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Current Question (ID: 20232)

Question:

$\text{A wire of } 1 \, \Omega \text{ has a length of } 1 \, \text{m. It is stretched till its length increases by } 25\%. \text{ The percentage change in resistance to the nearest integer is:}$

Options:

1. $56\%$
2. $25\%$
3. $12.5\%$
4. $76\%$

Solution:

$\text{Hint: } R = \frac{\rho \ell}{A}$ $R_0 = 1 \, \Omega \quad R_1 = ?$ $\ell_0 = 1 \, \text{m} \quad \ell_1 = 1.25 \, \text{m}$ $A_0 = A$ $\text{As the volume of wire remains constant so}$ $A_0 \ell_0 = A_1 \ell_1 \Rightarrow A_1 = \frac{\ell_0 \, A_0}{\ell_1}$ $\text{Now}$ $\text{Resistance } (R) = \frac{\rho \ell}{A}$ $\frac{R_0}{R_1} = \frac{\ell_0 / A_0}{\rho \ell_1 / A_1}$ $\frac{1}{R_1} = \frac{\ell_0}{A_0} \left( \frac{\ell_0 \, A_0}{\ell_1 \times \ell_1} \right)$ $R_1 = \frac{\ell_1^2}{\ell_0^2} = 1.5625 \, \Omega$ $\text{So } \% \text{ change in resistance}$ $= \frac{R_1 - R_0}{R_0} \times 100\%$ $= \frac{1.5625 - 1}{1} \times 100\%$ $= 56.25\%$

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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
}