The pressure is the force exerted per unit area. The SI unit of pressure is Newton per meter square.

The pressure exerted by air or atmosphere is called atmospheric pressure.

As we know that a sharp nail covers less area but the blunt one covers more area. Since the area is less in the sharper nail so the pressure is more but the blunt one has less pressure as it covers more area, so a sharp nail pierce a wooden block easily whereas a blunt one does not.

S.N. Air pressure Atmospheric pressure
1. It is measured by air gauge. It is measured by barometer.
2. Air pressure is local pressure, such as in a tire or balloon. Atmospheric pressure is the weight per unit area of 5 miles height of air piled up and acting at the Earth's surface.
3. Air pressure includes any air that is pressurized. Atmospheric pressure is the pressure of the air in the atmosphere.

S.N. Force Pressure
1. Force is the product of mass and acceleration. Pressure is the force acting per unit area.
2. F= P x A P = F/A
3. Unit=N i.e Newton Unit= Pa i.e Pascal

The pointed heel of a girl has less surface area than the leg of the elephant. As pressure is inversely proportional to the area, so the pointed heel of a girl exerts more pressure.

The foundation of building is made wider than walls in order to reduce the pressure exerted by its ceiling on the base.

Truck and buses have back wheels in pairs to reduce the pressure exerted by heavy loads on the wheels of buses and trucks.

Solution,

\begin{align*} \text {Given,} \\ \text {area} \:(A) = 0.1\:m^2 \\ \text {pressure}\:(p) = 1000\: Nm^2 \\ \text {force} \:(F) = ? \\ \text {we know,} \\ P &= \frac FA \\ \text {or,}\: F &= P\times A \\ &= 1000 \times 0.1 \\ &= 100\: N \\ \therefore \text {The force applied is } 100\: N\end{align*}

Solution,

\begin{align*} \text {Given,} \\ \text {Depth} \: (h) = 0.5\:m \\ \text {pressure} \:(p) = 5000\: N/m^2 \\ \text {Acceleration due to gravity} \:(g) = 10 m/s^2 \\ \text {Density} \:(d) =\: ? \\ \text {We know that,} \\ P &= h\times d\times g \\ \text {or,}\: 5000 &= 0.5 \times d\times 10 \\ \text {or,}\: 5 \times d &= 5000 \\ \text {or,}\: d &= \frac {5000}{5} \\ \therefore \: \text {Density is equal to} \: 1000\: kg/m^3 \end{align*}

Solution,

\begin{align*} \text {Given,} \\ \text {Force}\: (f) = 40\: N \\ \text {Pressure}\: (p)\: =\: 8000 \:N/m^2 \\ \text {Area} \:(A)\: =\: ? \\ \text {We know that} \\ \text {Pressure}\:(p) &= \frac FA \\\text {or,} \: A &= \frac FP \\ \text {or,} \: A &= \frac {40}{8000} \\ \therefore \text {The area is }\: 0.005\:m^2. \\ \end{align*}

Solution,

\begin{align*} \text {Given,}\:\text {Mass}\: (m)\: =\: 60\: kg \\ \text {Area}\: (A)\: =\; 120\: cm^2 \\ &= \frac {120}{100 \times 100} \ \\ &= 0.012\:m^2 \\ \text {Pressure}\: (P)\: =\:? \\ \text {Force}\: (F)\: &=\: m \times g \\ &=60 \times 10 \\ &=600\: N \\ \text {We know,}\\ P &= \frac FA \\ &= \frac { 600}{0.012} \\ \text {Now,} \\ \ &=5 \times 10^4 N/m^2 \\\text {His pressure on the floor is} \:5\times10^4 N/m^2. \\ \text {Now,} \\ \text {Area}\: (A)\: =\: 15000\: cm^2 \\ &=\frac {15000}{100 \times 100} \\ &=1.5 m^2 \\\text {Force}\: (F)\: =\: 600\: N\\ \text {Pressure}\: (P)\: &= \frac FA \\ &= 600/1.5 \\ &= 400 N/m^2. \\ \text {His pressure on the floor after sitting is} \: 400N/m^2.\\ \end{align*}

Barometer is an instrument used for measuring the atmospheric pressure. An Italian scientist named Torricelli invented this. It is made by one-meter long glass closed at its one end, which is taken and filled completely with mercury. Then, by closing the mouth of tube, it is inverted in a trough containing mercury. We can see that the level of the Hg in the glass tube decreases at first and finally becomes constant. The empty space above  the surface of Hg in the tube is collected in Tarrecellian vacuum.

Solution,

Let us consider a liquid in the beaker. Let 'A' be the area of the beaker and 'h' be the height upto which the liquid is filled. If 'd' be the density of the liquid and volume= A \(\times\) h, then we have,

\begin{align*} P & = \frac {\text {weight of the liquid}}{ \text {base area of a container}} \\ \text {or,}\: P &=\frac WA \dots (i) \\ \text {or,}\: P &= \frac {m \times g}{A} \dots (ii) \\ \text {Here, weight of the liquid}\: (w)\: &= mg \\ \text {We can also say that,} \\P &= \frac {dwg}{ A} \dots (iii)\:\:\:\: [\text {since,}\: d =\frac mv \: \text {or} m=d \times v] \\ \text {or.}\: p &= d \times \frac {(A \times h)g}{ A} [since, v= A x h]
\\ \therefore P &= dhg,\: \text { proved.} \\ \end{align*}