Import Question JSON

Current Question (ID: 17525)

Question:

$\text{The focal length of a glass } (\mu = 1.5) \text{ lens in air is } 20 \text{ cm. If it is dipped in water } (\mu = \frac{4}{3}) \text{ its focal length in water will be:}$

Options:

1. $80 \text{ cm}$
2. $40 \text{ cm}$
3. $60 \text{ cm}$
4. $20 \text{ cm}$

Solution:

$\text{Hint: } \frac{1}{f_w} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ $\text{Step: Find the focal length in water.}$ $\text{The refractive indexes for glass } \mu_g = 1.5, \text{ for water } \mu_w = 1.33 \text{ and } \mu_a = 1$ $\text{If } f_a \text{ is the focal length of the lens in the air then}$ $\frac{1}{f_a} = \left( \frac{\mu_g}{\mu_a} - 1 \right) \times \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \ldots (1)$ $\text{If } f_w \text{ is the focal length of the lens in water then}$ $\frac{1}{f_w} = \left( \frac{\mu_g}{\mu_w} - 1 \right) \times \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \ldots (2)$ $\text{Dividing equation (1) by (2) we get,}$ $\frac{1}{f_a} = \left( \frac{\mu_g - 1}{\mu_a} \right)$ $\frac{1}{f_w} = \left( \frac{\mu_g - 1}{\mu_w} \right)$ $f_w = \left( \frac{\mu_g - \mu_a}{\mu_a} \times \frac{\mu_w}{\mu_g - \mu_w} \right) f_a$ $f_w = \left( \frac{1.5 - 1}{1} \times \frac{1.33}{1.5 - \frac{4}{3}} \right) \times 20 = 78.23 \text{ cm} \sim 80 \text{ cm}$ $\text{Hence, option (1) is the correct answer.}$

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
}