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Current Question (ID: 19029)

Question:

$\text{The velocity } (v) \text{ and time } (t) \text{ graph of a body in a straight-line motion is shown in the figure.}$ $\text{The point } S \text{ occurs at } 4.333 \text{ seconds.}$ $\text{The total distance covered by the body in } 6 \text{ s is:}$

Options:

1. $12 \text{ m}$
2. $\frac{49}{12} \text{ m}$
3. $11 \text{ m}$
4. $\frac{37}{3} \text{ m}$

Solution:

$\text{Hint: The area under } v-t \text{ graph represents the distance traveled.}$ $\text{Step: Find the total distance covered by the body in } 6 \text{ s.}$ $\text{The distance } OS \text{ is given by;} \ OS = 4 + \frac{1}{3} = \frac{13}{3}$ $\text{The distance } SD \text{ is given by;} \ SD = 2 - \frac{1}{3} = \frac{5}{3}$ $\text{The area of the trapezium } OABS \text{ is given by;} \ A_1 = \frac{1}{2} \left[ \frac{13}{3} + 1 \right] \cdot 4 = \frac{32}{3}$ $\text{The area of the triangle } SCD \text{ is given by;} \ A_2 = \frac{1}{2} \times \frac{5}{3} \times 2 = \frac{5}{3}$ $\text{The total distance covered by the body is given by;} \ \text{Distance} = A_1 + A_2 = \frac{32}{3} + \frac{5}{3} = \frac{37}{3}$ $\text{Hence, option } (4) \text{ is the correct answer.}$

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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
}