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Current Question (ID: 9770)

Question:

$\text{A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60, while the kinetic coefficient of friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If } g = 10 \text{ m/s}^2\text{, the resulting acceleration of the slab will be:}$

Options:

1. $1.0 \text{ m/s}^2$
2. $1.47 \text{ m/s}^2$
3. $1.52 \text{ m/s}^2$
4. $6.1 \text{ m/s}^2$

Solution:

$\text{Let us assume that blocks move together.}$ $\text{Then } a = \frac{F}{m} = \frac{100}{50}$ $\Rightarrow a = 2 \text{ m/s}^2$ $f_{\text{max}} = \mu_s \times N = \mu_s \times m_1 g$ $= 0.60 \times 10 \times 10 = 60 \text{ N}$ $\text{Here, } f_{\text{max}} < \text{applied force.}$ $\text{So, there is no relative motion between block and slab.}$ $\text{So, } F = \mu mg = 0.4 \times 10 \times 10 = 40 \text{ N}$ $\text{The resulting acceleration of the block,}$ $a = \frac{40}{40} = 1.0 \text{ m/s}^2$

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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
}