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Current Question (ID: 8822)

Question:

$\text{If the velocity of a particle is } v = At + Bt^2\text{, where } A \text{ and } B \text{ are constants, then the distance travelled by it between 1 s and 2 s is:}$

Options:

1. $3A + 7B$
2. $\frac{3}{2}A + \frac{7}{3}B$
3. $\frac{A}{2} + \frac{B}{3}$
4. $\frac{3A}{2} + 4B$

Solution:

$\text{Hint: The distance can be found by integrating the velocity w.r.t. time.}$ $\text{Step 1: Find the relation between the velocity and the displacement.}$ $\text{As we know, the velocity is defined as:}$ $v = \frac{dx}{dt}$ $\Rightarrow \frac{dx}{dt} = At + Bt^2$ $\Rightarrow dx = (At + Bt^2) dt$ $\text{Step 2: Find the distance covered by the particle.}$ $\text{Integrating both sides, we get,}$ $\int_{x_1}^{x_2} dx = \int_1^2 (At + Bt^2) dt$ $\Delta x = x_2 - x_1 = A \int_1^2 t dt + B \int_1^2 t^2 dt$ $= A \left[\frac{t^2}{2}\right]_1^2 + B \left[\frac{t^3}{3}\right]_1^2$ $= \frac{A}{2}(2^2 - 1^2) + \frac{B}{3}(2^3 - 1^3)$ $= \frac{3A}{2} + \frac{7B}{3}$ $\text{Hence, option (2) is the correct answer.}$

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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
}