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Current Question (ID: 8868)

Question:

$\text{The velocity of a particle moving along a straight line with constant acceleration } a \text{ reduces to } \frac{1}{5} \text{ of its initial velocity in time } \tau\text{. The total time taken by the body till its velocity becomes zero is:}$

Options:

1. $\frac{4\tau}{3}$
2. $\frac{5\tau}{4}$
3. $\frac{4\tau}{5}$
4. $\frac{5\tau}{4}$

Solution:

$\text{Hint: Apply the first equation of motion to find the acceleration. i.e., } v = u + at\text{.}$ $\text{Step 1: Find the acceleration of the particle.}$ $\text{Let Initial velocity be } v\text{.}$ $\frac{v}{5} = v - a\tau \text{...(1)}$ $a = \frac{4v}{5\tau}$ $\text{Step 2: Use again the first equation of motion:}$ $0 = v - at$ $t = \frac{v}{a}$ $t = \frac{5\tau}{4}$ $\text{Hence, option (2) is the correct answer.}$

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
}