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Current Question (ID: 10469)

Question:
$\text{The water flows through a frictionless tube with a varying cross-section as shown in the figure. The variation of pressure } P \text{ at the point } x \text{ along the axis is roughly given by:}$
Options:
  • 1. $\text{Graph showing stepped pressure profile}$
  • 2. $\text{Graph showing constant pressure profile}$
  • 3. $\text{Graph showing pressure drop in narrow section}$
  • 4. $\text{Graph showing V-shaped pressure profile}$
Solution:
$\text{Hint: Apply Bernoulli's theorem.}$ $\text{Step: Find the correct } P-x \text{ graph.}$ $\text{According to the Bernoulli's theorem and continuity equation, we get;}$ $P + \frac{1}{2}\rho v^2 = \text{constant}$ $A_1 v_1 = A_2 v_2$ $\text{We can say that if the area of the cross-section decreases, then the velocity of the water increases and the pressure decreases and vice versa. In a uniform cross-section, the pressure will be constant.}$ $\text{The graph of the velocity versus distance } (v-x) \text{ and pressure versus distance } (P-x) \text{ is shown in the figure below:}$ $\text{Hence, option (3) is the correct answer.}$

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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
}