Pressure in Fluids and Atmospheric Pressure — ICSE Class 9 Physics Important Questions

Answer: (b) pascal.

Pressure $=\dfrac{\text{force}}{\text{area}}$=force/area, so its unit is $\text{N m}^{-2}$N m^-2, called the pascal ($\text{Pa}$Pa).

Answer: (c) the depth and density of the liquid.

$P=h\rho g$P=h g, so pressure depends only on depth $h$h, density $\rho$ and $g$g — not on the shape, area or total volume of the liquid.

Answer: (b) barometer.

A barometer (commonly the mercury barometer) measures atmospheric pressure; a manometer measures gas/fluid pressure and a hydrometer measures relative density.

Answer: (c) $10.3\text{ m}$10.3 m.

Same pressure: $h_w\rho_w=h_m\rho_m$h_w_w=h_m_m. $h_w=h_m\times\dfrac{\rho_m}{\rho_w}=0.76\times13.6=10.34\text{ m}$h_w=h_m×_m/_w=0.76×13.6=10.34 m.

Answer: (a) Since $P=h\rho g$P=h g, pressure is greatest at the base of the dam, so the wall must be thickest there to withstand it; R correctly explains A.

Thrust is the total force acting perpendicularly (normally) on a surface; its SI unit is the newton (N).

Pressure is the thrust acting per unit area, $P=\dfrac{\text{thrust}}{\text{area}}$P=thrust/area; its SI unit is the pascal (Pa), where $1\text{ Pa}=1\text{ N m}^{-2}$1 Pa=1 N m^-2.

Given thrust $F=200\text{ N}$F=200 N, area $A=0.5\text{ m}^2$A=0.5 m^2.

$P=\dfrac{F}{A}=\dfrac{200}{0.5}=\boxed{400\text{ Pa}}$P=F/A=200/0.5=400 Pa

Pressure due to the water column:
$P=h\rho g=5\times1000\times10=\boxed{50000\text{ Pa}}=5\times10^{4}\text{ Pa}$P=h g=5×1000×10=50000 Pa=5×10^4 Pa

Total pressure at that depth = atmospheric + liquid pressure
$=1.0\times10^{5}+0.5\times10^{5}=\boxed{1.5\times10^{5}\text{ Pa}}$=1.0×10^5+0.5×10^5=1.5×10^5 Pa

(i) Pressure on the base:
$P=h\rho g=3\times1000\times10=\boxed{30000\text{ Pa}}$P=h g=3×1000×10=30000 Pa

(ii) Area of base $A=2\times1.5=3\text{ m}^2$A=2×1.5=3 m^2.
$\text{Thrust}=P\times A=30000\times3=\boxed{90000\text{ N}}=9\times10^{4}\text{ N}$Thrust=P× A=30000×3=90000 N=9×10^4 N

Convert: $h=6\text{ cm}=0.06\text{ m}$h=6 cm=0.06 m, $A=300\text{ cm}^2=300\times10^{-4}=0.03\text{ m}^2$A=300 cm^2=300×10^-4=0.03 m^2.

(i) $P=h\rho g=0.06\times1000\times10=\boxed{600\text{ Pa}}$P=h g=0.06×1000×10=600 Pa.

(ii) $\text{Thrust}=P\times A=600\times0.03=\boxed{18\text{ N}}$Thrust=P× A=600×0.03=18 N.

(Check: this equals the weight of water, mass $=0.03\times0.06\times1000=1.8\text{ kg}$=0.03×0.06×1000=1.8 kg, weight $=18\text{ N}$=18 N.)

Derivation: Consider a point at depth $h$h below the free surface of a liquid of density $\rho$. Imagine a horizontal area $A$A at that depth. The column of liquid above it has volume $V=A\times h$V=A× h and mass $m=\rho V=\rho A h$m= V= A h.

Weight (thrust) on the area $=mg=\rho A h g$=mg= A h g.

Pressure $=\dfrac{\text{thrust}}{\text{area}}=\dfrac{\rho A h g}{A}=\boxed{h\rho g}$=thrust/area= A h g/A=h g.

Properties and applications:

1. Pressure increases with depth ($P\propto h$P h). Application: dam walls are built thicker at the base to withstand greater pressure.
2. Liquid pressure acts equally in all directions at a given depth and is the same at all points on a horizontal level. Application: the hydraulic press / brakes transmit pressure equally throughout the fluid.

Barometer: A long glass tube closed at one end is filled with mercury and inverted into a mercury trough. The mercury falls until the pressure due to the mercury column balances the atmospheric pressure acting on the mercury in the trough. There is a vacuum (Torricellian vacuum) above the column. The vertical height of the column then measures the atmospheric pressure: $P_{atm}=h\rho g$P_atm=h g.

Why mercury: it is very dense, so the column is short and manageable (about $76\text{ cm}$76 cm rather than about $10.3\text{ m}$10.3 m for water); it does not stick to glass, has negligible vapour pressure, and is opaque so the level is easy to read.

Calculation: $h=75\text{ cm}=0.75\text{ m}$h=75 cm=0.75 m.
$P=h\rho g=0.75\times13600\times9.8=\boxed{99960\text{ Pa}}\approx1.0\times10^{5}\text{ Pa}$P=h g=0.75×13600×9.8=99960 Pa1.0×10^5 Pa

(i) Water pressure $=h\rho g=20\times1000\times10=\boxed{2\times10^{5}\text{ Pa}}$=h g=20×1000×10=2×10^5 Pa.

(ii) Total pressure = atmospheric + water pressure
$=1.0\times10^{5}+2.0\times10^{5}=\boxed{3.0\times10^{5}\text{ Pa}}$=1.0×10^5+2.0×10^5=3.0×10^5 Pa

(iii) No. Liquid pressure $P=h\rho g$P=h g depends only on depth, density and $g$g; it does not depend on the surface area or total quantity of water in the lake.

(iv) Going deeper increases $h$h, so the pressure increases (in direct proportion to the depth).