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Current Question (ID: 10466)

Question:

The following figure shows the flow of liquid through a horizontal pipe. Three tubes $A$, $B$ and $C$ are connected to the pipe. The radii of the tubes $A$, $B$ and $C$ at the junction are respectively $2 ext{ cm}$, $1 ext{ cm}$ and $2 ext{ cm}$. It can be said that:

Options:

1. The height of the liquid in the tube $A$ is maximum.
2. The height of the liquid in the tubes $A$ and $B$ is the same.
3. The height of the liquid in all three tubes is the same.
4. The height of the liquid in the tubes $A$ and $C$ is the same.

Solution:

As cross-section areas of both the tubes $A$ and $C$ are same and the tube is horizontal. Hence according to equation of continuity $v_A = v_C$ and therefore according to Bernoulli's theorem $P_A = P_C$ i.e. height of liquid is same in both the tubes $A$ and $C$.

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
}