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Current Question (ID: 10457)

Question:
$\text{In a horizontal pipe of a non-uniform cross-section, water flows with a velocity of } 1 \text{ ms}^{-1} \text{ at a point where the diameter of the pipe is } 20 \text{ cm. The velocity of water (ms}^{-1}\text{) at a point where the diameter of the pipe is } 5 \text{ cm is:}$
Options:
  • 1. $8$
  • 2. $16$
  • 3. $24$
  • 4. $32$
Solution:
$\text{Hint: Apply continuity equation.}$ $\text{Step 1: The continuity equation is given as } A_1v_1 = A_2v_2$ $\text{Step 2: Area of pipe = } \pi r^2$ $\text{Thus,}$ $\frac{\pi d_1^2 v_1}{4} = \frac{\pi d_2^2 v_2}{4}$ $\Rightarrow v_2 = v_1 \times \left(\frac{d_1}{d_2}\right)^2$ $\text{Given: } v_1 = 1 \text{ m/s, } d_1 = 20 \text{ cm, } d_2 = 5 \text{ cm}$ $\Rightarrow v_2 = 1 \times \left(\frac{20}{5}\right)^2 = 1 \times (4)^2 = 1 \times 16 = 16 \text{ m/s}$ $\text{Therefore, the velocity at the point where diameter is 5 cm is 16 m/s.}$

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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
}