Correct Answer - A::B::C
a. `T=7.5xx10^-2N/m, costheta=1`
`(a)`. `h_A=(2Tcostheta)/(r_A.rhog)`
`=(2xx7.5xx10^-2)/(5xx10^-3xx1000xx10)`
`=3xx10^-2m=3cm`
`(b)`. `h_B=(2Tcostheta)/(r_B.rhog)`
`=(2xx7.5xx10^-2)/(5xx10^-3xx10^3xx10)`
`=15xx10^-xm=1.5cm`
c. `h_0=(2Tcostheta)/(r_Crhog)`
`=(2xx7.5xx10^-2)/(1.5xx10^-3xx10^3xx10)`
`=15/1.5xx10^-3=10^-2m=1cm`
The height of `H_2O` levle in A,B,C cappliareis are 3 cm 1.5 cm nd 1 cm respectivel.