Import Question JSON

Current Question (ID: 9773)

Question:

$\text{An object of mass } m \text{ is held against a vertical wall by applying horizontal force } F \text{ as shown in the figure. The minimum value of the force } F \text{ will be: (Consider friction between wall and object.)}$

Options:

1. $\text{Less than } mg$
2. $\text{Equal to } mg$
3. $\text{Greater than } mg$
4. $\text{Cannot determine}$

Solution:

$\text{Analysis using Free Body Diagram:}$ $\text{From the free body diagram, the forces acting on the object are:}$ $\text{- Horizontal force } F \text{ (applied force)}$ $\text{- Normal force } N \text{ from the wall (horizontal, opposite to } F \text{)}$ $\text{- Weight } mg \text{ (downward)}$ $\text{- Friction force } f \text{ (upward)}$ $\text{From equilibrium in horizontal direction:}$ $N = F$ $\text{From equilibrium in vertical direction:}$ $f = mg$ $\text{For the object to remain stationary, the friction force must not exceed the maximum static friction:}$ $f \leq \mu N$ $\text{Substituting the values:}$ $mg \leq \mu N$ $\text{Since } N = F \text{:}$ $mg \leq \mu F$ $\text{Therefore:}$ $F \geq \frac{mg}{\mu}$ $\text{Since the coefficient of friction } \mu \text{ is typically less than 1 for most material combinations, we have } \frac{mg}{\mu} > mg$ $\text{Therefore, the minimum value of force } F \text{ will be greater than } mg \text{.}$

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
}