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Current Question (ID: 9679)

Question:

$\text{A man can row a boat with a speed of 10 kmph in still water. The river flows at 6 kmph. If he crosses the river from one bank to the other along the shortest possible path, the time taken to cross the river of width 1 km is:}$

Options:

1. $\frac{1}{8} \text{ hr}$
2. $\frac{1}{4} \text{ hr}$
3. $\frac{1}{2} \text{ hr}$
4. $1 \text{ hr}$

Solution:

\text{Hint: } t = \frac{\text{Width of the river}}{\text{Effective speed across the river}} \text{Step: Find the time taken by the man to cross the river along the shortest possible path.} \text{To cross the river along the shortest path, the man should have zero velocity along the river.} v_m \sin \theta = v_r 10 \sin \theta = 6 \Rightarrow \sin \theta = \frac{3}{5} \text{The time taken by the man to cross the river along the shortest possible path is given by:} t = \frac{d}{v_m \cos \theta} = \frac{d}{10 \cos \theta} \text{Here from the diagram, the value of } \cos \theta \text{ is given by:} \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \text{Substituting in the equation we get:} t = \frac{d}{10 \times \frac{4}{5}} = \frac{d}{8} \text{If } d = 1 \text{ km, then } t = \frac{1}{8} \text{ hr} \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
}