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Consider a bottle as shown in the figure. The bottle can be modeled as having to 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 2H. If now the honey and the water parts are mixed together to form a homogeneous solution (Assume that total volume does not change)

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The pressure inside the bottle at the base will increase by an amount

(1/2) ρ gH

The pressure inside the bottle at the base will remain unaltered

The pressure inside the bottle at the base will decreases by an amount

(1/4) ρ gH

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

answer is B, C.

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Detailed Solution

The normal reaction on the bottle from the ground will remain unchanged.

Pressure inside the bottle before mixing honey with water

$P = P_{1} + P_{2} = (2 H) ρ g + H (2 ρ) g$

$P^{1} = 4 H ρ g$

Density of Honey and water after mixing $ρ = \frac{m_{1} + m_{2}}{v_{1} + v_{2}}$

$ρ = \frac{2 π R^{2} H ρ + 4 π R^{2} H ρ}{2 π R^{2} H + 2 π R^{2} H} = \frac{6 π R^{2} H ρ}{4 π R^{2} H} = \frac{3 ρ}{2}$

Pressure inside the bottle after mixing honey with water

$P^{| |} = (3H) (\frac{3 ρ}{2}) g = \frac{9}{2} H ρ g$

Increase in pressure

$= ρ^{H} - ρ^{1} = \frac{9}{2} H ρ g - 4 H ρ g$

$= \frac{1}{2} H ρ g$

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