$A$ massless rod is suspended by two identical strings $AB$ and $CD$ of equal length. $A$ block of mass $m$ is suspended from point $O$ such that $BO$ is equal to $x$. Further,it is observed that the frequency of the $1^{st}$ harmonic (fundamental frequency) in $AB$ is equal to the $2^{nd}$ harmonic frequency in $CD$. Then,the length of $BO$ is:

$A$ 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

Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent $(i)$ a travelling wave,$(ii)$ a stationary wave or $(iii)$ none at all:
$(a)$ $y = 2 \cos(3x) \sin(10t)$
$(b)$ $y = 2 \sqrt{x - vt}$
$(c)$ $y = 3 \sin(5x - 0.5t) + 4 \cos(5x - 0.5t)$
$(d)$ $y = \cos x \sin t + \cos 2x \sin 2t$

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 $v_s = 340 \,m/s$):

$A$ hollow pipe of length $0.8 \ m$ is closed at one end. At its open end,a $0.8 \ m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the string is $50 \ N$ and the speed of sound in air is $320 \ m/s$,the mass of the string is: (in $g$)

$A$ vibrating string of length $\ell$ under a tension $T$ resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length $75 \,cm$ inside a tube closed at one end. The string also generates $4$ beats per second when excited along with a tuning fork of frequency $n$. Now, when the tension of the string is slightly increased, the number of beats reduces to $2$ per second. Assuming the velocity of sound in air to be $340 \,m/s$, the frequency $n$ of the tuning fork in $Hz$ is: