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Current Question (ID: 9833)

Question:

$\text{The position-time } (x-t) \text{ graph of a particle of mass 2 kg is shown in the figure. Total work done on the particle from } t = 0 \text{ to } t = 4 \text{ s is:}$

Options:

1. $8 \text{ J}$
2. $4 \text{ J}$
3. $0 \text{ J}$
4. $\text{can't be determined}$

Solution:

\text{Hint: The slope of the x-t graph gives the magnitude of velocity.} \text{Step 1: Find the initial and final speeds.} \text{From the graph, we can determine the velocity at different time intervals by finding the slope:} \text{At } t = 0 \text{ to } t = 2\text{s: } v_i = \frac{2-0}{2-0} = 1 \text{ m/s} \text{At } t = 2 \text{ to } t = 4\text{s: } v_f = \frac{4-2}{4-2} = 1 \text{ m/s} \text{Note: The velocity magnitude is the same at } t = 0 \text{ and } t = 4\text{s, both equal to 1 m/s.} \text{Step 2: Use the work-energy theorem.} \text{The work-energy theorem states that the work done on an object equals the change in its kinetic energy:} W = \Delta K = K_f - K_i \text{Calculate initial kinetic energy:} K_i = \frac{1}{2}mv_i^2 = \frac{1}{2} \times 2 \times 1^2 = 1 \text{ J} \text{Calculate final kinetic energy:} K_f = \frac{1}{2}mv_f^2 = \frac{1}{2} \times 2 \times 1^2 = 1 \text{ J} \text{Therefore:} W = K_f - K_i = 1 - 1 = 0 \text{ J}

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
}