Import Question JSON

Current Question (ID: 8871)

Question:

$\text{A student is standing at a distance of 50 metres from the bus. As soon as the bus begins its motion with an acceleration of } 1 \text{ ms}^{-2}\text{, the student starts running towards the bus with a uniform velocity } u\text{. Assuming the motion to be along a straight road, the minimum value of } u\text{, so that the student is able to catch the bus is:}$

Options:

1. $5 \text{ ms}^{-1}$
2. $8 \text{ ms}^{-1}$
3. $10 \text{ ms}^{-1}$
4. $12 \text{ ms}^{-1}$

Solution:

$\text{Hint: } h = ut + \frac{1}{2}gt^2$ $\text{Step: Find the minimum initial speed of the student, so that the student is able to catch the bus.}$ $\text{Let the student will catch the bus after } t \text{ sec. So it will cover distance } = ut\text{.}$ $\text{Similarly, the distance travelled by the bus } = \frac{1}{2}at^2\text{.}$ $\text{For the given condition;}$ $ut = 50 + \frac{1}{2}at^2 \text{...............(i)}$ $\text{Also, to catch the bus, the velocity of the boy = the velocity of the bus after time } t$ $\Rightarrow u = at \text{...............(ii)}$ $\text{Put this in equation (i):}$ $at \times t = 50 + \frac{1}{2}at^2$ $\frac{1}{2}at^2 = 50$ $t^2 = 100 \Rightarrow t = 10 \text{ sec}$ $\text{At } t = 10 \text{ sec, } u = 10 \text{ m/s}$ $\text{Hence, option (3) is the correct answer.}$

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
}