If $T$ is the reverberation time of an auditorium of volume $V$,then:

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 $\Delta t = \frac{1}{|f_1-f_2|} \ s$.

$A$ stretched uniform wire of length $L$ under tension $T$ is vibrating with fundamental frequency $n$. $A$ closed pipe of the same length $L$ is also vibrating with the same fundamental frequency $n$. If the tension $T$ is increased by $16 \,N$, the wire resonates with the $2^{\text{nd}}$ harmonic of the same closed pipe. The initial tension in the wire is: (in $\,N$)

The frequency of a stretched uniform wire of length $L$ under tension is in resonance with the fundamental frequency of a closed pipe of same length. If the tension in the wire is increased by $8 \ N$,it is in resonance with the first overtone of the same closed pipe. The initial tension in the wire is (in $N$)

If a microwave and an ultrasonic sound wave have the same wavelength, the ratio of their frequencies is approximately:

$A$ vibrating string of certain length $l$ 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: