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Current Question (ID: 9764)

Question:

$\text{The upper half of an inclined plane of inclination } \theta \text{ is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom if the coefficient of friction between the block and the lower half of the plane is given by:}$

Options:

1. $\mu = 2/\tan \theta$
2. $\mu = 2 \tan \theta$
3. $\mu = \tan \theta$
4. $\mu = 1/\tan \theta$

Solution:

$\text{Hint: The change in kinetic energy of the block is zero.}$ $\text{Given: Upper half is smooth, lower half is rough, block starts from rest and comes to rest at bottom}$ $\text{Let } s \text{ be the length along the incline for each half}$ $\text{Force approach:}$ $\text{Step 1: Find the final velocity for the first half}$ $\text{For top half (smooth):}$ $a = g\sin\theta$ $v^2 = 0 + 2a\left(\frac{s}{2}\right) = gs\sin\theta \quad \text{...(1)}$ $\text{Step 2: Find the final velocity for the second half}$ $\text{For bottom half (rough):}$ $a = -(\mu g\cos\theta - g\sin\theta)$ $0 = v^2 + 2a\left(\frac{s}{2}\right) \quad \text{...(2)}$ $\text{From equations (1) and (2):}$ $\mu g\cos\theta = 2g\sin\theta$ $\mu = 2\tan\theta$ $\text{Energy approach:}$ $\text{Step 1: Apply work-energy theorem}$ $\text{Gain in kinetic energy = Loss of kinetic energy due to friction}$ $mg\left(s \times \sin\theta\right) = \left(\mu mg\cos\theta\right)\frac{s}{2}$ $\mu = 2\tan\theta$

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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
}