To solve the problem of finding the time taken by a wave pulse to travel from the lighter end to the heavier end of a non-uniform wire, we will follow these steps:
### Step 1: Understand the given parameters
We have:
- Length of the wire, \( L = 10 \, \text{m} \)
- Mass per unit length, \( \mu(x) = \mu_0 + ax \)
- Tension in the wire, \( T = 100 \, \text{N} \)
- Given values: \( \mu_0 = 10^{-2} \, \text{kg/m} \) and \( a = 9 \times 10^{-3} \, \text{kg/m}^2 \)
### Step 2: Write the expression for wave velocity
The velocity \( v \) of a wave in a medium is given by the formula:
\[
v = \sqrt{\frac{T}{\mu}}
\]
where \( \mu \) is the mass per unit length.
### Step 3: Substitute the expression for \( \mu \)
Since \( \mu \) varies with \( x \), we can write:
\[
v(x) = \sqrt{\frac{T}{\mu_0 + ax}}
\]
### Step 4: Set up the integral for time
The time \( dt \) taken for an infinitesimal distance \( dx \) can be expressed as:
\[
dt = \frac{dx}{v(x)} = \frac{dx}{\sqrt{\frac{T}{\mu_0 + ax}}}
\]
Thus, the total time \( T \) taken to travel from \( x = 0 \) to \( x = L \) is:
\[
T = \int_0^L dt = \int_0^{10} \frac{dx}{\sqrt{\frac{T}{\mu_0 + ax}}}
\]
### Step 5: Simplify the integral
This can be rewritten as:
\[
T = \sqrt{\frac{1}{T}} \int_0^{10} \sqrt{\mu_0 + ax} \, dx
\]
Now, we need to evaluate the integral:
\[
\int_0^{10} \sqrt{\mu_0 + ax} \, dx
\]
### Step 6: Calculate the integral
Using the substitution \( u = \mu_0 + ax \), we have:
- When \( x = 0 \), \( u = \mu_0 \)
- When \( x = 10 \), \( u = \mu_0 + 10a \)
The differential \( dx = \frac{du}{a} \), thus:
\[
T = \sqrt{\frac{1}{T}} \int_{\mu_0}^{\mu_0 + 10a} \sqrt{u} \frac{du}{a}
\]
This integral evaluates to:
\[
\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}
\]
Therefore:
\[
T = \sqrt{\frac{1}{T}} \cdot \frac{1}{a} \cdot \left[ \frac{2}{3} \left( (\mu_0 + 10a)^{3/2} - \mu_0^{3/2} \right) \right]
\]
### Step 7: Substitute the values
Substituting \( \mu_0 = 10^{-2} \) and \( a = 9 \times 10^{-3} \):
- \( \mu_0 + 10a = 10^{-2} + 10 \times 9 \times 10^{-3} = 10^{-2} + 0.09 = 0.1 \)
- \( \mu_0^{3/2} = (10^{-2})^{3/2} = 10^{-3} \)
Now substituting these into the equation gives:
\[
T = \sqrt{\frac{1}{100}} \cdot \frac{1}{9 \times 10^{-3}} \cdot \frac{2}{3} \left[ 0.1^{3/2} - 10^{-3} \right]
\]
### Step 8: Calculate the final time
Calculating \( 0.1^{3/2} = 0.031622 \) and substituting gives:
\[
T = \frac{1}{10} \cdot \frac{1}{9 \times 10^{-3}} \cdot \frac{2}{3} \left[ 0.031622 - 0.001 \right]
\]
This simplifies to:
\[
T = \frac{1}{10} \cdot \frac{1}{9 \times 10^{-3}} \cdot \frac{2}{3} \cdot 0.030622
\]
Calculating this gives approximately \( T \approx 0.2268 \) seconds.
### Final Answer
The time taken by the pulse to travel from the lighter end to the heavier end is approximately:
\[
\boxed{0.227 \, \text{seconds}}
\]
