Import Question JSON

Current Question (ID: 9743)

Question:

$\text{Three identical masses, each of mass 4 kg, are connected by massless inextensible strings. The string joining } A \text{ and } B \text{ passes over a massless frictionless pulley as shown in the figure. The tension in the string connecting mass } B \text{ and } C \text{ is:}$

Options:

1. $40\text{ N}$
2. $20\text{ N}$
3. $26.67\text{ N}$
4. $13.33\text{ N}$

Solution:

$\text{Hint: Acceleration will be the same in the magnitude of three masses.}$ $\text{Step 1: Find the acceleration of the blocks.}$ $\text{Since all masses are connected by inextensible strings, they all have the same acceleration.}$ $\text{Using Newton's second law for the entire system:}$ $a = \frac{40 + 40 - 40}{12}$ $a = \frac{40}{12} = \frac{10}{3}\text{ m/s}^2$ $\text{Step 2: Find tension in the string between B and C.}$ $\text{Consider mass C alone. The forces acting on it are:}$ $\text{- Weight } 4g \text{ downward}$ $\text{- Tension } T \text{ upward}$ $\text{Applying Newton's second law to mass C:}$ $4g - T = 4a$ $\Rightarrow T = 4g - 4a$ $= 40 - 4 \times \frac{10}{3}$ $= 40 - \frac{40}{3}$ $= \frac{120 - 40}{3} = \frac{80}{3}$ $= 26.67\text{ N}$ $\text{Hence, option (3) is the correct answer.}$

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
}