$A$ 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

$A$ body is suspended from a string of length $1 \ m$ and mass $2 \ g$. The mass of the body required to produce a fundamental mode of $100 \ Hz$ frequency in the string is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of $A$ is equal to the second overtone of $B$. If the radius of $A$ is twice that of $B$,then the ratio of the lengths of the strings $l_A/l_B$ is:

$A$ sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same,the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?

The correct graph between the frequency $n$ and the square root of density $(\rho)$ of a wire,keeping its length,radius,and tension constant,is:

Two wires of the same material with radii $r$ and $2r$ respectively are welded together end to end. The combination is then used as a sonometer wire under tension $T$. The joint is kept midway between the two bridges. The ratio of the number of loops in the wires,given that the joint is a node,is: