To solve the problem step by step, we will break it down into two parts as per the question:
### Part (a): Time taken by the wave-pulse to reach the other end `R`
1. **Calculate the mass per unit length (μ) for both wires:**
- For wire `PQ`:
\[
\mu_1 = \frac{\text{mass of PQ}}{\text{length of PQ}} = \frac{0.06 \, \text{kg}}{4.8 \, \text{m}} = 0.0125 \, \text{kg/m}
\]
- For wire `QR`:
\[
\mu_2 = \frac{\text{mass of QR}}{\text{length of QR}} = \frac{0.2 \, \text{kg}}{2.56 \, \text{m}} = 0.078125 \, \text{kg/m}
\]
2. **Calculate the velocity of the wave in each segment:**
- The velocity of the wave in wire `PQ` (V1):
\[
V_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{80 \, \text{N}}{0.0125 \, \text{kg/m}}} = \sqrt{6400} = 80 \, \text{m/s}
\]
- The velocity of the wave in wire `QR` (V2):
\[
V_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{80 \, \text{N}}{0.078125 \, \text{kg/m}}} = \sqrt{1024} = 32 \, \text{m/s}
\]
3. **Calculate the time taken for the wave to travel from `P` to `R`:**
- Time taken from `P` to `Q` (tPQ):
\[
t_{PQ} = \frac{\text{length of PQ}}{V_1} = \frac{4.8 \, \text{m}}{80 \, \text{m/s}} = 0.06 \, \text{s}
\]
- Time taken from `Q` to `R` (tQR):
\[
t_{QR} = \frac{\text{length of QR}}{V_2} = \frac{2.56 \, \text{m}}{32 \, \text{m/s}} = 0.08 \, \text{s}
\]
- Total time taken (t):
\[
t = t_{PQ} + t_{QR} = 0.06 \, \text{s} + 0.08 \, \text{s} = 0.14 \, \text{s}
\]
### Part (b): Amplitude of the reflected and transmitted wave-pulse after crossing joint `Q`
1. **Calculate the amplitude of the reflected wave (AR):**
- The formula for the amplitude of the reflected wave is:
\[
A_R = \frac{V_2 - V_1}{V_2 + V_1} \times A_i
\]
- Substituting the values:
\[
A_R = \frac{32 - 80}{32 + 80} \times 3.5 \, \text{cm} = \frac{-48}{112} \times 3.5 \approx -1.5 \, \text{cm}
\]
- The negative sign indicates that the reflected wave is inverted.
2. **Calculate the amplitude of the transmitted wave (AT):**
- The formula for the amplitude of the transmitted wave is:
\[
A_T = \frac{2V_2}{V_2 + V_1} \times A_i
\]
- Substituting the values:
\[
A_T = \frac{2 \times 32}{32 + 80} \times 3.5 \, \text{cm} = \frac{64}{112} \times 3.5 \approx 2 \, \text{cm}
\]
### Final Answers:
- (a) The time taken by the wave-pulse to reach the other end `R` is **0.14 seconds**.
- (b) The amplitude of the reflected wave-pulse is **1.5 cm** (inverted), and the amplitude of the transmitted wave-pulse is **2 cm**.
