A bottle is kept on the ground as shown in the figure. The bottle can be modelled as having two cylindrical zones. The lower zone of the bottle has a $$cross-sectional$$ radius of $$R2$$ and is filled with honey of density $$2$$. The upper zone of the bottle is filled with the water of density ρ and has a $$cross-sectional$$ radius R. The height of the lower zone is H while that of the upper zone is $$2H$$. Now the honey and the water parts are mixed together to form a homogenous solution (Assume$$ that total volume does not $$change).Now match the given columns and select the correct option from the codes given below.

Question

A bottle is kept on the ground as shown in the figure. The bottle can be modelled as having two cylindrical zones. The lower zone of the bottle has a $$cross-sectional$$ radius of $$R2$$ and is filled with honey of density $$2$$. The upper zone of the bottle is filled with the water of density ρ and has a $$cross-sectional$$ radius R. The height of the lower zone is H while that of the upper zone is $$2H$$. Now the honey and the water parts are mixed together to form a homogenous solution (Assume$$ that total volume does not $$change).Now match the given columns and select the correct option from the codes given below.

Infinity Learn · Accepted Answer

Initial pressure at the $$bottom=$$ ρg × $$2H$$ $$+$$ $$2$$ρ×g×$$H=$$ $$4$$ρgHBase area of lower zone will be double of upper zone.Final density of the homogenous mixture: $$=$$ ρ$$1V1+$$ρ$$2V2V1+V2=$$ ρ×A×$$2H+2$$ρ×$$2A$$×HA×$$2H+2A$$×H $$=$$ $$32$$ρFinal pressure $$=$$ $$32$$ρ×g×$$3H$$ $$=$$ $$92$$ρgHpressure at P $$=$$ $$32$$ρ×g×$$2H$$ $$=$$ $$3$$ρgH

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