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Current Question (ID: 9755)

Question:

$\text{A system consists of three masses } m_1\text{, } m_2\text{, and } m_3 \text{ connected by a string passing over a pulley P. The mass } m_1 \text{ hangs freely, and } m_2 \text{ and } m_3 \text{ are on a rough horizontal table (the coefficient of friction } = \mu\text{.) The pulley is frictionless and of negligible mass. The downward acceleration of mass } m_1 \text{ is:}$ $\text{(Assume } m_1 = m_2 = m_3 = m \text{ and } g \text{ is the acceleration due to gravity.)}$

Options:

1. $\frac{g(1-g\mu)}{9}$
2. $\frac{2g\mu}{3}$
3. $\frac{g(1-2\mu)}{3}$
4. $\frac{g(1-2\mu)}{2}$

Solution:

$\text{Hint: On a slipping block, kinetic friction acts.}$ $\text{Step: Find the downward acceleration of mass } m_1$ $\text{From the force analysis diagram, we can identify the forces:}$ $\text{For mass } m_1 \text{ (hanging):}$ $\text{- Weight: } mg \text{ (downward)}$ $\text{- Tension: } T_2 \text{ (upward)}$ $\text{- Net force: } F_{net} = ma$ $\text{For masses } m_2 \text{ and } m_3 \text{ (on table):}$ $\text{- Normal reaction: } N = mg \text{ for each mass}$ $\text{- Friction force: } F_1 = F_2 = \mu mg \text{ for each mass}$ $\text{- Tension forces: } T_1 \text{ (connecting } m_2 \text{ and } m_3\text{)}$ $\text{Applying Newton's second law to the system:}$ $\text{Net force equation: } m_1g - \mu(m_2 + m_3)g = (m_1 + m_2 + m_3)a$ $\text{Substituting } m_1 = m_2 = m_3 = m\text{:}$ $mg - \mu(m + m)g = (m + m + m)a$ $mg - 2\mu mg = 3ma$ $g - 2\mu g = 3a$ $\Rightarrow a = \frac{g - 2\mu g}{3} = \frac{g(1-2\mu)}{3}$ $\text{Hence, option (3) is the correct answer.}$

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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
}