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Current Question (ID: 19443)

Question:

$\text{On the } x\text{-axis at a distance } x \text{ from the origin, the gravitational field}$ $\text{due to a mass distribution is given by } \frac{Ax}{(x^2+a^2)^{3/2}} \text{ in the } x\text{-direction.}$ $\text{The magnitude of gravitational potential on the } x\text{-axis at a distance } x,$ $\text{taking its value to be zero at infinity, is:}$

Options:

1. $\frac{1}{(x^2+a^2)^{3/2}}$
2. $A(x^2+a^2)^{1/2}$
3. $A(x^2+a^2)^{3/2}$
4. $\frac{A}{(x^2+a^2)^{1/2}}$

Solution:

$\text{Hint: } \vec{E} = -\frac{dV}{dr}$ $\text{Given } E_G = \frac{Ax}{(x^2+a^2)^{3/2}}, \quad V_\infty = 0$ $\int_{V_\infty}^{V_x} dV = -\int_{\infty}^{x} \vec{E}_G \cdot dx$ $V_x - V_\infty = -\int_{\infty}^{x} \frac{Ax}{(x^2+a^2)^{3/2}} dx$ $\text{put } x^2 + a^2 = z$ $2x \, dx = dz$ $V_x - 0 = -\int_{\infty}^{x} \frac{Adz}{2(z)^{3/2}} = \left[ \frac{A}{z^{1/2}} \right]_{\infty}^{x} = \left[ \frac{A}{(x^2+a^2)^{1/2}} \right]_{\infty}^{x}$ $V_x = \frac{A}{(x^2+a^2)^{1/2}} - 0 = \frac{A}{(x^2+a^2)^{1/2}}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}