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Current Question (ID: 19597)

Question:

$\text{Water is flowing inside the conical type tube having a ratio of area of cross-section } 6 : 1. \text{ If the speed of the water outlet through a smaller area is } 60 \text{ m/s, then the pressure difference across these two cross-sections is:}$ $\text{(assume incompressible fluid, density of water } = 1000 \text{ kg/m}^3 )$

Options:

1. $175 \times 10^4 \text{ Pa}$
2. $175 \times 10^3 \text{ Pa}$
3. $250 \times 10^4 \text{ Pa}$
4. $250 \times 10^3 \text{ Pa}$

Solution:

$\text{Hint: } P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}$ $\text{Step 1: Find the velocity at the bigger cross-section.}$ $\text{By using the continuity equation we get, } A_1 v_1 = A_2 v_2$ $\text{Given, } \frac{A_1}{A_2} = 6 \text{ and } v_2 = 60 \text{ m/s is the velocity at the smaller cross-section.}$ $\Rightarrow 6v_1 = 60$ $\Rightarrow v_1 = 10 \text{ ms}^{-1}$ $\text{Step 2: Find the pressure difference across the two cross-sections.}$ $\text{Bernoulli's equation relates the pressure, velocity, and height between two points in a fluid flow and is given by;}$ $P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$ $\text{Here, we assume that the tube is horizontal, so the gravitational potential energy terms } (\rho gh) \text{ cancel out. Thus, Bernoulli's equation becomes:}$ $P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$ $\text{Rearranging the equation to solve for the pressure difference } (P_1 - P_2): P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$ $\text{Substitute the values in the given equation we get,}$ $\Delta P = \frac{1}{2} \rho [(60)^2 - (10)^2] \text{ Pa}$ $\rho = 1000 \text{ kg/m}^3 \text{ is the density of water.}$ $\Delta P = \frac{1}{2} \times 1000 \times [(60)^2 - (10)^2] \text{ Pa}$ $\Delta P = 175 \times 10^4 \text{ Pa}$ $\text{Hence, option (1) is the correct answer.}$

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{
  "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
}