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Current Question (ID: 10455)

Question:

$\text{A metallic sphere with an internal cavity weighs 40 gwt in air and 20 gwt in water. If the density of the material with the cavity is } 8 \text{ gm/cm}^3\text{, then the volume of the cavity is:}$

Options:

1. $0 \text{ cm}^3$
2. $5 \text{ cm}^3$
3. $20 \text{ cm}^3$
4. $15 \text{ cm}^3$

Solution:

$\text{Hint: Loss in weight = Buoyant Force}$ $\text{Step 1: Find the buoyant force.}$ $F_b = W_{\text{air}} - W_{\text{water}}$ $\Rightarrow F_b = 40 - 20$ $\Rightarrow F_b = 20 \text{ gwt}$ $\text{Step 2: Find the volume of the body and the volume of the material.}$ $F_b = (\rho)_w V_b g$ $\Rightarrow 20 \times g = 1 \times V_b \times g$ $\Rightarrow V_b = 20 \text{ cm}^3$ $\text{The mass of the sphere is equal to its weight in air (since 1 gwt = 1 g mass): } M_{\text{sphere}} = 40 \text{ g}$ $\text{The density of the material is given as } 8 \text{ gm/cm}^3\text{, and the volume of the material (without the cavity) is given by;}$ $V_{\text{material}} = \frac{M_{\text{sphere}}}{\rho}$ $\Rightarrow V_{\text{material}} = \frac{40 \text{ g}}{8 \text{ g/cm}^3} = 5 \text{ cm}^3$ $\text{Step 3: Find the volume of the cavity.}$ $V_c = V_b - V_m$ $\Rightarrow V_c = 20 - 5 = 15 \text{ cm}^3$ $\text{Hence, option (4) is the correct answer.}$

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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
}