Question
A 10x1.5x0.2 cm3 glass plate weighs 8.6 gm in air. Now it is immersed half in water with longest side vertical. What will be its apparent weight? (surface tension of water is 70 dyne /cm.)
Solution
The different forces acting on the plate are:
(i) Weight W vertically downward,
(ii) Upward thrust U vertically upward,
(iii) Surface tension force T in downward direction .
These forces are given by,
Weight, W=8.6 x 980 = 8428 dyne
Upward Thrust, U=weight of water displaced
=Volume of water displaced x density of water x g
Let,
T = Total length of the water in touch with plate x surface tension
=2(1.5 + 0.2) x 70 = 238 dyne
Now the apparent weight,
Wa + W + T - U = 8428 + 238 - 1470
= 7196 dyne
= 7.434 gf
