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Current Question (ID: 10168)

Question:

$\text{On a rough horizontal surface (coefficient of friction } \mu\text{), a cubical block of side 'a' and mass m is projected horizontally. The net torque on the block about its centre of mass till the block stops is equal to:}$

Options:

1. $\text{zero}$
2. $\frac{1}{2}\mu mga$
3. $\mu mga$
4. $mga$

Solution:

$\text{Hint: } \vec{\tau} = \vec{r} \times \vec{F}$ $\text{Step 1: Draw the free body diagram for the block}$ $\text{The forces acting on the block are:}$ $\text{- Normal force N (upward)}$ $\text{- Weight Mg (downward)}$ $\text{- Friction force f (opposing motion)}$ $\text{Here, } \tau_C = \tau_N$ $\text{Step 2: Identify the angular acceleration about center of mass}$ $\text{Since the block is sliding without rotation, the angular acceleration } \alpha = 0$ $\text{Therefore, } \tau_C = 0$ $\text{The net torque about the center of mass is zero because all forces either pass through the center of mass or have equal and opposite moments that cancel out.}$

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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
}