To increase the frequency of transverse oscillations of a stretched string by $40\%$, the tension must be increased by (in $\%$)

$A$ string with a mass density of $4 \times 10^{-3} \, kg/m$ is under a tension of $360 \, N$ and is fixed at both ends. One of its resonance frequencies is $375 \, Hz$. The next higher resonance frequency is $450 \, Hz$. The mass of the string is:

$A$ tuning fork gives $5 \text{ beats per second}$ with a $33 \text{ cm}$ length of sonometer wire. If the length of the wire is shortened by $1 \text{ cm}$, the number of beats is still the same. The frequency of the fork is (in $\text{ Hz}$)

$A$ second harmonic has to be generated in a string of length $l$ stretched between two rigid supports. The points where the string has to be plucked and touched are respectively

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:

An iron load of $2 \ kg$ is suspended in air from the free end of a sonometer wire of length $1 \ m$. $A$ tuning fork of frequency $256 \ Hz$ is in resonance with $\frac{1}{\sqrt{7}}$ times the length of the sonometer wire. If the load is immersed in water, the length of the wire in metres that will be in resonance with the same tuning fork is (Specific gravity of iron $= 8$).