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Current Question (ID: 9903)

Question:
$\text{A body of mass } m \text{ accelerates uniformly from rest to } v_1 \text{ in time } t_1\text{. As a function of time } t\text{, the instantaneous power delivered to the body will be:}$
Options:
  • 1. $\frac{mv_1t}{t_1}$
  • 2. $\frac{mv_1^2t}{t_1}$
  • 3. $\frac{mv_1t^2}{t_1}$
  • 4. $\frac{mv_1^2t}{t_1^2}$
Solution:
\text{Given:} \text{Body starts from rest (initial velocity } u = 0\text{)} \text{Final velocity } = v_1 \text{Time taken } = t_1 \text{Motion is uniformly accelerated} \text{For uniformly accelerated motion:} \text{Acceleration } a = \frac{v_1 - 0}{t_1} = \frac{v_1}{t_1} \text{Velocity at time } t: v = u + at = 0 + \frac{v_1}{t_1} \cdot t = \frac{v_1t}{t_1} \text{Instantaneous power is given by:} P = F \cdot v = ma \times v \text{Substituting the values:} P = m \times \frac{v_1}{t_1} \times \frac{v_1t}{t_1} = m \times \frac{v_1^2t}{t_1^2} \text{Therefore, the instantaneous power delivered to the body as a function of time is:} P = \frac{mv_1^2t}{t_1^2}

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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
}