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Current Question (ID: 8719)

Question:
$\text{The velocity } v \text{ of a particle at time } t \text{ is given by } v = at + \frac{b}{t+c}\text{. The dimensions of } a\text{, } b\text{, and } c \text{ are respectively:}$
Options:
  • 1. $[LT^{-2}], [L], [T]$
  • 2. $[L^2], [T] \text{ and } [LT^2]$
  • 3. $[LT^2], [LT] \text{ and } [L]$
  • 4. $[L], [LT], \text{ and } [T^2]$
Solution:
$\text{Hint: Recall the dimension analysis.}$ $\text{Step: Find the dimension of } a\text{, } b \text{ and } c$ $\text{The given function is } v = at + \frac{b}{t+c}$ $\text{According to dimensions analysis, dimensions of LHS = dimension of RHS}$ $\text{Dimension of } [v] = [at] \text{ i.e., } [LT^{-1}] = [a][T] \Rightarrow [a] = [LT^{-2}]$ $\text{Dimension of } [c] \text{ is equal to time i.e., } [t] = [c] = [T]$ $\text{And } [v] = \frac{[b]}{[t]} \Rightarrow [b] = [LT^{-1} T] = [L]$ $\text{Therefore, the dimensions of } [a], [b] \text{ and } [c] \text{ are } [LT^{-2}], [L], [T] \text{ respectively.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}