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Current Question (ID: 10520)

Question:

There are four cylindrical vessels identical in dimensions. If these are filled with equal masses of four different liquids P, Q, R, and S such that their densities are $\rho_P > \rho_Q > \rho_R > \rho_S$, then pressure at the base of vessel will be:

Options:

1. $P_P = P_Q = P_R = P_S$
2. $P_P > P_Q > P_R > P_S$
3. $P_P < P_Q < P_R < P_S$
4. Data is insufficient to predict the relation

Solution:

The pressure at the base of each vessel will be equal ($P_P = P_Q = P_R = P_S$). This is because pressure is force per unit area, and since the mass (and thus weight, which is the force) is equal, and the base area is identical, the pressure will be the same regardless of density.

Import JSON File

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
}