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Current Question (ID: 10450)

Question:

A solid is suspended from an independent support in a liquid placed on a weighing machine, due to which the weight of the liquid:

Options:

1. Increases
2. Decreases
3. Remain unchanged
4. May increase or decrease

Solution:

Hint: Draw FBD for liquid. Step 1: Draw FBD for block & liquid [Diagram showing a solid suspended in a liquid with forces T (tension), $F_B$ (buoyant force), and W (weight) acting on the solid, and a separate diagram showing the liquid on a weighing machine with forces N (normal force), $F_B$ (buoyant force acting on the liquid by the solid), and W (weight of the liquid) acting on it.] $N = W + F_B$ Based on the free-body diagram for the liquid and the weighing machine, the normal force N from the weighing machine on the liquid is equal to the weight of the liquid (W) plus the buoyant force ($F_B$) exerted on the liquid by the solid. Since the solid is submerged, it displaces some volume of liquid, and the liquid exerts an upward buoyant force on the solid. By Newton's third law, the solid exerts an equal and opposite downward force on the liquid. This downward force increases the effective weight measured by the weighing machine. Therefore, the weight of the liquid appears to increase.

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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
}