Import Question JSON

Current Question (ID: 19382)

Question:

$\text{Match the physical quantities given in Column-I with the physical dimensions in Column-II:}$ $\begin{array}{|c|c|} \hline \text{Column-I} & \text{Column-II} \\ \hline \text{(A) Torque} & \text{(P)} \ [ML^{-1}T^{-2}] \\ \text{(B) Stress} & \text{(Q)} \ [ML^2T^{-2}] \\ \text{(C) Pressure Gradient} & \text{(R)} \ [ML^{-2}T^{-2}] \\ \text{(D) Angular momentum} & \text{(S)} \ [ML^{-2}T^{-1}] \\ \hline \end{array}$ $\text{Choose the correct option from the given ones:}$

Options:

1. $\text{A - S, B - P, C - R, D - Q}$
2. $\text{A - Q, B - P, C - R, D - S}$
3. $\text{A - P, B - S, C - R, D - Q}$
4. $\text{A - Q, B - P, C - S, D - R}$

Solution:

$\text{Hint: } \tau = \vec{r} \times \vec{F}$ $\text{Step: Identify the correct match.}$ $(A) \text{ Torque: The torque is given by; } \tau = \vec{r} \times \vec{F}$ $[\tau] = [r][F] = [L][MLT^{-2}] = [ML^2T^{-2}]$ $(B) \text{ Stress: The Stress is given by; } \text{Stress} = \frac{[F]}{[A]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$ $(C) \text{ Pressure Gradient: The Pressure Gradient is given by; } [\text{Pressure Gradient}] = \frac{[P]}{[Z]} = \frac{[ML^{-1}T^{-2}]}{[L]} = [ML^{-2}T^{-2}]$ $(D) \text{ Angular momentum: The Angular momentum is given by; } [\text{Angular Momentum}] = [\tau][t] = [ML^2T^{-2}][T] = [ML^2]$ $\text{Therefore the correct match is A - Q, B - P, C - R, D - S.}$ $\text{Hence, option (2) is the correct answer.}$

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
}