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Current Question (ID: 9630)

Question:

$\text{A projectile is fired from the surface of the earth with a velocity of 5 m/s and at an angle } \theta \text{ with the horizontal. Another projectile fired from another planet with a velocity of 3 m/s at the same angle follows a trajectory that is identical to the trajectory of the projectile fired from the Earth. The value of the acceleration due to gravity on the other planet is: (given } g = 9.8 \text{ m/s}^2\text{)}$

Options:

1. $3.5 \text{ m/s}^2$
2. $5.9 \text{ m/s}^2$
3. $16.3 \text{ m/s}^2$
4. $110.8 \text{ m/s}^2$

Solution:

$\text{Hint: } s = ut + \frac{1}{2}at^2$ $\text{Step: Find the value of the acceleration due to gravity on the other planet.}$ $\text{The angle of projection is the same for both projectiles.}$ $\text{Let the acceleration due to gravity on the planet = a}$ $\text{The trajectory is the same, so max. vertical distance travelled is also the same.}$ $\text{The initial velocity on earth = v}_e$ $\text{The initial velocity on the planet = v}_p$ $\text{The final velocities for both projectiles = 0}$ $\text{Using the second equation of motion:}$ $v_e^2\sin^2\theta = 2gh$ $v_p^2\sin^2\theta = 2ah$ $\frac{v_e^2\sin^2\theta}{v_p^2\sin^2\theta} = \frac{g}{a}$ $\frac{25}{9} = \frac{9.8}{a}$ $a = \frac{9.8 \times 9}{25} = 3.528 \text{ ms}^{-2}$ $\text{Hence, option (1) is the correct answer.}$

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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
}