$A$ wire of linear mass density $\mu = 9.8 \times 10^{-3} \, kg \, m^{-1}$ passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle of $30^{\circ}$ with the horizontal. Masses $m$ and $M$ are tied at the two ends of the wire such that $m$ rests on the plane and $M$ hangs freely vertically downwards. The entire system is in equilibrium and a transverse wave propagates along the wire with a velocity of $100 \, m \, s^{-1}$. Choose the correct option for $m$ in $kg$.

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:

$A$ pipe closed at one end has a length of $0.8 \,m$. At its open end, a $0.5 \,m$ long uniform string is vibrating in its $2^{nd}$ harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \,N$ and the speed of sound is $320 \,m/s$, what is the mass of the string (in $\,g$)?

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

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

Two uniform strings of mass per unit length $\mu$ and $4 \mu$,and length $L$ and $2 L$,respectively,are joined at point $O$,and tied at two fixed ends $P$ and $Q$,as shown in the figure. The strings are under a uniform tension $T$. If we define the frequency $v_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$,which of the following statement$(s)$ is(are) correct?
$(A)$ With a node at $O$,the minimum frequency of vibration of the composite string is $v_0$
$(B)$ With an antinode at $O$,the minimum frequency of vibration of the composite string is $2 v_0$
$(C)$ When the composite string vibrates at the minimum frequency with a node at $O$,it has $6$ nodes,including the end nodes
$(D)$ No vibrational mode with an antinode at $O$ is possible for the composite string