An object of density $2000 \ kg \ m^{-3}$ is hung from a thin light wire. The fundamental frequency of the transverse waves in the wire is $200 \ Hz$. If the object is immersed in water such that half of its volume is submerged,then the fundamental frequency of the transverse waves in the wire is (in $Hz$)

$A$ 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:

Four different independent waves are represented by $(i)$ $y_1 = a_1 \sin \omega t$,$(ii)$ $y_2 = a_2 \sin 2\omega t$,$(iii)$ $y_3 = a_3 \cos \omega t$,and $(iv)$ $y_4 = a_4 \sin (\omega t + \frac{\pi}{3})$. With which two waves is interference possible?

An auditorium has a volume of $10^5 \ m^3$ and a surface area of absorption of $2 \times 10^4 \ m^2$. Its average absorption coefficient is $0.2$. The reverberation time of the auditorium in seconds is:

An air column in a tube $32 \,cm$ long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length $66 \,cm$ is in resonance with another tuning fork. When these two tuning forks are sounded together, they produce $8$ beats per second. Then the frequencies of the two tuning forks are (Consider fundamental frequencies only):

The superposing waves are represented by the following equations: ${y_1} = 5\sin 2\pi (10t - 0.1x)$ and ${y_2} = 10\sin 2\pi (20t - 0.2x)$. The ratio of intensities $\frac{I_{\max}}{I_{\min}}$ will be: