Dear student Considering equilibrium of water in the cone Weight of water=mass×g=volume×density×g=13πR2Hρgforce exerted by the bottom on water in the upward direction is=Pressure at the bottom×area of the bottom=Hρg×πR2as mg is lesser than Hρg×πR2 so curved part of the cone applies a force in the downward direction Hρg×πR2=13Hρg×πR2+FF=(1-13) Hρg×πR2=23 Hρg×πR2Regards

Solve this:

Solve this: A heavy hollow cone of radius R and height h is placed on a horizontal table surface, with its flat base on the table. The whole volume inside the cone is filled with water of density p. The circular of the cone's base has a watertight seal with the table's surface and the top apex of the cone has a small hole. Neglecting atmospheric pressure find the total upward force exerted by water on the cone is (d) None (c) ltR2hpg (b) (I / 3)TtR2hpg (a) (2/3)nR2hpg

Dear student

C o n s i d e r i n g e q u i l i b r i u m o f w a t e r i n t h e c o n e W e i g h t o f w a t e r = m a s s \times g = v o l u m e \times d e n s i t y \times g = \frac{1}{3} π R^{2} H ρ g f o r c e e x e r t e d b y t h e b o t t o m o n w a t e r i n t h e u p w a r d d i r e c t i o n i s = P r e s s u r e a t t h e b o t t o m \times a r e a o f t h e b o t t o m = H ρ g \times π R^{2} a s m g i s l e s s e r t h a n H ρ g \times π R^{2} s o c u r v e d p a r t o f t h e c o n e a p p l i e s a f o r c e i n t h e d o w n w a r d d i r e c t i o n H ρ g \times π R^{2} = \frac{1}{3} H ρ g \times π R^{2} + F F = (1 - \frac{1}{3}) H ρ g \times π R^{2} = \frac{2}{3} H ρ g \times π R^{2} R e g a r d s

View Full Answer