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Current Question (ID: 19627)

Question:

$\text{When the temperature of a metal wire is increased from } 0^\circ \text{C to } 10^\circ \text{C, its length increases by } 0.02\%. \text{ What is the percentage decrease in its mass density?}$ $1.\ 0.008\%$ $2.\ 0.06\%$ $3.\ 0.8\%$ $4.\ 2.3\%$

Options:

1. $0.008\%$
2. $0.06\%$
3. $0.8\%$
4. $2.3\%$

Solution:

$\text{Hint: } \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V}$ $\text{Step: Find the percentage decrease in its mass density.}$ $\text{Mass density is given by:}$ $\rho = \frac{M}{V}$ $\Rightarrow \frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V} \quad [\because M = \text{constant}]$ $\Rightarrow \frac{\Delta \rho}{\rho} = -\frac{\Delta l}{l} - \frac{\Delta A}{A} \quad [\because V = l \times A]$ $\Rightarrow \frac{\Delta \rho}{\rho} = -\frac{\Delta l}{l} - 2\alpha \Delta T$ $\Rightarrow \frac{\Delta \rho}{\rho} = -\frac{\Delta l}{l} - \frac{2\Delta l}{l} \quad [\because \frac{\Delta l}{l} = \alpha \Delta T]$ $\Rightarrow \frac{\Delta \rho}{\rho} = -\frac{3\Delta l}{l}$ $\Rightarrow \frac{\Delta \rho}{\rho} = -3 \times 0.02\% \quad [\text{Given; } \frac{\Delta l}{l} = 0.02\%]$ $\Rightarrow \frac{\Delta \rho}{\rho} = -0.06\%$ $\text{Therefore, the percentage decrease in its mass density is } 0.06\%. \text{ Hence, option (2) is the correct answer.}$

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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
}