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Current Question (ID: 16618)

Question:

$\text{The resistance of a wire is } R \text{ ohm. If it is melted and stretched to } n \text{ times its original length, its new resistance will be:}$

Options:

1. $nR$
2. $\frac{R}{n}$
3. $n^2R$
4. $\frac{R}{n^2}$

Solution:

$\text{Hint: The volume of the wire remains the same.}$ $\text{Step: Find the resistance of the wire after melting and stretching.}$ $\text{The resistance of the wire is given by:}$ $R = \rho \frac{l_1}{A_1}$ $\text{As volume remains the same,}$ $A_1l_2 = A_2l_2 \Rightarrow \frac{A_1}{n} = A_2$ $\text{The resistance of the wire after stretching:}$ $R_2 = \rho \frac{l_2}{A_2} \Rightarrow R_2 = \rho \frac{n \times nl_1}{A_1} \Rightarrow R_2 = n^2 \rho \frac{l_1}{A_1}$ $\Rightarrow R_2 = n^2 \rho \frac{l_1}{A_1} = n^2R$ $\text{Hence, option (3) is the correct answer.}$

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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
}