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Current Question (ID: 19157)

Question:

$\text{A smooth circular groove has a smooth vertical wall as shown in the figure. A block of mass } m \text{ moves against the wall with a speed } v. \text{ Which of the following curve represents the correct relation between the normal reaction on the block by the wall } (N) \text{ and the speed of the block } (v)?$

Options:

1. $N \text{ vs } v^2 \text{ (parabolic)}$
2. $N \text{ vs } v \text{ (linear)}$
3. $N \text{ vs constant}$
4. $N \text{ vs } v \text{ (non-linear)}$

Solution:

$\text{Hint: } F_c = \frac{mv^2}{R}$ $\text{Step: Find the relation between the normal reaction } N \text{ and speed } v.$ $\text{The block moves in a circular path. Any object moving in a circular path experiences a centripetal force that keeps it moving along the curve. This centripetal force is provided entirely by the normal reaction force from the wall.}$ $\text{For an object moving with a speed } v \text{ in a circular path of radius } R, \text{ the centripetal force } F_c \text{ is given by: } F_c = \frac{mv^2}{R} \ldots (1)$ $\text{This centripetal force is supplied by the normal reaction force from the wall, so: } N = \frac{mv^2}{R} \ldots (2)$ $\text{From the above equation, we can clearly see that the normal reaction force } N \text{ is directly proportional to the square of the speed } v. \text{ The relation between } N \text{ and } v \text{ is: } N \propto v^2$ $\text{The relation between the normal reaction } N \text{ and speed } v \text{ is parabolic. The correct curve should show } N \text{ increasing quadratically with } v. \text{ Therefore, the graph of } N \text{ versus } v \text{ is an upward-opening parabola as shown in the diagram below;}$

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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
}