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Current Question (ID: 19144)

Question:

$\text{A particle is projected with velocity } v_0 \text{ along } x\text{-axis.}$ $\text{A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e, } ma = -\alpha x^2.$ $\text{The distance at which the particle stops:}$

Options:

1. $\left( \frac{3mv_0^2}{2\alpha} \right)^{\frac{1}{2}}$
2. $\left( \frac{2mv_0}{3\alpha} \right)^{\frac{1}{3}}$
3. $\left( \frac{2mv_0^2}{3\alpha} \right)^{\frac{1}{2}}$
4. $\left( \frac{3mv_0^2}{2\alpha} \right)^{\frac{1}{3}}$

Solution:

$\text{Hint: } a = v \frac{dv}{dx}$ $F = -\alpha x^2$ $ma = -\alpha x^2$ $a = \frac{-\alpha x^2}{m}$ $v \frac{dv}{dx} = -\frac{\alpha}{m} x^2$ $\int_{v_0}^{0} v dv = \int_{0}^{x} -\frac{\alpha}{m} x^2 \, dx$ $\left( \frac{v^2}{2} \right)_{v_0}^{0} = -\frac{\alpha}{m} \left( \frac{x^3}{3} \right)_{0}^{x}$ $-\frac{v_0^2}{2} = -\frac{\alpha}{m} \frac{x^3}{3}$ $x = \left( \frac{3mv_0^2}{2\alpha} \right)^{\frac{1}{3}}$

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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
}