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Current Question (ID: 10276)

Question:

$\text{In the figure given below, } O \text{ is the centre of an equilateral triangle } ABC \text{ and } \vec{F}_1, \vec{F}_2, \vec{F}_3 \text{ are three forces acting along the sides } AB, BC \text{ and } AC. \text{ What should be the magnitude of } \vec{F}_3 \text{ so that total torque about } O \text{ is zero?}$

Options:

1. $|\vec{F}_3| = |\vec{F}_1| + |\vec{F}_2|$
2. $|\vec{F}_3| = |\vec{F}_1| - |\vec{F}_2|$
3. $|\vec{F}_3| = F_1 + 2F_2$
4. $\text{Not possible}$

Solution:

$\text{From the centre distance of three sides equal}$ $\text{For an equilateral triangle, the perpendicular distance from the centre O to each side is equal. Let this distance be } x.$ $\text{The torque due to each force about point O:}$ $\text{Torque due to } \vec{F}_1: \tau_1 = F_1 \times x$ $\text{Torque due to } \vec{F}_2: \tau_2 = F_2 \times x$ $\text{Torque due to } \vec{F}_3: \tau_3 = F_3 \times x$ $\text{For zero net torque about O, considering the direction of rotation:}$ $F_1 x + F_2 x - F_3 x = 0$ $F_3 = F_1 + F_2$ $\text{Therefore: } |\vec{F}_3| = |\vec{F}_1| + |\vec{F}_2|$

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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
}