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Current Question (ID: 17579)

Question:

$\text{A microscope has an objective of focal length } 1.5 \text{ cm and an eyepiece of focal length } 2.5 \text{ cm.}$ $\text{If the distance between the objective and the eye-piece is } 25 \text{ cm,}$ $\text{what is the approximate value of magnification produced for the relaxed eye?}$

Options:

1. $75$
2. $110$
3. $140$
4. $25$

Solution:

$\text{Hint: } |M| = \frac{v_0}{u_0} \times \frac{D}{f_e}$ $\text{Step: Find the approximate value of magnification produced for the relaxed eye.}$ $\text{The length of the tube is } L = v_0 + f_0 + f_e$ $v_0 = L - f_e - f_0$ $v_0 = 25 - 2.5 - 1.5 = 21 \text{ cm}$ $\text{Now applying } \frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$ $\text{we have } \frac{1}{21} - \frac{1}{u_0} = \frac{1}{1.5}$ $\therefore |u_0| \approx 1.6 \text{ cm}$ $\therefore |M| = \frac{v_0}{u_0} \times \frac{D}{f_e}$ $|M| = \left( \frac{21}{1.6} \right) \left( \frac{25}{2.5} \right) \approx 140$ $\text{Hence, option (3) is the correct answer.}$

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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
}