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Current Question (ID: 19155)

Question:

$\text{Two masses, } M_1 \text{ and } M_2, \text{ are connected by a light, inextensible string that passes over a frictionless pulley.}$ $\text{When the mass } M_2 \text{ is twice that of } M_1, \text{ the system experiences an acceleration } a_1.$ $\text{Similarly, when } M_2 \text{ is three times the mass of } M_1, \text{ the system's acceleration becomes } a_2.$ $\text{The ratio } \frac{a_1}{a_2} \text{ will be:}$

Options:

1. $\frac{1}{3}$
2. $\frac{2}{3}$
3. $\frac{3}{2}$
4. $\frac{1}{2}$

Solution:

$\text{Hint: } F = ma$ $\text{Step: Find the ratio of acceleration } \frac{a_1}{a_2}.$ $\text{The acceleration of the two masses pulley system is given by:}$ $a = \frac{M_2 - M_1}{M_1 + M_2} g \quad [M_2 > M_1]$ $\text{The acceleration of the first case when } M_2 = 2M_1 \text{ is given by:}$ $a_1 = \left(\frac{2M_1 - M_1}{2M_1 + M_1}\right) g = \frac{g}{3}$ $\text{The acceleration of the second case when } M_2 = 3M_1 \text{ is given by:}$ $a_2 = \left(\frac{3M_1 - M_1}{3M_1 + M_1}\right) g = \frac{2g}{4} = \frac{g}{2}$ $\text{The ratio of two accelerations is given by:}$ $\frac{a_1}{a_2} = \frac{\frac{g}{3}}{\frac{g}{2}} = \frac{2}{3}$ $\text{Hence, option (2) is the correct answer.}$

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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
}