Import Question JSON

Current Question (ID: 9708)

Question:

$\text{A metal sphere is suspended from a wall by a string. The forces acting on the sphere are shown in the figure. Which of the following statements is NOT correct?}$

Options:

1. $\vec{N} + \vec{T} + \vec{W} = 0$
2. $T^2 = N^2 + W^2$
3. $T = N + W$
4. $N = W \tan \theta$

Solution:

\text{Hint: Net force on the sphere is zero} \text{Step 1: Draw FBD of the sphere.} \text{The sphere experiences three forces:} \text{- Tension T in the string (along the string)} \text{- Normal force N from the wall (horizontal)} \text{- Weight W downward} \text{Step 2: Balance all the forces i.e., } F_{\text{net}} = 0 N + T + W = 0 \text{ (vector addition) ... (1)} \text{This confirms option 1 is correct.} \text{Step 3: Balance force in x and y-direction.} \text{Horizontal equilibrium: } N = T \sin \theta \text{Vertical equilibrium: } W = T \cos \theta \text{Therefore: } N = W \tan \theta \text{ ... (2)} \text{This confirms option 4 is correct.} \text{From the force components:} N^2 + W^2 = T^2 \text{ ... (3)} \text{This confirms option 2 is correct.} \text{However, option 3 states } T = N + W\text{, which is algebraic addition.} \text{This is incorrect because forces are vectors and must be} \text{added vectorially, not algebraically.} \text{Hence, option (3) is the correct answer as it is the incorrect statement.}

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
}