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Current Question (ID: 12054)

Question:

The dimensions of gravitational constant G and the moment of inertia are respectively

Options:

1. $ML^3T^{-2};ML^2T^0$
2. $M^{-1}L^3T^{-2};ML^2T^0$
3. $M^{-1}L^3T^{-2};M^{-1}L^2T$
4. $ML^3T^{-2};M^{-1}L^2T$

Solution:

The dimension of gravitational constant G can be derived from Newton's law of gravitation, $F = G\frac{m_1m_2}{d^2}$. Rearranging, $G = \frac{Fd^2}{m_1m_2}$. So, $[G] = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$. The dimension of moment of inertia is $[I]=[Mr^2]=[ML^2]$. Thus, the dimensions are $[M^{-1}L^3T^{-2}]$ and $[ML^2T^0]$.

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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
}