$A$ uniform wire $20 \,m$ long and weighing $50 \,N$ hangs vertically. The speed of the wave at the midpoint of the wire is (acceleration due to gravity $= g = 10 \,ms^{-2}$)

If the initial tension on a stretched string is doubled,then the ratio of the initial and final speeds of a transverse wave along the string is:

Two strings with circular cross section and made of same material are stretched to have the same amount of tension. $A$ transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section $R$ is $v_1$,and that in the other string having radius of cross section $R/2$ is $v_2$. Then $\frac{v_2}{v_1} = $

$A$ string of mass $M$ is under a tension $T$. The length of the string is $L$. $A$ transverse wave starts at one end of the string. The time required for the disturbance to reach the other end 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:

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: