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Current Question (ID: 9756)

Question:

$\text{Two masses, } A \text{ and } B\text{, of mass 4 kg and 1 kg, respectively, are connected with the help of a massless inextensible string. Mass } A \text{ is placed on a rough horizontal table, and mass } B \text{ is suspended with the help of a string passing through a smooth hole at the centre of the table. For the system to be in equilibrium, what should be the minimum value of the coefficient of friction?}$

Options:

1. $0.5$
2. $0.25$
3. $2.5$
4. $0.125$

Solution:

$\text{Hint: } f_0 = \mu R$ $\text{Step 1: Draw a diagram to analyse the forces acting on two blocks.}$ $\text{From the force diagram, we can see that:}$ $\text{- Mass A (4 kg) experiences tension T horizontally and friction force } f = \mu R \text{ opposing motion}$ $\text{- Mass B (1 kg) experiences weight 10 N downward and tension T upward}$ $\text{- Normal reaction R = 40 N acts on mass A}$ $\text{Step 2: Find the minimum value of the coefficient of friction.}$ $\text{For equilibrium of block A:}$ $T = \mu R$ $\Rightarrow T = \mu \times 40 \quad \ldots(i)$ $\text{For equilibrium of block B:}$ $T = 10 \text{ N} \quad \ldots(ii)$ $\text{From equation (i) and (ii):}$ $\Rightarrow 10 = \mu \times 40$ $\mu = \frac{10}{40} = 0.25$

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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
}