Import Question JSON

Current Question (ID: 19159)

Question:

$\text{Given below are two statements:}$ $\text{Assertion (A):}$ \text{If the weight of the lift is equal to the tension force in the cable, then it moves with uniform velocity.}$ $\text{Reason (R):}$ \text{If the lift moves downward with an acceleration, the contact force between the boy's feet and the lift floor is more than the weight of the boy.}$

Options:

1. $\text{Both (A) and (R) are True and (R) is the correct explanation of (A).}$
2. $\text{Both (A) and (R) are True but (R) is not the correct explanation of (A).}$
3. $\text{(A) is True but (R) is False.}$
4. $\text{Both (A) and (R) are False.}$

Solution:

$\text{Hint: } T = mg$ $\text{Step: Analyse each statement one by one.}$ $\text{When the weight of the lift (gravitational force) equals the tension in the cable, the net force is zero, resulting in uniform velocity (either at rest or moving at a constant speed). Therefore, (A) is correct.}$ $\text{When the lift is accelerating downward, the contact force (normal force) on the boy would be less than his weight, not more i.e., } N = m(g - a). \text{ The boy would feel lighter as the lift accelerates downwards.}$ $\text{Therefore, (R) is incorrect.}$ $\text{Therefore, (A) is true but (R) is false.}$ $\text{Hence, option (3) is the correct answer.}$

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
}