A well acts as a closed organ pipe. The resonant frequencies are odd harmonics: \(f_n = (2n-1)f_1\). The difference between consecutive resonant frequencies is \(300 - 180 = 120\text{ Hz}\), which corresponds to \(2f_1\). Thus, \(f_1 = 60\text{ Hz}\).

Speed of a wave on a string is \( \sqrt{T/\mu} \) (a matches iii). Speed of a longitudinal wave in a metallic bar is \( \sqrt{Y/\rho} \) (b matches i). Speed of a longitudinal wave in a fluid is \( \sqrt{B/\rho} \) (c matches ii).

At a free end, there's no transverse force, so the slope \( \frac{\text{dy}}{\text{dx}} = 0 \). The resultant pulse is formed such that the free end is a displacement antinode. This implies zero slope, ensuring no transverse force. Both Assertion and Reason are true, and R correctly explains A.

For wind instruments, pitch \( f \) is proportional to the speed of sound in air \( v_{\text{air}} propto \sqrt{T_{\text{temp}}} \). As temperature rises, \( v_{\text{air}} \) increases, so pitch rises. For string instruments, \( f propto v_{\text{string}} = \sqrt{T_{\text{tension}}/\mu} \). As temperature rises, the string expands, reducing tension \( T_{\text{tension}} \), so \( v_{\text{string}} \) decreases and pitch falls. Reason R accurately explains this behavior for both cases.

The speed of sound in a gas is \( v = \sqrt{\gamma P/\rho} \). Moist air (humid air) has a lower average molecular mass and thus lower density \( \rho \) than dry air at the same pressure and temperature. A lower density leads to a higher speed of sound. Thus, A is true. Reason R is false; increasing humidity does not necessarily increase total pressure, and the primary factor for faster sound is reduced density.

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 frequency of a wave is determined by the source that generates it and remains constant as the wave propagates from one medium to another. The speed of a wave, however, is a characteristic property of the medium it is traveling through, not the source. Thus, Assertion A is true, but Reason R is false.

The speed of a transverse wave on a string is given by \( v = \sqrt{T/\mu} \), where \( T \) is the tension (elastic property) and \( \mu \) is the linear mass density (inertial property). If the string is uniform (constant \( \mu \)) and has uniform tension (constant \( T \)), then \( v \) must be constant for all waves propagating on it. Both A and R are true, and R correctly explains A.

Intensity of a sound wave is \( I = \frac{1}{2} \rho v \omega^2 A^2 \), where \( \omega = 2\pi f \). If \( I_1 = I_2 \) and the medium is the same (so \( \rho \) and \( v \) are constant), then \( \omega_1^2 A_1^2 = \omega_2^2 A_2^2 \). Given \( f_1 : f_2 = 1 : 2 \), so \( \omega_1 : \omega_2 = 1 : 2 \). This yields \( (1)^2 A_1^2 = (2)^2 A_2^2 \), so \( A_1^2 = 4 A_2^2 \), meaning \( A_1 : A_2 = 2 : 1 \). Thus, A is true. Reason R is false because intensity also depends on frequency (\( \omega \)), so waves with the same velocity and amplitude but different frequencies will have different intensities.

In a sinusoidal traveling wave, each particle of the string oscillates in simple harmonic motion (SHM) perpendicular to the direction of wave propagation. So, Assertion A is true. However, for a *traveling* wave, energy is continuously transmitted along the string. Therefore, the total energy within a small segment of the string is *not* conserved, as energy flows into and out of the segment. Reason R is false.