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Current Question (ID: 9707)

Question:
$\text{A mass } M \text{ of 100 kg is suspended with the use of strings } A, B \text{ and } C \text{ as shown in the figure. The tension in the string } B \text{ will be:}$
Options:
  • 1. $100g \text{ N}$
  • 2. $\text{zero}$
  • 3. $100\sqrt{2}g \text{ N}$
  • 4. $\frac{100}{\sqrt{2}}g \text{ N}$
Solution:
\text{Hint: Balance the forces.} \text{Step 1: Draw free-body diagram.} \text{At the junction point, three forces act:} \text{- Tension } T_A \text{ in string A (vertically downward)} \text{- Tension } T_B \text{ in string B (horizontally to the left)} \text{- Tension } T_C \text{ in string C (at 45° to the horizontal)} \text{Step 2: Find the tension in string B.} \text{Since string A supports the 100 kg mass directly:} T_A = 100g \text{ N} \text{For equilibrium at the junction, resolving forces:} \text{Vertical equilibrium: } T_C \sin 45° = T_A \text{Therefore: } T_C = \frac{100g}{0.707} = 141.4g \text{ N} \text{Horizontal equilibrium: } T_C \cos 45° = T_B \text{Therefore: } T_B = 141.4g \times 0.707 = 100g \text{ N} \text{Therefore, the tension in string B is } 100g \text{ N.} \text{Hence, option (1) is the correct answer.}

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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
}