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Current Question (ID: 9604)

Question:
$\text{Two inclined planes are placed as shown in the figure. A block is projected from the Point } A \text{ of the inclined plane } AB \text{ along its surface with a velocity just sufficient to carry it to the top point } B \text{ at a height of 10 m. After reaching the point } B \text{ the block slides down on the inclined plane } BC\text{. The time it takes to reach the point } C \text{ from the point } A \text{ is } t(\sqrt{2} + 1) \text{ s. The value of } t \text{ is: (Take } g = 10 \text{ m/s}^2)$
Options:
  • 1. $1$
  • 2. $2$
  • 3. $3$
  • 4. $4$
Solution:
\text{Hint: } a = g\sin\theta \text{Step 1: Find the time taken to reach from A to B} v = u + at 0 = u - \frac{g}{2}t_1 t_1 = \frac{2u}{g} \text{Step 2: Find the time taken to reach from B to C} v = u + at v = 0 + \frac{g}{2}t_2 t_2 = \frac{2v}{g} \text{Step 3: Find the initial velocity of the block.} \text{By applying conservation of energy:} mg(10) = \frac{1}{2}mu^2 u = \sqrt{200} \text{ m/s} \text{Step 4: Find the total time taken.} t_{\text{total}} = t_1 + t_2 \text{Step 5: Find the value of } t\text{.} \text{The time taken to reach point C from point A is } t(\sqrt{2} + 1) \text{ s.} \text{Therefore, } t = 2\text{.}

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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
}