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Current Question (ID: 8752)

Question:

$\text{A small steel ball of radius } r \text{ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity } \eta. \text{ After some time the velocity of the ball attains a constant value known as terminal velocity } v_T. \text{ The terminal velocity depends on (i) the mass of the ball } m \text{ (ii) } \eta \text{ (iii) } r \text{ and (iv) acceleration due to gravity } g. \text{ Which of the following relations is dimensionally correct:}$

Options:

1. $v_T \propto \frac{mg}{\eta r}$
2. $v_T \propto \frac{\eta r}{mg}$
3. $v_T \propto \eta r mg$
4. $v_T \propto \frac{mgr}{\eta}$

Solution:

\text{Hint: } v_T = \frac{2a^2(\rho - \sigma)g}{9\eta} \text{Step: Checking the dimensional formula for the terminal velocity.} \text{By substituting the dimensions of each quantity in RHS in option (1) we get,} \frac{mg}{\eta r} = \frac{[M \times LT^{-2}]}{[ML^{-1}T^{-1} \times L]} = [LT^{-1}] \text{So, option (1) gives the dimensions of velocity.} \text{Therefore, } v_T \propto \frac{mg}{\eta r} \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
}