A uniformly tapering circular bar has its diameter ranging from (D-a) to (D+a) along its length L. The bar is subjected to an axial force P and young's modulus is Y. The percentage error in the calculation of elongation of bar if mean diameter is used in calculation instead of actual diameter.

Options

A. `(2a^2)/(D^2)xx100`
B. `((10a)/D)^2`
C. `((a)/(10D))^2`
D. `a/Dxx100`

Answer

To solve the problem of calculating the percentage error in the elongation of a uniformly tapering circular bar when using the mean diameter instead of the actual diameter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the diameters**:
- Let the larger diameter be \( D_1 = D + a \) and the smaller diameter be \( D_2 = D - a \).
- The mean diameter \( D_m \) can be calculated as:
\[ D_m = \frac{D_1 + D_2}{2} = \frac{(D + a) + (D - a)}{2} = D \]

2. **Calculate the change in length for the tapering bar**:
- The change in length \( \Delta L \) for an element of the bar can be expressed as:
\[ \Delta L = \frac{P \cdot dx}{A \cdot Y} \]
- The area \( A \) of the element with diameter \( d \) is:
\[ A = \frac{\pi d^2}{4} \]
- Since the diameter varies linearly from \( D_1 \) to \( D_2 \), we can express the diameter \( d \) at a distance \( x \) from \( D_1 \) as:
\[ d = D_1 - \left( \frac{D_1 - D_2}{L} \cdot x \right) = D_1 - \left( \frac{2a}{L} \cdot x \right) \]

3. **Integrate to find total elongation**:
- Substitute \( d \) into the expression for \( \Delta L \) and integrate from \( 0 \) to \( L \):
\[ \Delta L = \int_0^L \frac{P \cdot dx}{\left(\frac{\pi}{4} \cdot d^2\right) \cdot Y} \]
- This leads to:
\[ \Delta L = \frac{4P}{\pi Y} \int_0^L \frac{dx}{d^2} \]
- After performing the integration, we find:
\[ \Delta L = \frac{4PL}{\pi Y} \cdot \frac{1}{D_1 D_2} \]

4. **Calculate elongation using mean diameter**:
- If we consider the bar to have a uniform diameter \( D_m \), the elongation \( \Delta L' \) would be:
\[ \Delta L' = \frac{4PL}{\pi Y} \cdot \frac{1}{D_m^2} \]
- Given that \( D_m = D \), we have:
\[ \Delta L' = \frac{4PL}{\pi Y} \cdot \frac{1}{D^2} \]

5. **Calculate the percentage error**:
- The percentage error in elongation when using the mean diameter instead of the actual diameter is given by:
\[ \text{Percentage Error} = \left( \frac{\Delta L - \Delta L'}{\Delta L} \right) \times 100 \]
- Substituting the values:
\[ \text{Percentage Error} = \left( \frac{\frac{4PL}{\pi Y} \cdot \frac{1}{D_1 D_2} - \frac{4PL}{\pi Y} \cdot \frac{1}{D^2}}{\frac{4PL}{\pi Y} \cdot \frac{1}{D_1 D_2}} \right) \times 100 \]
- Simplifying this expression leads to:
\[ \text{Percentage Error} = \left( \frac{D^2 - D_1 D_2}{D^2} \right) \times 100 \]
- Finally, substituting \( D_1 = D + a \) and \( D_2 = D - a \):
\[ \text{Percentage Error} = \left( \frac{D^2 - (D^2 - a^2)}{D^2} \right) \times 100 = \left( \frac{a^2}{D^2} \right) \times 100 \]

### Final Result:
The percentage error in the calculation of elongation of the bar when using the mean diameter instead of the actual diameter is:
\[ \text{Percentage Error} = \frac{a^2}{D^2} \times 100 \]