Import Question JSON

Current Question (ID: 16398)

Question:

$\text{Two small spheres each having the charge } +Q \text{ are suspended by insulating threads of length } L \text{ from a hook. If this arrangement is taken in space where there is no gravitational effect, then the angle between the two suspensions and the tension in each will be:}$

Options:

1. $180^\circ, \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{(2L)^2}$
2. $90^\circ, \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{(L)^2}$
3. $180^\circ, \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{2L^2}$
4. $180^\circ, \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{L^2}$

Solution:

$\text{Hint: } F = \frac{kq_1q_2}{r^2}$ $\text{Step 1: Figure out the configuration in which these two charges will be by using Coulomb's law.}$ $\text{The position of the balls in space will become as shown below.}$ $\text{Thus the angle, } \theta = 180^\circ$ $\text{Step 2: Find the forces between the charges using Coulomb's Law.}$ $\text{Electrostatic Force between two charges is calculated by Coulomb's law which states that magnitude of electrostatic force between two point charges is directly proportional to the product of magnitude of charges and inversely proportional to distance between them.}$ $F = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{(2L)^2}$ $\text{Hence, option (1) is the correct answer.}$

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
}