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Current Question (ID: 9763)

Question:

$\text{Block A has a mass of 10 kg. Between block A and the table, the coefficient of static friction is 0.2, and the coefficient of kinetic friction is also 0.2. The required mass of B to start the motion will be:}$

Options:

1. $2 \text{ kg}$
2. $2.2 \text{ kg}$
3. $4.8 \text{ kg}$
4. $200 \text{ gm}$

Solution:

$\text{Given:}$ $\text{Mass of block A, } m_A = 10 \text{ kg}$ $\text{Coefficient of static friction, } \mu_s = 0.2$ $\text{Coefficient of kinetic friction, } \mu_k = 0.2$ $\text{Analysis:}$ $\text{For block A (horizontal forces):}$ $T - \mu \times m_A g = 0$ $T = \mu \times m_A g$ $T = 0.2 \times 10 \times 10 = 20 \text{ N}$ $\text{For block B (vertical forces):}$ $m_B g - T = 0$ $m_B g = T$ $m_B \times 10 = 20$ $m_B = 2 \text{ kg}$

Import JSON File

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
}