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Current Question (ID: 8737)

Question:

$\text{The position of a particle at time } t \text{ is given by the relation } x(t) = \left(\frac{v_0}{\alpha}\right)\left(1 - e^{-\alpha t}\right)\text{, where } v_0 \text{ is a constant and } \alpha > 0\text{. The dimensions of } v_0 \text{ and } \alpha \text{ are respectively:}$

Options:

1. $[M^0L^1T^{-1}] \text{ and } [T^{-1}]$
2. $[M^0L^1T^0] \text{ and } [T^{-1}]$
3. $[M^0L^1T^{-1}] \text{ and } [LT^{-1}]$
4. $[M^0L^1T^{-1}] \text{ and } [T]$

Solution:

\text{Hint: Input of an exponential function is dimensionless} \text{Step: Find the dimensions of } \alpha \text{ and } v_0 \text{The dimensions of } \alpha t \text{ are } [M^0L^0T^0] [\alpha t] = [M^0L^0T^0] [\alpha] = [M^0L^0T^{-1}] = [T^{-1}] \text{Since } x(t) \text{ represents position, } [x(t)] = [L]. \text{ The term } (1 - e^{-\alpha t}) \text{ is dimensionless.} \left[\frac{v_0}{\alpha}\right] = [M^0L^1T^0] = [L] [v_0] = [\alpha][L] = [T^{-1}][L] = [M^0L^1T^{-1}] \text{Therefore, } [v_0] = [M^0L^1T^{-1}] \text{ and } [\alpha] = [T^{-1}] \text{Hence, option (1) is the correct answer.}

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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
}