A bottle is kept on the ground as shown in the figure. The bottle can be modeled 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 $2 ρ$ . 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 $2 H$ . If now the honey and the water parts are mixed together to form a homogeneous solution : (Assume that total volume does not change)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The pressure inside the bottle at the base will increase by an amount (1/4) $ρ g H$

The pressure inside the bottle at the base will remain unaltered.

The normal reaction on the bottle from the ground will remain unaltered

The pressure inside the bottle at the base will increase by an amount (1/2) $ρ g H$

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Mass of liquid in upper cylinder $m_{1} = ρ π R^{2} (2 H) = 2 ρ R^{2} H$
Mass of liquid in lower cylinder $m_{2} = 2 ρ π {(R \sqrt{2})}^{2} 4 = 4 ρ π R^{2} H$
$∴ m = m_{1} + m_{2} = 6 ρ π R^{2} h$
Normal reaction from ground on the bottle ${N=mg=6ρπR}^{2} hg=constant$ . Since total mass is constant.
Density of homogeneous mixture
$ρ^{1} = \frac{m}{π R^{2} 2 H + π {(\sqrt{2 R})}^{2} H} = \frac{6 ρ π R^{2} h}{4 π R^{2} H} = \frac{3 ρ}{2}$
Pressure on bottom before mixing
$P_{1} = P_{0} + ρ g (2 H) + 2 ρ g H = P_{0} + 4 ρ g H$
Pressure on bottom after mixing
$P_{2} = P_{0} + (\frac{3 ρ}{2}) o f (3 H) = P_{0} + \frac{9}{2} ρ g h$
$∴ P_{2} - P_{1} = \frac{1}{2} ρ g H$
Hence (b) and (c) are correct.