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Current Question (ID: 10166)

Question:

$\text{A meter stick is balanced on a knife edge at its center. When two coins, each of the mass 5 gm are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the meter stick?}$

Options:

1. $\text{66 gm}$
2. $\text{56 gm}$
3. $\text{76 gm}$
4. $\text{79 gm}$

Solution:

$\text{Hint: Apply the concept of rotational equilibrium.}$ $\text{Step: Balance the torque about the point of equilibrium.}$ $\text{Mass of meter stick = } m'$ $\text{The mass of each coin = 5 g}$ $\tau_N = \tau_W$ $\text{we know that } \tau = r \times F$ $\text{Setting up the torque equation about the 45 cm mark:}$ $\text{Torque due to coins = Torque due to meter stick}$ $10g \times (45 - 12) = m'g \times (50 - 45)$ $10g \times 33 = m'g \times 5$ $m' = \frac{10 \times 33}{5} = 66 \text{ gm}$ $\text{Hence, option (1) is the correct answer.}$

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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
}