Import Question JSON

Current Question (ID: 9817)

Question:
$\text{A simple pendulum hangs from the roof of a train moving on horizontal rails. If the string is inclined towards the front of the train, then the train is:}$
Options:
  • 1. $\text{moving with constant velocity.}$
  • 2. $\text{in accelerated motion.}$
  • 3. $\text{in retarded motion.}$
  • 4. $\text{at rest.}$
Solution:
$\text{When a pendulum is in a non-inertial frame (accelerating train), it experiences a pseudo force.}$ $\text{Let's analyze the forces on the pendulum bob:}$ $\text{1. Weight } mg \text{ acting downward}$ $\text{2. Tension } T \text{ in the string}$ $\text{3. Pseudo force } ma \text{ acting backward (opposite to train's acceleration)}$ $\text{For the pendulum to be in equilibrium in the inclined position:}$ $\text{The net force must be zero in the train's reference frame.}$ $\text{If the string is inclined towards the front of the train, it means the pendulum bob is displaced backward relative to the train.}$ $\text{This backward displacement occurs due to the pseudo force, which acts opposite to the train's acceleration.}$ $\text{Therefore, the train must be accelerating forward.}$ $\text{If the train were:}$ $\text{- At rest or moving with constant velocity: No pseudo force, pendulum would hang vertically}$ $\text{- Decelerating (retarded motion): Pseudo force would act forward, pendulum would lean backward}$ $\text{Since the string leans forward, the train is accelerating forward.}$

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
}