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Current Question (ID: 19136)

Question:

$\text{A block of mass } 5 \text{ kg is (i) pushed in case } (A) \text{ and (ii) pulled in case } (B), \text{ by a force } F = 20 \text{ N, making an angle of } 30^\circ \text{ with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is } \mu = 0.2. \text{ The difference between the accelerations of the blocks, in case } (B) \text{ and case } (A) \text{ will be: } (g = 10 \text{ ms}^{-2})$

Options:

1. $3.2 \text{ ms}^{-2}$
2. $0.8 \text{ ms}^{-2}$
3. $0 \text{ ms}^{-1}$
4. $0.4 \text{ ms}^{-1}$

Solution:

$\text{Hint: } f = \mu N$ $\text{Case-A}$ $\mu = 0.2$ $N_1 = mg + 10$ $(fs)_{\text{max}} = 0.2 \times (60) = 12 \text{ N}$ $a_A = \frac{10\sqrt{2}}{5} = \frac{17.32 - 12}{5}$ $a_A = \frac{5.32}{5}$ $\text{Case-B}$ $N_2 = mg - 10 = 50 - 10 = 40 \text{ N}$ $(fs)_{\text{max}} = 0.2 \times 40 = 8 \text{ N}$ $\text{difference between acceleration } a_B - a_A = \frac{1}{5}$ $\Delta a = 0.8 \text{ m/s}^2$

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
}