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Current Question (ID: 9761)

Question:

$\text{A plank with a box on it at one end is gradually raised at the other end. As the angle of inclination with the horizontal reaches } 30°, \text{ the box starts to slip and slides } 4.0 \text{ m down the plank in } 4.0 \text{ s. The coefficients of static and kinetic friction between the box and the plank, respectively, will be:}$

Options:

1. $0.6 \text{ and } 0.6$
2. $0.6 \text{ and } 0.5$
3. $0.5 \text{ and } 0.6$
4. $0.4 \text{ and } 0.3$

Solution:

\text{Given: A plank with a box on it. One end is gradually raised until the angle of inclination is } 30°\text{, the box starts to slip and slides down } 4\text{ m of the plank in } 4\text{ sec as shown in the figure.} \text{The coefficient of static friction: } \mu_s = \tan 30° = \frac{1}{\sqrt{3}} \approx 0.577 \text{For the sliding motion, using kinematic equation:} s = ut + \frac{1}{2}at^2 \text{Here, } u = 0 \text{ and } a = g(\sin\theta - \mu_k \cos\theta) \text{Substituting values:} 4 = \frac{1}{2} \times g(\sin 30° - \mu_k \cos 30°) \times (4)^2 4 = \frac{1}{2} \times 10 \times \left(\frac{1}{2} - \mu_k \times \frac{\sqrt{3}}{2}\right) \times 16 4 = 80 \times \left(\frac{1}{2} - \frac{\sqrt{3}}{2}\mu_k\right) \frac{4}{80} = \frac{1}{2} - \frac{\sqrt{3}}{2}\mu_k 0.05 = 0.5 - \frac{\sqrt{3}}{2}\mu_k \frac{\sqrt{3}}{2}\mu_k = 0.5 - 0.05 = 0.45 \mu_k = \frac{0.45 \times 2}{\sqrt{3}} = \frac{0.9}{\sqrt{3}} \approx 0.52 \text{Therefore, the coefficient of kinetic friction between the box and the plank is approximately } 0.52

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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
}