Import Question JSON

Current Question (ID: 19441)

Question:

$\text{The height } 'h' \text{ at which the weight of a body will be the same as that at the same depth } 'h' \text{ from the surface of the earth is (Radius of the earth is } R \text{ and effect of the rotation of the earth is neglected):}$

Options:

1. $\frac{\sqrt{5}R - R}{2}$
2. $\frac{\sqrt{5}}{2} R - R$
3. $\frac{R}{2}$
4. $\frac{\sqrt{3}R - R}{2}$

Solution:

$\text{Hint: } g = \frac{g_0}{\left(1 + \frac{h}{R}\right)^2}$ $\text{* } M = \text{mass of earth}$ $M_1 = \text{mass of shaded portion}$ $R = \text{Radius of earth}$ $\text{* } M_1 = \frac{M}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi (R - h)^3 = \frac{M(R-h)^3}{R^3}$ $\text{* Weight of body is same at P and Q}$ $\text{i.e. } mg_P = mg_Q$ $g_P = g_Q$ $\frac{GM_1}{(R-h)^2} = \frac{GM}{(R+h)^2}$ $\frac{GM(R-h)^3}{(R-h)^2 R^3} = \frac{GM}{(R+h)^2}$ $\left(R-h\right)\left(R+h\right)^2 = R^3$ $R^3 - hR^2 - h^2R - h^3 + 2R^2h - 2Rh^2 = R^3$ $R^2 - Rh^2 - h^3 = 0$ $R^2 - Rh - h^2 = 0$ $h^2 + Rh - R^2 = 0 \Rightarrow h = \frac{-R \pm \sqrt{R^2 + 4R^2}}{2}$ $i.e. \ h = \frac{-R + \sqrt{5}R}{2} = \left(\frac{\sqrt{5} - 1}{2}\right)R$

Import JSON File

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
}