$A$ sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same,by what factor will the frequency change?

Two strings $(A, B)$ having linear densities $\mu_{A} = 2 \times 10^{-4} \ kg/m$ and $\mu_{B} = 4 \times 10^{-4} \ kg/m$ and lengths $L_{A} = 2.5 \ m$ and $L_{B} = 1.5 \ m$ respectively are joined. Free ends of $A$ and $B$ are tied to two rigid supports $C$ and $D$, respectively, creating a tension of $500 \ N$ in the wire. Two identical pulses, sent from $C$ and $D$ ends, take time $t_1$ and $t_2$, respectively, to reach the joint. The ratio $t_1 / t_2$ is:

$A$ uniform rope of length $12 \ m$ and mass $6 \ kg$ hangs vertically from a rigid support. $A$ block of mass $2 \ kg$ is attached to the free end of the rope. $A$ transverse pulse of wavelength $0.06 \ m$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is (in $m$)

$A$ transverse wave is travelling with velocity $V$ through a metal wire of length $L$ and density $\rho$. The tensile stress in the wire is

$A$ composite string is made up by joining two strings of different mass per unit length,$\mu$ and $4\mu$. The composite string is under the same tension $T$. $A$ transverse wave pulse,$Y = (6 \text{ mm}) \sin(5t + 40x)$,where $t$ is in seconds and $x$ is in meters,is sent along the lighter string towards the joint. The joint is at $x = 0$. The equation of the wave pulse reflected from the joint is:

$A$ uniform rope of length $L$ and mass $m_1$ hangs vertically from a rigid support. $A$ block of mass $m_2$ is attached to the free end of the rope. $A$ transverse pulse of wavelength $\lambda_1$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is $\lambda_2$. The ratio $\lambda_2 / \lambda_1$ is: