$A$ narrow tube is bent in the form of a circle of radius $R,$ as shown in the figure. Two small holes $S$ and $D$ are made in the tube at positions right-angled to each other. $A$ source placed at $S$ generates a wave of intensity $I_0$ which is equally divided into two parts: one part travels along the longer path,while the other travels along the shorter path. Both the waves meet at the point $D$ where a detector is placed. The maximum intensity produced at $D$ is given by

Select the correct alternative$(s)$ :-
$(A)$ Number of nodes equals to number of antinodes in closed organ pipe.
$(B)$ In open organ pipe,if number of antinodes is $m$,then number of nodes will be $m-1$.
$(C)$ If frequency of $4^{\text{th}}$ harmonic of open organ pipe is $400 \ Hz$,then frequency of $2^{\text{nd}}$ overtone of closed organ pipe of same length is $250 \ Hz$.
$(D)$ Time interval between successive maxima or minima (for superposition of two waves) is $\Delta t = \frac{1}{|f_1-f_2|} \ s$.

In the Kundt's tube experiment (shown in fig. $(i)$),the rod is clamped at the center. In the modified experiment (shown in fig. $(ii)$),the rod is clamped at the end. It is known that the speed of sound in air is $330\ m/s$,the powder piles up at successive distances of $0.6\ m$,and the length of the rod used is $1\ m$. Calculate the speed of sound in the rod in $m/s$.

The persistence of sound in a room after the source of sound is turned off is called reverberation. The measure of reverberation time is the time required for sound intensity to decrease by $60 \,dB$. It is given that the intensity of sound falls off as $I = I_0 \exp(-c_1 \alpha)$,where $I_0$ is the initial intensity,$c_1$ is a dimensionless constant with value $1/4$. Here,$\alpha$ is a positive constant which depends on the speed of sound $v_s$,volume of the room $V$,reverberation time $t$,and the effective absorbing area $A_e$. The value of $A_e$ is the product of the absorbing coefficient and the area of the room. For a concert hall of volume $V = 600 \,m^3$,the value of $A_e$ (in $m^2$) required to give a reverberation time of $t = 1 \,s$ is closest to (speed of sound in air $v_s = 340 \,m/s$):

Match the following List-$I$ with List-$II$.

List-$I$	List-$II$
$(A)$ Transverse wave	$(i)$ Vibrations parallel to the direction of propagation
$(B)$ Longitudinal wave	$(ii)$ Vibrations perpendicular to the direction of propagation
$(C)$ Beats	$(iii)$ Superposition of waves travelling in the opposite directions
$(D)$ Stationary waves	$(iv)$ Superposition of waves travelling in same direction

The correct answer is

The ratio of intensities between two coherent sound sources is $4:1$. The difference of loudness in $dB$ between maximum and minimum intensities when they interfere in the space is ..........