Correct Answer - (i) `(1)/(1089pi)m` , (ii) `(1)/(1089)kg//m^(3)`
Given length of pipe `= 3m`
Third harmonic Implies that `(3lambda)/(2) , lambda = (2l)/(3), = (2 xx 3)/(3) = 2 m`
`k = (2pi)/(lambda) = pi`
`BkS = P`
`BpiS = 100`
`B = rhoV^(2) = 330^(2) xx 1`
`S = (100)/(piB)`
`S = (100)/(pi330^(2))`
`S = (1)/(1089pi)`
Bulk moduls of elasticity
`B = (-dp)/((dv//v))`, Volume `= (mass)/(density) = (m)/(rho)`
`dv = (-m)/(rho^(2)) drho = (-Vdrho)/(rho) rArr (dv)/(V) = - (drho)/(rho) , dP = B(drho)/(rho)`
`Deltarho_(max) = (rho)/(B)DeltaP_(max) = (DeltaP_(max))/(v^(2)) = (10^(2))/((330)^(2)) , (1)/(1089) kg//m^(3)`