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Current Question (ID: 10452)

Question:

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is $36 \text{ g}$ and its density is $9 \text{ g/cm}^3$. If the mass of the other is $48 \text{ g}$, its density in $(\text{g/cm}^3)$ will be:

Options:

1. $\frac{4}{3}$
2. $\frac{3}{2}$
3. $3$
4. $5$

Solution:

\text{Apparent weight} = V(\rho - \sigma)g = \frac{m}{\rho}(\rho - \sigma)g \text{where } m = \text{mass of the body}, \rho = \text{density of the body} \sigma = \text{density of water} \text{If two bodies are in equilibrium then their apparent weight must be equal.} \therefore \frac{m_1}{\rho_1}(\rho_1 - \sigma) = \frac{m_2}{\rho_2}(\rho_2 - \sigma) \Rightarrow \frac{36}{9}(9 - 1) = \frac{48}{\rho_2}(\rho_2 - 1) \text{By solving we get } \rho_2 = 3.

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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
}