Import Question JSON

Current Question (ID: 10475)

Question:

$\text{Find the rate of flow of glycerin of density } 1.25 \times 10^3 \text{ kg/m}^3 \text{ through the conical section of a pipe, if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is 10 N/m}^2\text{.}$

Options:

1. $3.14 \times 10^{-4} \text{ m}^3\text{/s}$
2. $6.28 \times 10^{-4} \text{ m}^3\text{/s}$
3. $12.56 \times 10^{-4} \text{ m}^3\text{/s}$
4. $1.57 \times 10^{-4} \text{ m}^3\text{/s}$

Solution:

$\text{Hint: Apply Bernoulli's equation.}$ $\text{Step 1: Apply the continuity equation to calculate the relation between the velocities at the two ends.}$ $\text{By equation of continuity,}$ $A_1 v_1 = A_2 v_2$ $v_2 = \frac{25}{4} v_1$ $\text{Given:}$ $\Delta P = P_1 - P_2 = 10 \text{ N/m}^2$ $\rho = 1.25 \times 10^3 \text{ kg/m}^3$ $\text{Step 2: Apply Bernoulli's equation to calculate } v_1\text{.}$ $\text{Bernoulli's equation:}$ $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ $10 = \frac{1}{2}\rho \left[\left(\frac{25}{4}v_1\right)^2 - v_1^2\right]$ $v_1 = 0.02 \text{ ms}^{-1}$ $\text{So, Volumetric flow rate = } A_1 v_1 =$ $\pi r_1^2 v_1 = 3.14 \times (0.1)^2 \times 0.02$ $= 6.28 \times 10^{-4} \text{ m}^3\text{/s}$

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
}