Question

A glass plate of length 10 cm and breadth 1.54 cm and thickness 0.20 cm weighs 8.2 gm in the air. It is held vertically with the long side horizontal and the lower half under water. Find the apparent weight of the plate. [Surface tension of water = 73 dyne per cm g = 980 cm/$se{c}^2$]

Answer 1

Volume of the portion of the plate immersed in water is
$10\times \frac{1}{2}\left(1.54\right)\times 0.2=1.54c{m}^3$
Therefore if the density of water is taken a 1 then upthrust = Wt. of the water displaced = 1.54 x 1 x 980 = 1509.2 dynes
Now the total length of the plate in contact with the water surface is 2 (10 + 0.2 ) = 20.4 cm
$\therefore$ downward pull upon the plate due to surface tension = 20.4 x 73 = 1489.2 dynes $\therefore$ resultant upthrust
= 1509.2 - 1489.2 = 20.0 dynes = $\frac{20}{980}=0.0204gm-wt$
$\therefore$ apparent weight of the plate in water = weight of the plate in air - resultant upthrust = 8.2 - 0.0204 = 8.1796 gm