At the open end of an organ pipe, the pressure is approximately atmospheric pressure, meaning there is no excess pressure variation, hence it is a pressure node. When a pressure wave reflects from an open end (a boundary to a less dense medium), it undergoes a \( \pi \) (180 degrees) phase shift. This phase shift causes the incident and reflected waves to destructively interfere at the open end, creating a pressure node. Both A and R are true, and R explains A.

The fundamental frequency of an open organ pipe is \(f = \frac{v}{2L}\). Velocity of sound \(v\) increases with temperature as \(v \propto \sqrt{T}\). While the pipe's length \(L\) also increases with temperature, the increase in \(v\) is proportionally greater than \(L\). Thus, \(f\) increases. Both A and R are true, and R correctly explains A.

Our ears perceive sound based on pressure variations. At a displacement node, particles have zero displacement, but pressure variations are maximum (a pressure antinode). Therefore, maximum sound is heard at a displacement node. Both (A) and (R) are true, and (R) provides the direct physical reason for (A).

For a closed organ pipe, the resonance lengths are in the ratio \(L_1 : L_2 : L_3 = 1 : 3 : 5\). If \(L_1 = 60 text{ cm}\), then \(L_2 = 3 times 60 \text{ cm} = 180 \text{ cm}\). So (A) is true. The resonant frequencies for a closed pipe are \(f_n = (2n-1)f_1\), thus the second resonance (third harmonic) is \(f_2 = 3f_1\). (R) is true and correctly explains (A).

At an open end, the pressure must remain atmospheric, establishing a pressure node. An incident high pressure pulse reflects as a low pressure pulse ((pi) phase change) to maintain this. Thus (A) is true. At a pressure node, particles have maximum displacement, which is a displacement antinode. Thus (R) is true and explains the reflection in (A).

Assertion (A) is true. If a transverse wave pulse propagates along the rotating hoop in a direction opposite to the hoop's rotation, and its speed relative to the string is equal to the string's linear speed, the wave appears stationary to a ground observer. Reason (R) provides this exact condition: if \(v_{\text{wave relative to string}} = -v_{\text{string}}\), then the net velocity relative to the lab frame is zero. Both are true and (R) correctly explains (A).

Assertion (A) is true by definition. Wavelength (\(lambda\)) in a longitudinal wave is the distance between two consecutive compressions or rarefactions. Reason (R) is false; a standing wave is a form of wave resulting from the superposition of two progressive waves. While it does not transfer net energy, it still represents a wave phenomenon with energy oscillation within its segments.

In a harmonic traveling wave, all particles oscillate with the same amplitude but their phases vary with position. So, (A) is true. In a stationary wave, particles between two successive nodes vibrate in phase, but particles on either side of a node are \(180^\circ\) out of phase. Their amplitudes also vary with position. So, (R) is false.

An acoustic guitar uses the resonance of its hollow body to amplify and shape the sound produced by vibrating strings, so (A) is true. An electric guitar, being a solid-body instrument, generates sound via electromagnetic pickups sensing string vibrations, which are then amplified electronically. It does not rely on acoustic body resonance for sound production. Thus, Reason (R) is false.