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Current Question (ID: 8730)

Question:

$\text{When units of mass, length, and time are taken as 10 kg, 60 m and 60 s respectively, the new unit of energy becomes } x \text{ times the initial SI unit of energy. The value of } x \text{ will be:}$

Options:

1. $10$
2. $20$
3. $60$
4. $120$

Solution:

$\text{Hint: Use the relation } n_1u_1 = n_2u_2\text{.}$ $\text{Step: Find the new units of energy i.e., } x\text{.}$ $\text{Given quantities, } 1 \text{ m}' = 60 \text{ m}, 1 \text{ s}' = 60 \text{ s}, 1 \text{ (kg)}' = 10 \text{ kg}$ $\text{The dimensions of energy are given by } [ML^2T^{-2}]$ $1 \text{ J} = [1 \text{ kg}]^1[1 \text{ m}]^2[1 \text{ s}]^{-2} \ldots (1)$ $1 \text{ J}' = [1 \text{ (kg)}']^1[1 \text{ m}']^2[1 \text{ s}']^{-2} \ldots (2)$ $\text{From (1) and (2) we get;}$ $1 \text{ J}' = \frac{10 \text{ kg} \times (60 \text{ m})^2}{(60 \text{ s})^2} = 10 \text{ J}$ $\text{Hence, option (1) is the correct answer.}$

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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
}