One end of thin aluminium rod (cross -sectional Area `=10^(-6)m^(2)`) is bult welded to the end of copper rod with same diameter. Both rods are very long and under uniform tension of `900N` Given that
`Y_((Cu))=1.6xx10^(11)Nm^(-2), Y_((Al))=0.9xx10^(11)Nm^(-2)`
`rho_((Cu))=8.1xx10^(3)kgm^(-3), rho_((Al))=2.5xx10^(3)Kgm^(-3)`

Options

A. The value of relection factor `(A_(r))/(A_(i))` for longitudinal pulse approacing the junction along copper rod is 0.15
B. The value of reflection factor `(A_(r))/(A_(i))` for longitudinal pulse approachig the junction along copper rod is approximately 0.3
C. The value of reflection factor `(A_(r))/(A_(i))` or transverse pulse approaching the junction along copper rod is `0.15`
D. The value of reflection factor `(A_(r))/(A_(i))` for transverse pulse approaching the junction along copper rod is approximately 0.3

Explanation

To solve the problem, we need to calculate the relaxation factor for both transverse and longitudinal waves in the aluminum and copper rods.

### Step 1: Calculate the speed of transverse waves in aluminum (V_al)
The formula for the speed of transverse waves is given by:

\[ V_t = \sqrt{\frac{F}{\mu}} \]

Where:
- \( F \) is the force (900 N)
- \( \mu \) is the mass per unit length, given by \( \mu = \rho \cdot A \)

For aluminum:
- Cross-sectional area \( A = 10^{-6} \, m^2 \)
- Density \( \rho_{Al} = 2.5 \times 10^3 \, kg/m^3 \)

Calculating \( \mu_{Al} \):

\[ \mu_{Al} = \rho_{Al} \cdot A = (2.5 \times 10^3) \cdot (10^{-6}) = 2.5 \times 10^{-3} \, kg/m \]

Now, substituting into the speed equation:

\[ V_{Al} = \sqrt{\frac{900}{2.5 \times 10^{-3}}} \]

Calculating \( V_{Al} \):

\[ V_{Al} = \sqrt{360000} = 600 \, m/s \]

### Step 2: Calculate the speed of transverse waves in copper (V_cu)
Using the same method for copper:
- Density \( \rho_{Cu} = 8.1 \times 10^3 \, kg/m^3 \)

Calculating \( \mu_{Cu} \):

\[ \mu_{Cu} = \rho_{Cu} \cdot A = (8.1 \times 10^3) \cdot (10^{-6}) = 8.1 \times 10^{-3} \, kg/m \]

Now, substituting into the speed equation:

\[ V_{Cu} = \sqrt{\frac{900}{8.1 \times 10^{-3}}} \]

Calculating \( V_{Cu} \):

\[ V_{Cu} = \sqrt{111111.11} \approx 333.33 \, m/s \]

### Step 3: Calculate the relaxation factor for transverse waves (AR)
The relaxation factor is given by:

\[ AR = \frac{V_{Al} - V_{Cu}}{V_{Al} + V_{Cu}} \]

Substituting the values:

\[ AR = \frac{600 - 333.33}{600 + 333.33} = \frac{266.67}{933.33} \approx 0.286 \]

### Step 4: Calculate the speed of longitudinal waves in aluminum (V_al_long)
The formula for the speed of longitudinal waves is given by:

\[ V_L = \sqrt{\frac{Y}{\rho}} \]

Where \( Y \) is Young's modulus.

For aluminum:
- \( Y_{Al} = 0.9 \times 10^{11} \, N/m^2 \)

Calculating \( V_{Al\_long} \):

\[ V_{Al\_long} = \sqrt{\frac{0.9 \times 10^{11}}{2.5 \times 10^3}} = \sqrt{36000000} = 6000 \, m/s \]

### Step 5: Calculate the speed of longitudinal waves in copper (V_cu_long)
For copper:
- \( Y_{Cu} = 1.6 \times 10^{11} \, N/m^2 \)

Calculating \( V_{Cu\_long} \):

\[ V_{Cu\_long} = \sqrt{\frac{1.6 \times 10^{11}}{8.1 \times 10^3}} = \sqrt{19753086.42} \approx 4444.44 \, m/s \]

### Step 6: Calculate the relaxation factor for longitudinal waves (AR_long)
Using the same formula for the relaxation factor:

\[ AR_{long} = \frac{V_{Al\_long} - V_{Cu\_long}}{V_{Al\_long} + V_{Cu\_long}} \]

Substituting the values:

\[ AR_{long} = \frac{6000 - 4444.44}{6000 + 4444.44} = \frac{1555.56}{10444.44} \approx 0.149 \]

### Final Answers:
- Relaxation factor for transverse waves \( AR \approx 0.286 \)
- Relaxation factor for longitudinal waves \( AR_{long} \approx 0.149 \)