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Current Question (ID: 18955)

Question:

$\text{Which of the following equations is dimensionally consistent according to the principle of homogeneity, where } T \text{ is the time period, } G \text{ is the gravitational constant, } M \text{ is mass, and } r \text{ is the orbital radius?}$

Options:

1. $T^2 = \frac{4\pi^2 r^3}{GM}$
2. $T^2 = \frac{4\pi^2 r^2}{GM}$
3. $T^2 = 4\pi^2 r^3$
4. $T^2 = \frac{4\pi^2 r}{GM^2}$

Solution:

$\text{Hint: Use dimensional analysis. LHS and RHS must have the same dimensions.}$ $\text{Step: Examine each option.}$ $\text{(1) } T^2 = \frac{4\pi^2 r^3}{GM}$ $\Rightarrow [T^2] = \frac{[L^3]}{[M^{-1}L^3T^{-2}][M]}$ $\Rightarrow [T^2] = [T^2]$ $\text{(2) } T^2 = \frac{4\pi^2 r^2}{GM}$ $\Rightarrow [T^2] = \frac{[L^3]}{[M^{-1}L^2T^{-2}][M]}$ $\Rightarrow [T^2] \neq [LT^2]$ $\text{(3) } T^2 = 4\pi^2 r^3$ $\Rightarrow [T^2] \neq [L^3]$ $\text{(4) } T^2 = \frac{4\pi^2 r}{GM^2}$ $\Rightarrow [T^2] = \frac{[L]}{[M^{-1}L^2T^{-2}][M]}$ $\Rightarrow [T^2] \neq [L^{-1}T^2]$ $\text{Hence, option (1) is the correct answer.}$

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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
}