The frequency of a sonometer wire is $100 \ Hz$. When the weights producing the tension are completely immersed in water,the frequency becomes $80 \ Hz$,and on immersing the weights in a certain liquid,the frequency becomes $60 \ Hz$. The specific gravity of the liquid is:

$A$ 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 $100 \ Hz$.

$A$ sonometer wire stretched by weight '$w$' is in unison with a tuning fork. The corresponding resonating length is '$L_1$'. If the weight is increased by '$3w$',the corresponding resonating length of the sonometer in unison with the tuning fork becomes '$L_2$'. The ratio $\left(\frac{L_1}{L_2}\right)$ is:

The tension of a stretched string is increased by $69\%$. In order to keep its frequency of vibration constant,its length must be increased by .... $\%$

$A$ wire of length $2L$ is made by joining two wires $A$ and $B$ of same lengths but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$,then the ratio $p : q$ is

When a stretched wire of fundamental frequency $f$ is divided into three segments,the fundamental frequencies of these three segments are $f_1, f_2$ and $f_3$ respectively. Then the relation among $f_1, f_2, f_3$ and $f$ is (Assume tension is constant).