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Current Question (ID: 20309)

Question:

$\text{A wire of resistance } R_1 \text{ is drawn out so that its length is increased by twice its original length. The ratio of new resistance to original resistance is:}$

Options:

1. $9 : 1$
2. $1 : 9$
3. $4 : 1$
4. $3 : 1$

Solution:

$\text{Hint: } R = \frac{\rho l}{A}$ $\text{When the length is increased by twice, the new length } l_2 = 3l_1.$ $\text{Since volume remains constant, } A_1l_1 = A_2l_2.$ $\text{Thus, } A_2 = \frac{A_1l_1}{3l_1} = \frac{A_1}{3}.$ $\text{New resistance } R_2 = \frac{\rho l_2}{A_2} = \frac{\rho (3l_1)}{\frac{A_1}{3}} = 9 \times \frac{\rho l_1}{A_1} = 9R_1.$ $\text{Therefore, the ratio is } 9:1.$

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
}