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Current Question (ID: 10240)

Question:

$\text{Two loads } P_1 \text{ and } P_2 \text{ (}P_1 > P_2\text{) are connected by a string passing over a fixed pulley. The center of gravity of loads are initially at the same height. Find the acceleration of the center of gravity of the system:}$

Options:

1. $\left(\frac{P_1 - P_2}{P_1 + P_2}\right)^2 g$
2. $\left(\frac{P_1 - P_2}{P_1 + P_2}\right) g$
3. $\left(\frac{P_1 - P_2}{P_1 + P_2}\right)^2 g$
4. $\left(\frac{P_1 + P_2}{P_1 - P_2}\right) g$

Solution:

$\text{Common acceleration of the masses:}$ $a_1 = a_2 = \left[\frac{p_1 - p_2}{p_1 + p_2}\right] g = a$ $\text{Acceleration of centre of mass:}$ $\vec{a}_{cm} = \frac{p_1 \vec{a}_1 + p_2 \vec{a}_2}{p_1 + p_2}$ $\text{Since the masses move in opposite directions with the same magnitude of acceleration:}$ $a_{cm} = \frac{(p_1 - p_2) a}{p_1 + p_2} = \left[\frac{p_1 - p_2}{p_1 + p_2}\right]^2 g$

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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
}