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Current Question (ID: 10385)

Question:

A force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2R$ will be:

Options:

1. $\frac{F}{2}$
2. $2F$
3. $4F$
4. $\frac{F}{4}$

Solution:

The breaking force of a wire is directly proportional to its cross-sectional area. The breaking stress ($\sigma$) is a material property and remains constant for a given material. Breaking Force = $\sigma \times \text{Area of the cross-section of wire}$ For a wire with radius $R$, the area of the cross-section is $A = \pi R^2$. So, the initial breaking force is $F = \sigma \times \pi R^2$. If the radius of the wire is doubled to $2R$, the new area of the cross-section will be: $A' = \pi (2R)^2 = \pi (4R^2) = 4 \pi R^2 = 4A$ The new breaking force $F'$ will be: $F' = \sigma \times A' = \sigma \times (4A) = 4 (\sigma \times A) = 4F$ Therefore, if the radius of the wire is doubled, the breaking force will become four times.

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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
}