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Current Question (ID: 10453)

Question:

$\text{A solid sphere of density } \eta \text{ (} > 1\text{) times lighter than water is suspended in a water tank by a string tied to its base as shown in fig. If the mass of the sphere is m, then the tension in the string is given by:}$

Options:

1. $\left(\frac{\eta - 1}{\eta}\right)mg$
2. $\eta mg$
3. $\frac{mg}{\eta}$
4. $(\eta - 1)mg$

Solution:

\text{Density of the solid sphere, } \rho = \frac{\sigma}{\eta} \Rightarrow \sigma = \rho \times \eta \text{where } \sigma \text{ is the density of the water.} \text{Tension in the string } T = \text{upthrust} - \text{weight of sphere} T = V\sigma g - V\rho g = V\eta\rho g - V\rho g \quad (\text{As } \sigma = \eta\rho) T = (\eta - 1)V\rho g = (\eta - 1)mg \text{Explanation:} \text{The sphere is denser than water since } \eta > 1\text{, so it would sink without the string.} \text{Forces acting on the sphere:} \text{1. Weight (downward): } mg \text{2. Buoyant force (upward): } V\sigma g = V\eta\rho g = \eta mg \text{3. Tension (upward): } T \text{For equilibrium: } T + mg = \text{Buoyant force} T + mg = \eta mg \text{Therefore: } T = \eta mg - mg = (\eta - 1)mg

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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
}