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Current Question (ID: 9813)

Question:

$\text{The kinetic energy } K \text{ of a particle moving in a circular path varies with the distance covered } S \text{ as } K = aS^2 \text{ where } a \text{ is constant. The angle between the tangential force and the net force acting on the particle is: (}R \text{ is the radius of the circular path)}$

Options:

1. $\tan^{-1}\left(\frac{S}{R}\right)$
2. $\tan^{-1}\left(\frac{R}{S}\right)$
3. $\tan^{-1}\left(\frac{a}{R}\right)$
4. $\tan^{-1}\left(\frac{R}{a}\right)$

Solution:

\text{Note: Solution image not provided. Based on physics analysis:} \text{Given: } K = aS^2 \text{ where } a \text{ is constant} \text{From } K = \frac{mv^2}{2} = aS^2\text{, we get } v^2 = \frac{2aS^2}{m} \text{Taking derivative: } 2v\frac{dv}{dt} = \frac{4aS}{m}\frac{dS}{dt} = \frac{4aSv}{m} \text{Therefore: } \frac{dv}{dt} = \frac{2aS}{m} \text{Tangential acceleration: } a_t = \frac{dv}{dt} = \frac{2aS}{m} \text{Centripetal acceleration: } a_c = \frac{v^2}{R} = \frac{2aS^2}{mR} \text{The angle } \theta \text{ between tangential force and net force:} \tan \theta = \frac{F_c}{F_t} = \frac{a_c}{a_t} = \frac{\frac{2aS^2}{mR}}{\frac{2aS}{m}} = \frac{S}{R} \text{Therefore: } \theta = \arctan\left(\frac{S}{R}\right)

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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
}