A glass plate of length `10cm`, breath `1.54 cm`and thickness `0.20cm` weigh `8.2 gm` in air. It is held vertically with the long side horizontal and the lower half under water. Find the apperent weight of the plate. Surface tension of water `=73` dyne per cm, `g=980 cm//sec^(2)`.

Explanation

The force acting on the plate are
a. Weight of plate (`mg`) acting vertically downward.
b. force of surface tension (`TL`) acting vertically downward.
c. Upthrust.

Weight of glass plate `=8.2g=8.2xx10^(-3)kg`
Total length of plate is contact in with water surface is
`l=2(10+0.2)cm=20.4xx10^(-2)m`
Therefore downard pull upon the plate due to surface tension
`TL=7.xx10^(-2)xx20.4xx10^(2)=14.892xx10^(-3)N`
`=(14.892xx10^(-3))/9.8=1.5196xx10^(-3)kgwt`
Volume of plate immersed in water is
`v=10xx1/2(1.54)xx0.2=1.54.10^(-6)m^(3)`
Therefore, upthrust `=vsigmag=1.54xx10^(-6)x10^(3)xx9.8N`
`=(1.54xx10^(-3)xx9.8)/9.8kgwt=1.54xx10^(-3)kgwt`
Therefore apparent weight of plate in water `=` weight of plate in air `+` Force of surface tension `-`Upthrust
`=8.2xx10^(-3)+1.5196xx10^(-3)-1.54xx10^(-3)`
`=8.1796xx10^(-6)kgw=8.1796kgwt`