Question

damstty plund DAbloek f Hoat face down in a lens p(20oo tThe block 800m theus has height H= 6.0 cm ) By what deft s the block Submarpa If block is fully submarsl find the aceleaton te blace.

Answer 1

Sol- from Question we have Block density f = 8ookg/m3 Eluid density of 1200kg/m3 The block height H=6cm. (a) Let A is area of bottom of block So volume of block = Area xheight V =AH mass of Block = POV So in this case Volume of displaced fluid=Ah Ah= Volume of that part of object beneath the surface of fluid we have Buoyancy fore = weight of Block Pf g Ah Povg Р ЯАЖ = Pb A Ha h = P6 H Sf h=800x6 12.00 h=4cm Hence h=4cm (6) If Block is fully submerged then we have V = displaced fluid Volume = Object Block Volume Buoyant force=Pq gv . weight of Block = f g v so we het force (Po-Pogv=ma Where m= mass of block a= acceleration m=PV so we have (Pf-8) gv=Pova ilf-Polg = 9 P6 (1200-800x10=a g=10m/s2 5 m/s²=a Hence acceleration of Block = 5m/s2