A glass plate of length 50 cm, breadth 40 cm and thickness 5 cm weighs 3 kg in air. If it is held vertically with long side horizontal and the plate half immersed in a liquid, what will be its apparent weight (in N)? Surface tension of the liquid = 50 x 10-2 N/m.

Here, l = 50 cm; b = 40 cm and t = 5 cm;        T=50×10−3N/mThe glass plate is under the action of the following forces (i)   weight mg acting vertically downwards, (ii) force due to surface tension acting vertically downwards and (iii) upthrust experienced due to the half-immersed part of the plate. Thus, apparent weight of the glass plate,            mg′=mg+2(l+t)×T−l×t×b2ρg=3×10+2(50+5)×10−2×0.5−50×5×402×10−6×103×10=30+0.55−25=5.55N