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Current Question (ID: 8750)

Question:

$\text{The density of a material in a CGS system of units is 4 grams/cm}^3. \text{ In a system of units in which the unit of length is 10 cm and the unit of mass is 100 grams, the value of the density of the material will be:}$

Options:

1. $0.04$
2. $0.4$
3. $40$
4. $400$

Solution:

$\text{Hint: } \rho = \frac{M}{V}$ $\text{Step: Find the value of the density of the material.}$ $\text{By the concept of unit conversion.}$ $\Rightarrow n_1\rho_1 = n_2\rho_2$ $\Rightarrow 4 \times (\text{gm cm}^{-3}) = n_2(100 \text{ gm} \times (10 \text{ cm})^{-3})$ $\Rightarrow 4 \times (\text{gm cm}^{-3}) = n_2(10^2\text{gm} \times 10^{-3}(\text{cm})^{-3})$ $\Rightarrow n_2 = \frac{4}{10^{-1}} = 40$ $\text{Hence, option (3) is the correct answer.}$

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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
}