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The damped air contains 10% vapour by volume. This means the mixture consists of 10 parts by volume of vapours and 90 parts by volume of dry air. Let the densities of the mixtures, water vapour and dry air be `p_(m), p_(w) and p_(d)," Let "V_(m)` be the velocity of sound in mixture and `V_d` velocity in the dry air. `p_(d)/p_(w)=1.6, p_(w)=(p_(d))/(1.6)` <br> Let the volume of the mixture be 100V. <br> Volume of dry air in the mixture =90V <br> Volume of water vapour in the mixture =10V <br> `100 V xx p_(m)=90 V xx p_(d) +10 V xx p_(w)` <br> `10p_(m)=9p_(d) divide p_(w)=9p_(d)+p_(d)/(1.6) =(154)/(16)p_(d)` <br> `p_(d)/p_(m)=(160)/(154) V_(m)=V_(d) sqrt(p_d/p_m)=340 sqrt((160)/(154))` <br> `V_m=346.56ms^(-1).`