Import Question JSON

Current Question (ID: 17507)

Question:

$\text{A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels up, to the surface of the liquid and moves along its surface (see figure).}$ $\text{How fast is the light traveling in the liquid?}$

Options:

1. $1.8 \times 10^8 \text{ m/s}$
2. $2.4 \times 10^8 \text{ m/s}$
3. $3.0 \times 10^8 \text{ m/s}$
4. $1.2 \times 10^8 \text{ m/s}$

Solution:

$\text{Hint: } \sin \theta_c = \frac{1}{\mu}$ $\text{Step: Find the speed of the light traveling in the liquid.}$ $\text{Given that, the coin is at the bottom of the liquid container.}$ $\text{We know, that the critical angle is the angle of incidence in the denser}$ $\text{medium for which the angle of refraction in the rarer medium is } 90^\circ.$ $\text{As shown in the figure, a light ray from the coin will not emerge out of the}$ $\text{liquid, if the incidence angle } (i) > \text{ critical angle } (\theta_c).$ $\text{Therefore, the minimum radius } R \text{ corresponds to } i = \theta_c.$ $\text{In } \triangle SAB$ $\text{We know,}$ $\sin \theta_c = \frac{1}{\mu}$ $\frac{R}{H} = \tan \theta_c$ $\text{Or } R = H \tan \theta_c$ $\text{Or } R = H \frac{1}{\sqrt{\mu^2 - 1}}$ $\text{Given, } R = 3 \text{ cm, } h = 4 \text{ cm}$ $\text{Hence, } \frac{3}{4} = \frac{1}{\sqrt{\mu^2 - 1}}$ $\text{Or } \mu^2 = \frac{25}{9} \text{ or } \mu = \frac{5}{3}$ $\text{But } \mu = \frac{c}{v} \text{ or } v = \frac{c}{\mu}$ $v = \frac{3 \times 10^8}{5/3}$ $\Rightarrow v = 1.8 \times 10^8 \text{ m/s}$ $\text{Hence, option (1) is the correct answer.}$

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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
}