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Current Question (ID: 19102)

Question:
$\text{Two particles } A \text{ and } B \text{ are projected with speeds } 40 \text{ m/s and } 60 \text{ m/s at angles } 30^\circ \text{ and } 60^\circ \text{ above the horizontal, respectively.}$ $\text{What is the ratio of the range of } A \text{ to the range of } B?$
Options:
  • 1. $\sqrt{2} : 3$
  • 2. $\sqrt{3} : 2$
  • 3. $4 : 9$
  • 4. $2 : 1$
Solution:
$\text{Hint: } R = \frac{u^2 \sin 2\theta}{g}$ $\text{Step 1: Find the expression for the range of the particle.}$ $R = \frac{u^2 \sin 2\theta}{g}$ $\text{Step 2: Find the ratio of range of particle } A \text{ and } B.$ $\frac{R_A}{R_B} = \frac{u_A^2 \sin 2\theta_A}{u_B^2 \sin 2\theta_B}$ $\Rightarrow \frac{R_A}{R_B} = \frac{40^2 \sin(2 \times 30^\circ)}{60^2 \sin(2 \times 60^\circ)}$ $\Rightarrow \frac{R_A}{R_B} = \frac{40^2 \sin(60^\circ)}{60^2 \sin(120^\circ)}$ $\Rightarrow \frac{R_A}{R_B} = \frac{4}{9}$ $\text{Hence, option (3) is the correct answer.}$

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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
}