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 $R\sqrt{2}$ and is filled with honey of density $2\rho$. The upper zone of the bottle is filled with water of density $\rho$ and has a cross-sectional radius R. The height of the lower zone is H and that of the upper zone is 2H. If now the honey and water parts are mixed to form a homogeneous solution (Assume that the total volume does not change)

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

Now match the given columns and select the correct option from the codes given below

• A. i - p, ii - r, iii - q, iv - s
• B. i - p, ii - s, iii - q, iv - r
• C. i - s, ii - r, iii - p, iv - q
• D. i - p, ii - q, iii - r, iv -s

Answer 1

The correct option is A i - p, ii - r, iii - q, iv - s

Initial pressure at bottom = $\rho g\times 2H+2\rho \times g\times H=4\rho gH$
Base area of lower zone will be double of upper zone.
Final density of the homogeneous mixture:
$=\frac{{\rho }_1{V}_1+{\rho }_2{V}_2}{V_1+V_2}=\frac{\rho \times A\times 2H+2\rho \times 2A\times H}{A\times 2H+2A\times H}=\frac{3}{2}\rho$
Final pressure = $\frac{3}{2}\rho \times g\times 3H=\frac{9}{2}\rho gH$
Pressure at P = $\frac{3}{2}\rho \times g\times 2H=3\rho gH$