_A R_V = G^2 \text{ and } \frac{R_A}{R_V} = \frac{I_g}{I_0 - I_g}$
2. $R_A R_V = G^2 \left( \frac{I_g}{I_0 - I_g} \right) \text{ and } \frac{R_A}{R_V} = \left( \frac{I_0 - I_g}{I_g} \right)^2$
3. $R_A R_V = G^2 \left( \frac{I_g}{I_0 - I_g} \right) \text{ and } \frac{R_A}{R_V} = \frac{I_g}{I_0 - I_g}$
4. $R_A R_V = G^2 \text{ and } \frac{R_A}{R_V} = \left( \frac{I_g}{I_0 - I_g} \right)^2$

Solution:

$\text{Hint: } I_g G = (I_0 - I_g) R_A$ $\text{Galvanometer is converted into ammeter of range } 0 \text{ to } I_0$ $I_g G = (I_0 - I_g) R_A$ $R_A = \frac{I_g G}{(I_0 - I_g)} \quad \ldots \ldots (1)$ $\text{Galvanometer is converted into voltmeter of range } 0 \text{ to } V$ $V = I_g (G + R_V)$ $G I_0 = I_g (G + R_V)$ $R_V = \frac{G I_0}{I_g} - G$ $R_V = \frac{G (I_0 - I_g)}{I_g} \quad \ldots \ldots (2)$ $\text{So from } (1) \text{ and } (2)$ $R_A R_V = G^2$ $\frac{R_A}{R_V} = \left( \frac{I_g}{I_0 - I_g} \right)^2$

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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
}