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Current Question (ID: 9733)

Question:
$\text{Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. The acceleration of the masses and the tension in the string when the masses are released are:}$
Options:
  • 1. $2 \text{ ms}^{-2}, 69 \text{ N}$
  • 2. $1 \text{ ms}^{-2}, 69 \text{ N}$
  • 3. $2 \text{ ms}^{-2}, 96 \text{ N}$
  • 4. $1 \text{ ms}^{-2}, 96 \text{ N}$
Solution:
$\text{Hint: If both blocks and string are taken as a system then tension in the string is acting as an internal force.}$ $\text{Step 1: Draw the FBD and find acceleration of the system.}$ $F_{\text{net}} = (m_1 + m_2)a$ $\Rightarrow (m_2 - m_1)g = (m_1 + m_2)a$ $\Rightarrow a = \left(\frac{m_2 - m_1}{m_2 + m_1}\right)g = \left(\frac{12 - 8}{12 + 8}\right) \times 10 = 2 \text{ m/s}^2$ $\text{Step 2: Find the tension in the string.}$ $T = \frac{2m_1m_2g}{m_1 + m_2} = \frac{2 \times 8 \times 12 \times 10}{8 + 12} = \frac{1920}{20} = 96 \text{ N}$

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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
}