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Current Question (ID: 9768)

Question:

$\text{If } \mu \text{ between block A and inclined plane is 0.5 and that between block B and the inclined plane is 0.8, then the normal reaction between blocks A and B will be:}$

Options:

1. $180 \text{ N}$
2. $216 \text{ N}$
3. $0$
4. $\text{none of these}$

Solution:

$\text{Hint: Check if both the blocks will move.}$ $\text{Step 1: Find the limiting friction for both blocks.}$ $\text{The normal force acting on the block is given by;}$ $N = mg \cos \theta$ $\text{The limiting friction for the block A is given by;}$ $f_A = \mu_A N$ $f_A = 0.5 \times \frac{4}{5} m_A g$ $f_A = 0.4m_A g$ $\text{Similarly, the limiting friction for the block B is given by;}$ $f_B = \mu_B N$ $f_B = 0.64m_B g$ $\text{Step 2: Check if both the blocks will move.}$ $\text{In the case of block B:}$ $0.6m_B g < 0.64m_B g$ $\text{If the limiting value of friction is greater than the downward force the block B will not move.}$ $\text{In the case of block A:}$ $0.6m_A g > 0.4m_A g$ $\text{Here, the limiting value of friction is lesser than the downward force of the block so block A will move.}$ $\text{Hence, there will be no contact between block A and block B and the normal reaction between blocks A and B will be equal to zero.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}