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Current Question (ID: 18949)

Question:

$\text{The density } (\rho) \text{ of a body depends on the force applied } (F), \text{ its speed, and time of motion } (t) \text{ according to the relation:}$ $\rho = KF^a v^b t^c,$ $\text{where } K \text{ is a dimensionless constant. Then the values of } a, b, \text{ and } c \text{ are:}$

Options:

1. $a = 1, \ b = -4, \ c = -2$
2. $a = 2, \ b = -4, \ c = -1$
3. $a = -1, \ b = -4, \ c = 2$
4. $a = 1, \ b = 4, \ c = -2$

Solution:

$\text{Hint: Use dimensional analysis.}$ $\text{Step: Find the values of } a, b, \text{ and } c.$ $\rho = KF^a v^b t^c$ $\text{According to the principle of homogeneity, in any physical equation,}$ $\text{the dimensions of all terms on both sides of the equation must be}$ $\text{consistent. Hence:}$ $[ML^{-3}] = [MLT^{-2}]^a [LT^{-1}]^b [T]^c$ $\Rightarrow [ML^{-3}] = [M]^a [L]^a [T]^{-2a} [L]^b [T]^{-b} [T]^c$ $\Rightarrow [ML^{-3}] = [M]^a [L]^{a+b} [T]^{-2a-b+c}$ $\Rightarrow a = 1, \ a + b = -3, \ -2a - b + c = 0$ $\Rightarrow a = 1, \ b = -4, \ c = -2$ $\text{Hence, option (1) is the correct answer.}$

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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
}