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Current Question (ID: 8748)

Question:

$\text{The dimensions of resistivity in terms of } M, L, T, \text{ and } Q \text{ where } Q \text{ stands for the dimensions of charge, will be:}$

Options:

1. $[ML^3T^{-1}Q^{-2}]$
2. $[ML^3T^{-2}Q^{-1}]$
3. $[ML^2T^{-1}Q^{-1}]$
4. $[MLT^{-1}Q^{-1}]$

Solution:

$\text{Hint: } \rho = \frac{RA}{l}$ $\text{Step: Find the dimensional formula of resistivity of the material.}$ $\text{The resistivity of the material is given by, } \rho = \frac{RA}{l}$ $\text{The dimensions of resistance are given by Ohm's law i.e., } R = \frac{V}{I}$ $R = \frac{V}{I} = \frac{\text{Work}}{\text{Charge}} \times \frac{1}{\text{Current}} = \frac{[ML^2T^{-2}]}{[Q]} \times \frac{[T]}{[Q]} = \frac{[ML^2T^{-1}]}{[Q]^2}$ $\text{The dimensions of resistivity are given by;}$ $\rho = \frac{[ML^2T^{-1}]}{[Q]^2} \times \frac{[L]^2}{[L]} = \frac{[ML^3T^{-1}]}{[Q]^2}$ $\text{Therefore, the dimensions of resistivity are } [ML^3T^{-1}Q^{-2}].$ $\text{Hence, option (1) is the correct answer.}$

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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
}