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Current Question (ID: 20382)

Question:

$\text{In a moving coil galvanometer, if the number of turns in the coil increase by } 25\%, \text{ the change in voltage sensitivity is:}$

Options:

1. $\text{zero}$
2. $1\%$
3. $25\%$
4. $50\%$

Solution:

$\text{Hint: Voltage sensitivity } = \frac{NAB}{KR}$ $\text{Step: Find the change in the voltage sensitivity.}$ $\text{The voltage sensitivity } (S_V) \text{ of a moving coil galvanometer is given by:}$ $S_V = \frac{NAB}{KR}$ $\text{where, } N = \text{Number of turns, } B = \text{Magnetic field, } A = \text{Area of the coil, } K = \text{Torsional constant, and } R = \text{Resistance of the coil.}$ $\text{The current sensitivity of the coil is given by:}$ $S_I = \frac{\theta}{I} = \frac{NBA}{K}$ $\text{The voltage sensitivity is related to current sensitivity by the expression as:}$ $S_V = \frac{S_I}{R}$ $\text{If the number of turns } N \text{ increases by } 25\%, \text{ it would increase the current sensitivity } (S_I) \text{ proportionally.}$ $\text{However, the resistance } R \text{ of the coil also increases proportionally to } N \text{ (since more turns mean longer wire).}$ $\text{Since } S_V = \frac{S_I}{R} \text{ and both } S_I \text{ and } R \text{ increase proportionally, the voltage sensitivity remains unchanged.}$ $\text{Therefore, the change in the voltage sensitivity is zero.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}