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Current Question (ID: 8863)

Question:

$\text{A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms}^{-1} \text{ to 20 ms}^{-1} \text{ while covering a distance of 135 m in } t \text{ seconds. The value of } t \text{ is:}$

Options:

1. $10$
2. $1.8$
3. $12$
4. $9$

Solution:

$\text{Hint: } v^2 = u^2 + 2as$ $\text{Step: Find the value of the time } t\text{.}$ $\text{According to the equation of motion, we can write:}$ $v^2 - u^2 = 2as \text{ (i)}$ $\text{By putting the respective values, we get:}$ $(20)^2 - (10)^2 = 2a \times 135$ $\text{Then, by solving the above equation: } a = 1.11\text{ms}^{-2}$ $\text{Now, by using the equation of motion, we know:}$ $v = u + at$ $\text{Or } t = \frac{v-u}{a}$ $\text{By putting the values, } v=20\text{ms}^{-1} \text{ and } u = 10\text{ms}^{-1} \text{ and } a = 1.11\text{ms}^{-2}$ $\text{And } t = \frac{20-10}{1.11} = 9.00 \text{ sec.}$ $\text{Therefore, the value of the time } t \text{ is 9 sec.}$ $\text{Hence, option (4) is the correct answer.}$

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
}