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Current Question (ID: 19223)

Question:

$\text{A point particle of mass } m, \text{ moves along the uniformly rough track } PQR \text{ as shown in the figure. The coefficient of friction, between the particle and the rough track equals } \mu. \text{ The particle is released, from rest, from the point } P \text{ and it comes to rest at a point } R. \text{ The energies, lost by the ball, over the parts, } PQ \text{ and } QR, \text{ of the track, are equal to each other, and no energy is lost when the particle changes direction from } PQ \text{ to } QR. \text{ The values of the coefficient of friction } \mu \text{ and the distance } x \ (= QR), \text{ are, respectively close to:}$

Options:

1. $0.2 \text{ and } 6.5 \text{ m}$
2. $0.2 \text{ and } 3.5 \text{ m}$
3. $0.29 \text{ and } 3.5 \text{ m}$
4. $0.29 \text{ and } 6.5 \text{ m}$

Solution:

$\text{Hint: Apply work-energy theorem.}$ $\text{From work-energy theorem}$ $W_g + W_f + W_{f_2} = \Delta K$ $mgh + 2W = 0 \quad [\therefore W_{f_1} = W_f]$ $W = -\frac{mgh}{2}$ $\text{and } W_{f_1} = -\frac{mgh}{2}$ $-\mu mg \cos 30^\circ (4) = -\frac{mgh}{2}$ $\mu = 0.29$ $\text{and } W_{f_2} = +\frac{mgh}{2}$ $-\mu mgx = -\frac{mgh}{2}$ $x = \frac{h}{2\mu} = 3.5 \text{ m}$

Import JSON File

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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
}