Solution

Given,

Length of wire (l_w) = 40 cm = 0.4 m
Diameter of wire (d) = 0.25 mm = 0.25 x 10^-3 m
Length of pipe (l_p) = 60 cm = 0.6 m
Density of steel (ρ) = 7800 kg/m^3

Let, 

v₁ = 332 m/s
T₁ = 0C = 273 K
v₂ = ?
T₂ = 27C = 300 K

Now using, the velocity temperature relation, we get

\(\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}\)
Or, \(\frac{332}{v_2}=\sqrt{\frac{273}{300}}\)
∴ v₂ = 348.03 m/s

Then, the fundamental frequency of vibration of wire is given as

\(f=\frac1{2l_w}\sqrt{\frac T\mu}\)
Or, \(f=\frac1{2l_w}\sqrt{\frac T{V\times\rho}}\) , where V is the volume per unit length
Or, \(f=\frac1{2l_w}\sqrt{\frac T{\displaystyle\frac{\pi d^2l}{4l}\times\rho}}=\frac1{2l_w}\sqrt{\frac{4T}{\pi d^2\rho}}=\frac1{2l_w}\times\frac2d\sqrt{\frac T{\pi\rho}}=\frac1{l_wd}\sqrt{\frac T{\pi\rho}}\)

Again, the fundamental frequency of vibration in an open pipe is given as:

\(f'=\frac v{2l_0}\)

When steel wire and open pipe vibrates in unison, we have

f = f'
Or, \(\frac1{l_wd}\sqrt{\frac T{\pi\rho}}=\frac v{2l_0}\)
Or, \(\frac1{0.4\times0.25\times10^{-3}}\sqrt{\frac T{\pi\times7800}}=\frac{348.03}{2\times0.6}\)
Or, \(\sqrt{\frac T{7800\pi}}=0.029\)
Or, \(\frac{\displaystyle T}{\displaystyle7800\pi}=8.41\times10^{-4}\)
∴ T = 20.6 N

Hence, the required tension is 20.6 N