A wire can be broken by applying a load of 20kg wt. Then the force required to break the wire of twice the diameter and same length, same material, is

Options

A. 5 kg wt
B. 160 kg wt
C. 80 kg wt
D. 20 kg wt

Explanation

To solve the problem, we need to determine the breaking force required for a wire that has twice the diameter of the original wire, while keeping the same length and material. Here’s the step-by-step solution:

### Step 1: Understand the relationship between breaking force and area
The breaking stress (σ) is defined as the breaking force (F) divided by the cross-sectional area (A) of the wire:
\[ \sigma = \frac{F}{A} \]
Since both wires are made of the same material, they will have the same breaking stress.

### Step 2: Define the areas of the wires
Let’s denote:
- The original wire's radius as \( R_1 \) and its area as \( A_1 \).
- The new wire's radius as \( R_2 \) and its area as \( A_2 \).

The area of a wire is given by:
\[ A = \pi R^2 \]
Thus, for the original wire:
\[ A_1 = \pi R_1^2 \]
For the new wire, since the diameter is doubled, the radius \( R_2 \) will be:
\[ R_2 = 2R_1 \]
Therefore, the area of the new wire is:
\[ A_2 = \pi R_2^2 = \pi (2R_1)^2 = \pi (4R_1^2) = 4\pi R_1^2 \]

### Step 3: Set up the equation using breaking stress
Since the breaking stress is the same for both wires, we can write:
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]
Where:
- \( F_1 = 20 \, \text{kg wt} \) (the breaking force of the original wire)
- \( A_1 = \pi R_1^2 \)
- \( F_2 \) is the breaking force for the new wire
- \( A_2 = 4\pi R_1^2 \)

### Step 4: Substitute the values into the equation
Substituting the areas into the equation gives:
\[ \frac{20}{\pi R_1^2} = \frac{F_2}{4\pi R_1^2} \]

### Step 5: Simplify the equation
We can cancel \( \pi R_1^2 \) from both sides:
\[ \frac{20}{1} = \frac{F_2}{4} \]
This simplifies to:
\[ F_2 = 20 \times 4 = 80 \, \text{kg wt} \]

### Conclusion
The force required to break the wire of twice the diameter and the same length and material is:
\[ \boxed{80 \, \text{kg wt}} \]

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