Import Question JSON

Current Question (ID: 9701)

Question:

$\text{A steam boat goes across a lake and comes back (a) On a quiet day when the water is still and (b) On a rough day when there is a uniform air current so as to help the journey onward and to impede the journey back. If the speed of the launch on both the days was the same, in which case will the steam boat complete the journey in lesser time:}$

Options:

1. $\text{Case (a)}$
2. $\text{Case (b)}$
3. $\text{Same in both cases}$
4. $\text{Nothing can be predicted based on given data}$

Solution:

\text{If the breadth of the lake is } l \text{ and the velocity of the boat is } v_b\text{, then:} \text{Time in going and coming back on a quiet day:} t_Q = \frac{l}{v_b} + \frac{l}{v_b} = \frac{2l}{v_b} \text{ .....(i)} \text{Now if } v_a \text{ is the velocity of air-current, then:} \text{Time taken in going across the lake:} t_1 = \frac{l}{v_b + v_a} \text{ [As current helps the motion]} \text{Time taken in coming back:} t_2 = \frac{l}{v_b - v_a} \text{ [As current opposes the motion]} \text{Therefore:} t_R = t_1 + t_2 = \frac{l}{v_b + v_a} + \frac{l}{v_b - v_a} t_R = \frac{l(v_b - v_a) + l(v_b + v_a)}{(v_b + v_a)(v_b - v_a)} = \frac{2lv_b}{v_b^2 - v_a^2} t_R = \frac{2l}{v_b} \times \frac{1}{1 - (v_a/v_b)^2} \text{ .....(ii)} \text{From equations (i) and (ii):} \frac{t_R}{t_Q} = \frac{1}{1-(v_a/v_b)^2} > 1 \text{Since } 1 - \frac{v_a^2}{v_b^2} < 1\text{, we have } t_R > t_Q \text{Therefore, time taken on a quiet day is less than on a rough day.}

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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
}