Import Question JSON

Current Question (ID: 11285)

Question:

$\text{The time period of a spring mass system at the surface of the earth is } 2 \text{ s.} \\ \text{What will be the time period of this system on the moon where the acceleration due to gravity is } \frac{1}{16}^{\text{th}} \text{ of the value of } g \text{ on the earth's surface?}$

Options:

1. $\frac{1}{\sqrt{6}}\text{ s}$
2. $2\sqrt{6}\text{ s}$
3. $2\text{ s}$ (Correct)
4. $12\text{ s}$

Solution:

$\text{The time period } \text{T} \text{ of a spring-mass system is given by the formula } \text{T} = 2\pi\sqrt{\frac{m}{k}}, \text{ where } m \text{ is the mass of the object and } k \text{ is the spring constant. This formula shows that the time period of a spring-mass system depends only on the mass of the object and the spring constant. It does not depend on the acceleration due to gravity } g. \text{The restoring force in a spring-mass system is provided by the spring, not by gravity. Therefore, the time period of the system will remain the same on the Moon as it is on Earth, regardless of the change in gravitational acceleration. Since the time period on Earth is } 2 \text{ s}, \text{ the time period on the Moon will also be } 2 \text{ s. Hence, option (3) is the correct answer.}$

Import JSON File

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
}