**Solution**

Given,

Frequency (f_{1}) = 440 Hz

Temperature (T_{1}) = 27 + 273 = 300 K

Frequency (f_{2}) = ?

Temperature (T_{2}) = 0 + 273 = 273 K

For an open organ pipe, we have

\(f_1=\frac{v_1}{2l}\) ------ (i)

and, \(f_2=\frac{v_2}{2l}\) ------ (ii)

Dividing equation (ii) by equation (ii) we get

\(\frac{f_2}{f_1}=\frac{v_2}{2l}\times\frac{2l}{v_1}=\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}\)

Or, \(f_2=\sqrt{\frac{273}{300}}\times440=419.73\) Hz

Hence, thefrequency at 0°C is 419.73 Hz