Import Question JSON

Current Question (ID: 18947)

Question:

$\text{Match Column-I with Column-II and choose the correct option.}$ $\begin{array}{|c|c|} \hline \text{Column-I} & \text{Column-II} \\ \hline \text{I. Torque} & (a) \ [M^0LT^{-2}] \\ \text{II. Stress} & (b) \ [ML^{-1}T^{-2}] \\ \text{III. Coefficient of viscosity} & (c) \ [ML^{-1}T^{-1}] \\ \text{IV. Gravitational potential gradient} & (d) \ [ML^2T^{-2}] \\ \hline \end{array}$ $\text{Choose the correct option from the given ones:}$

Options:

1. $\text{I} \rightarrow (a), \text{II} \rightarrow (c), \text{III} \rightarrow (b), \text{IV} \rightarrow (d)$
2. $\text{I} \rightarrow (d), \text{II} \rightarrow (b), \text{III} \rightarrow (c), \text{IV} \rightarrow (a)$
3. $\text{I} \rightarrow (d), \text{II} \rightarrow (c), \text{III} \rightarrow (b), \text{IV} \rightarrow (a)$
4. $\text{I} \rightarrow (a), \text{II} \rightarrow (c), \text{III} \rightarrow (d), \text{IV} \rightarrow (b)$

Solution:

$\text{Hint: Recall the dimensional analysis.}$ $\text{Step: Find the dimensions of the given physical quantity one by one.}$ $\text{Torque} = r \times F = [ML^2 T^{-2}]$ $\text{Stress} = \frac{F}{A} = [ML^{-1} T^{-2}]$ $\text{Coefficient of viscosity, } \eta = \frac{F}{A (dv/dx)} = [ML^{-1}T^{-1}]$ $\text{Gravitational potential gradient} = \frac{dV}{dr} = [M^0LT^{-2}]$ $\text{Therefore, the correct match is; I} \rightarrow \text{d; II} \rightarrow \text{c; III} \rightarrow \text{b; IV} \rightarrow \text{a.}$ $\text{Hence, option (3) is the correct answer.}$

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
}