Import Question JSON

Current Question (ID: 8747)

Question:

\text{On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is/are not correct.} \text{(a) } y = a \sin\left(\frac{2\pi t}{T}\right) \text{(b) } y = a \sin(vt) \text{(c) } y = \frac{a}{T} \sin\left(\frac{t}{a}\right) \text{(d) } y = a\sqrt{2} \left(\sin\left(\frac{2\pi t}{T}\right) - \cos\left(\frac{2\pi t}{T}\right)\right) \text{(Symbols have their usual meanings.)} \text{Choose the correct option:}

Options:

1. $\text{(a), (c)}$
2. $\text{(a), (b)}$
3. $\text{(b), (c)}$
4. $\text{(a), (d)}$

Solution:

$\text{Hint: Use the principle of homogeneity of dimensions.}$ $\text{Step: Analyse each option one by one.}$ $\text{Now, by using the principle of homogeneity of dimensions LHS and RHS. of (a) and (d) will be the same, and is } L$ $\text{For the option (c), } [\text{LHS}] = L$ $[\text{RHS}] = \frac{L}{T} = [LT]^{-1}$ $[\text{LHS}] \neq [\text{RHS}]$ $\text{Hence, (c) is not the correct option.}$ $\text{In option (b) the dimension of the angle is } vt \text{ i.e., } L$ $\Rightarrow \text{RHS} = L \cdot L = L^2 \text{ and LHS} = L$ $\Rightarrow \text{LHS} \neq \text{RHS}$ $\text{So, option (b) is also not correct.}$ $\text{Therefore, options b and c are not correct.}$ $\text{Hence, option (3) is the correct answer.}$

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
}