Import Question JSON

Current Question (ID: 19079)

Question:

$\text{A particle is projected with a speed of } 2 \text{ m/s from the base of a plane inclined at } 30^\circ \text{ to the horizontal.}$ $\text{The direction of projection makes an angle of } 15^\circ \text{ above the inclined plane as shown in the figure.}$ $\text{If } g = 10 \text{ m/s}^2, \text{ what is the distance along the plane from the point of projection to the point where the particle strikes the plane?}$

Options:

1. $14 \text{ cm}$
2. $28 \text{ cm}$
3. $20 \text{ cm}$
4. $36 \text{ cm}$

Solution:

$\text{Hint: } y = x \tan \phi - \frac{gx^2}{2u^2 \cos^2 \phi}$ $\text{Step 1: Find the } x\text{-coordinate of the point where the particle strikes the plane.}$ $y = x \tan \phi - \frac{gx^2}{2u^2 \cos^2 \phi}$ $\Rightarrow \frac{y}{x} = \tan \phi - \frac{gx}{2u^2 \cos^2 \phi}$ $\Rightarrow \tan 30^\circ = \tan(30^\circ + 15^\circ) - \frac{gx}{2u^2 \cos^2(30^\circ + 15^\circ)}$ $\Rightarrow \tan 30^\circ = 1 - \frac{10x}{2 \times 2^2 \times \frac{1}{2}}$ $\Rightarrow \frac{1}{\sqrt{3}} = 1 - \frac{10x}{4}$ $\Rightarrow \frac{1}{\sqrt{3}} = 1 - \frac{10x}{4}$ $\Rightarrow \frac{1.73}{3} = 1 - \frac{10x}{4}$ $\Rightarrow 0.6 = 1 - \frac{10x}{4}$ $\Rightarrow x = 0.16 \text{ m}$ $\text{Step: Find the distance along the plane from the point of projection to the point where the particle strikes the plane.}$ $x = L \cos \alpha$ $\Rightarrow L = \frac{x}{\cos \alpha}$ $\Rightarrow L = \frac{0.16}{\cos 30^\circ}$ $\Rightarrow L = \frac{0.16 \times 2}{\sqrt{3}}$ $\Rightarrow L = 0.188 \text{ m}$ $\Rightarrow L = 18.8 \text{ cm} \approx 20 \text{ cm}$ $\text{Hence, option (3) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}