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

Answer: (b) Pressure $=\dfrac{\text{thrust}}{\text{area}}$=thrust/area. The thrust (the person's weight) is shared by hundreds of nail tips, so the total area is large and the pressure at each point is too small to pierce the skin.

Answer: (c) $1.25h$1.25h.

$P=h\rho g$P=hρ g, so for the same pressure $h_1\rho_1=h_2\rho_2$h₁ρ₁=h₂ρ₂. The liquid is $0.8$0.8 times as dense as water, so $h_2=\dfrac{h\times1}{0.8}=1.25h$h₂=h×1/0.8=1.25h. A less dense liquid needs a deeper column to give the same pressure.

Answer: (a) Since $P=h\rho g$P=hρ 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 m².

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

(i) $P=h\rho g=5\times1000\times10=\boxed{5\times10^{4}\text{ Pa}}$P=hρ g=5×1000×10=5×10⁴ Pa

(ii) $P\propto h$P∝ h in the same liquid, so double the pressure needs double the depth: $\boxed{10\text{ m}}$10 m.

(iii) $P=h\rho g=5\times800\times10=\boxed{4\times10^{4}\text{ Pa}}$P=hρ g=5×800×10=4×10⁴ Pa The same depth gives a smaller pressure because kerosene is less dense than water.

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

(ii) Area of base $A=2\times1.5=3\text{ m}^2$A=2×1.5=3 m².
$\text{Thrust}=P\times A=30000\times3=\boxed{90000\text{ N}}=9\times10^{4}\text{ N}$Thrust=P× A=30000×3=90000 N=9×10⁴ N

(i) Liquid pressure $P=h\rho g$P=hρ g becomes very large at great depths. The strong suit withstands this pressure, which would otherwise squeeze the diver's body and make breathing difficult.

(ii) The pressure at a tap depends on the height $h$h of the water surface above it. A tank on the roof gives a large $h$h, so water reaches every floor with enough pressure to flow out.

(iii) Pressure increases with depth ($P\propto h$P∝ h), so the deeper the swimmer goes, the greater the pressure of the water on the eardrums.

Derivation: Consider a point at depth $h$h below the free surface of a liquid of density $\rho$ρ. 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=ρ V=ρ A h.

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

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

(a) Pressure applied at any point of a liquid enclosed in a vessel is transmitted equally and undiminished in all directions throughout the liquid.

(b) Pressure $=\dfrac{F}{A}=\dfrac{50}{10\times10^{-4}}=5\times10^4\text{ Pa}$=F/A=5010×10^-4=5×10⁴ Pa. The same pressure acts on the large piston:
$F=P\times A=5\times10^4\times500\times10^{-4}=2500\text{ N}.$F=P× A=5×10⁴×500×10^-4=2500 N.
(The force is multiplied $50$50 times, the ratio of the areas.)

(c) Hydraulic brakes and the hydraulic jack (or hydraulic lift).

(i) Water pressure $=h\rho g=20\times1000\times10=\boxed{2\times10^{5}\text{ Pa}}$=hρ g=20×1000×10=2×10⁵ Pa.

(ii) Thrust $=P\times A=2\times10^{5}\times0.02=\boxed{4000\text{ N}}$=P× A=2×10⁵×0.02=4000 N, which is why the mask glass must be thick.

(iii) No. Liquid pressure $P=h\rho g$P=hρ 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).