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Current Question (ID: 18933)

Question:

$\text{Velocity } (v) \text{ and acceleration } (a) \text{ in two systems of units } 1 \text{ and } 2 \text{ are related as } v_2 = \frac{n}{m^2} v_1 \text{ and } a_2 = \frac{a_1}{mn} \text{ respectively. Here } m \text{ and } n \text{ are constants. The relations for distance and time in the two systems, respectively, are:}$

Options:

1. $\frac{n^3}{m^3} L_1 = L_2 \text{ and } \frac{n^2}{m} T_1 = T_2$
2. $L_1 = \frac{n^4}{m^2} L_2 \text{ and } T_1 = \frac{n^2}{m} T_2$
3. $L_1 = \frac{n^2}{m} L_2 \text{ and } T_1 = \frac{n^4}{m^2} T_2$
4. $\frac{n}{m} L_1 = L_2 \text{ and } \frac{n^4}{m^2} T_1 = T_2$

Solution:

$\text{Hint: } L = vT$ $\text{Step: Find the relations for distance and time in the two systems.}$ $\text{We have; } v_2 = \frac{n}{m^2} v_1$ $\text{Then we can write; } [L_2 T_2^{-1}] = \frac{n}{m^2} [L_1 T_1^{-1}] \quad \ldots (1)$ $\text{Similarly; } [L_2 T_2^{-2}] = \frac{[L_1 T_1^{-2}]}{mn} \quad \ldots (2)$ $\text{Divide equation (1) by (2).}$ $\Rightarrow \frac{[L_2 T_2^{-1}]}{[L_2 T_2^{-2}]} = \frac{\frac{n}{m^2} [L_1 T_1^{-1}]}{\frac{[L_1 T_1^{-2}]}{mn}}$ $\Rightarrow \frac{n^2}{m} T_1 = T_2$ $\text{Which only matches with option (1).}$ $\text{Hence, option (1) is the correct answer.}$

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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
}