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Current Question (ID: 10163)

Question:

$\text{A rod of weight } w \text{ is supported by two parallel knife edges, } A \text{ and } B \text{, and is in equilibrium in a horizontal position. The knives are at a distance } d \text{ from each other. The centre of mass of the rod is at a distance } x \text{ from } A \text{. The normal reaction on } A \text{ is:}$

Options:

1. $\frac{wx}{d}$
2. $\frac{wd}{x}$
3. $\frac{w(d-x)}{x}$
4. $\frac{w(d-x)}{d}$

Solution:

$\text{As the weight } w \text{ balances the normal reactions.}$ $\text{For equilibrium: } N_1 + N_2 = w \text{ ... (i)}$ $\text{Now balancing torque about the COM (centre of mass),}$ $\text{anti-clockwise moment = clockwise moment}$ $N_2(d - x) = N_1 x$ $\text{Putting the value of } N_2 \text{ from Eq.(i) we get,}$ $\Rightarrow N_1 x = (w - N_1)(d - x)$ $\Rightarrow N_1 d = w(d - x)$ $\Rightarrow N_1 = \frac{w(d-x)}{d}$ $\text{Note: Balancing torque about A and B will also give the same result.}$

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
}