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Current Question (ID: 19580)

Question:
$\text{A hydraulic press can lift } 100 \text{ kg when a mass } m \text{ is placed on the smaller piston.}$ $\text{If the diameter of the larger piston is increased by a factor of } 4 \text{ and the diameter of the smaller piston is reduced by a factor of } 4,$ $\text{while keeping the same mass } m \text{ on the smaller piston, the press can lift:}$
Options:
  • 1. $2500 \text{ kg}$
  • 2. $50000 \text{ kg}$
  • 3. $25600 \text{ kg}$
  • 4. $550000 \text{ kg}$
Solution:
$\text{Apply Pascal's law}$ $\Delta P_1 = \Delta P_2$ $\Rightarrow \frac{100 \times g}{A_2} = \frac{mg}{A_1} \quad \cdots (1)$ $\text{Let } m \text{ mass can lift } M_0 \text{ in the second case then:}$ $\frac{M_0 g}{16 A_2} = \frac{mg}{A_1/16} \left[ A = \pi \frac{d^2}{4} \right] \quad \cdots (2)$ $\text{From equations } (1) \text{ and } (2) \text{ we get:}$ $\frac{M_0}{16 \times 16} = 100$ $\Rightarrow M_0 = 25600 \text{ Kg}$ $\text{Hence, option } (3) \text{ is the correct answer.}$

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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
}