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Current Question (ID: 20461)

Question:
$\text{A compass needle of oscillation magnetometer oscillates 20 times per minute at a place } P \text{ of dip } 30^\circ. \text{ The number of oscillations per minute becomes 10 at another place } Q \text{ of } 60^\circ \text{ dip. The ratio of the total magnetic field at the two places: } (B_Q : B_P) \text{ is:}$
Options:
  • 1. $\sqrt{3} : 4$
  • 2. $4 : \sqrt{3}$
  • 3. $\sqrt{3} : 2$
  • 4. $2 : \sqrt{3}$
Solution:
$\text{Hint: } T = 2\pi \sqrt{\frac{I}{B_H M}}$ $\text{Step 1: Find the magnetic field at a given point.}$ $T = 2\pi \sqrt{\frac{I}{M B_H}}$ $B_H = \frac{4\pi^2 I}{M T^2}$ $B_H = B \cos \theta$ $B = \frac{4\pi^2 I f^2}{M \cos \theta}$ $\text{Step 2: Find the ratio of the magnetic fields at } P \text{ and } Q.$ $\frac{B_P}{B_Q} = \frac{\cos 60^\circ \times 20^2}{\cos 30^\circ \times 10^2}$ $= \frac{1}{2} \times \frac{2}{\sqrt{3}} \times \frac{400}{100}$ $\frac{B_P}{B_Q} = \frac{4}{\sqrt{3}}$ $\frac{B_Q}{B_P} = \frac{\sqrt{3}}{4}$

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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
}