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 R $\sqrt{2}$ 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).
Column-I Column-II
i. Net force on bottle in horizontal direction p. zero
ii. Pressure at base of bottle before mixing of water and honey q. $\frac{9}{2} ρgH$
iii. Preesure at the base after mixing r. 4 $ρgH$
iv. Pressure at point P (figure) after making s. 3 $ρgH$
Now match the given columns and select the correct option from the codes given below.

Moderate

Solution

Initial pressure at the bottom
= $ρg \times 2 H + 2 ρ \times g \times H$
= $4 ρgH$
Base area of lower zone will be double of upper zone.
Final density of the homogenous mixture:

$= \frac{ρ_{1} V_{1} + ρ_{2} V_{2}}{V_{1} + V_{2}}$
= $\frac{ρ \times A \times 2 H + 2 ρ \times 2 A \times H}{A \times 2 H + 2 A \times H} = \frac{3}{2} ρ$
Final pressure = $\frac{3}{2} ρ \times g \times 3 H = \frac{9}{2} ρgH$
pressure at P = $\frac{3}{2} ρ \times g \times 2 H = 3 ρgH$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME

	Column-I		Column-II
i.	Net force on bottle in horizontal direction	p.	zero
ii.	Pressure at base of bottle before mixing of water and honey	q.	$\frac{9}{2} ρgH$
iii.	Preesure at the base after mixing	r.	4 $ρgH$
iv.	Pressure at point P (figure) after making	s.	3 $ρgH$