$A$ tuning fork gives $5$ beats per second with $40 \ cm$ length of a sonometer wire. If the length of the wire is shortened by $1 \ cm$,the number of beats is still the same. The frequency of the fork is (in $Hz$)

$A$ piano wire with a diameter of $0.90 \ mm$ is replaced by another wire of diameter $0.93 \ mm$ of the same material. If the tension of the wire is kept the same,then the percentage change in the frequency of the fundamental tone is

The frequency of a sonometer wire is $n$. If its tension is increased $4$ times and its length is doubled,then the new frequency will be

$A$ horizontal stretched string,fixed at two ends,is vibrating in its fifth harmonic according to the equation,$y(x, t) = (0.01 \ m) \sin[(62.8 \ m^{-1}) x] \cos[(628 \ s^{-1}) t]$. Assuming $\pi = 3.14$,the correct statement$(s)$ is (are) :
$(A)$ The number of nodes is $5$.
$(B)$ The length of the string is $0.25 \ m$.
$(C)$ The maximum displacement of the midpoint of the string from its equilibrium position is $0.01 \ m$.
$(D)$ The fundamental frequency is $100 \ Hz$.

$A$ stone is hung in air from a wire which is stretched over a sonometer. The bridges of the sonometer are $L \, cm$ apart when the wire is in unison with a tuning fork of frequency $N$. When the stone is completely immersed in water,the length between the bridges is $l \, cm$ for re-establishing unison. The specific gravity of the material of the stone is:

$A$ string vibrates in its fundamental mode when a tension $T_1$ is applied to it. If the length of the string is decreased by $25 \%$ and the tension applied is changed to $T_2$, the fundamental frequency of the string increases by $100 \%$, then $\frac{T_2}{T_1} =$ (Linear density of the string is constant)