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**Hint:**Use the formula for the pressure acting on an object. This formula gives the relation between the pressure acting on the object, force acting on the object and the surface area on which the force is acting. The force acting on the ground is the weight of the girl in the downward direction. Calculate the pressure exerted by the girl on the ground when she stands on her two feet and one foot using this formula.

**Formula used:**

The pressure \[P\] acting on an object is given by

\[P = \dfrac{F}{A}\] …… (1)

Here, \[F\] is the force acting on the object and \[A\] is the surface area on which the force is acting.

**Complete step by step answer:**

We have given that the weight of the girl is \[550\,{\text{N}}\].

\[mg = 550\,{\text{N}}\]

Here, \[m\] is the mass of the girl.

The area of contact of one flat shoe with the ground is \[160\,{\text{c}}{{\text{m}}^2}\].

\[A = 160\,{\text{c}}{{\text{m}}^2}\]

Convert the unit of the area of the one flat show in contact with the ground to the SI system of units.

\[A = \left( {160\,{\text{c}}{{\text{m}}^2}} \right)\left( {\dfrac{{{{10}^{ - 4}}\,{{\text{m}}^2}}}{{1\,{\text{c}}{{\text{m}}^2}}}} \right)\]

\[ \Rightarrow A = 160 \times {10^{ - 4}}\,{{\text{m}}^2}\]

Hence, the area of the one flat shoe in contact with the ground is \[160 \times {10^{ - 4}}\,{{\text{m}}^2}\].

(a) In the first case, the girl is standing on the ground with her two feet.

Hence, the area \[A'\] of the shoe in contact with the ground is twice the area of contact of the one shoe with the ground.

\[A' = 2A\]

The force acting on the two feet of the girl is her own weight. Let us calculate the pressure \[{P_1}\] exerted by the girl on the ground when she is standing on her two feet.

Substitute \[{P_1}\] for \[P\], \[mg\] for \[F\] and \[A'\] for \[A\] in equation (1).

\[{P_1} = \dfrac{{mg}}{{A'}}\]

Substitute \[2A\] for \[A'\] in the above equation.

\[{P_1} = \dfrac{{mg}}{{2A}}\]

Substitute \[550\,{\text{N}}\] for \[mg\] and \[160 \times {10^{ - 4}}\,{{\text{m}}^2}\] for \[A\] in the above equation.

\[{P_1} = \dfrac{{550\,{\text{N}}}}{{2\left( {160 \times {{10}^{ - 4}}\,{{\text{m}}^2}} \right)}}\]

\[ \Rightarrow {P_1} = 1.7187 \times {10^4}\,{\text{N/}}{{\text{m}}^2}\]

**Hence, the pressure exerted by the girl on the ground when she stands on her two feet is \[1.7187 \times {10^4}\,{\text{N/}}{{\text{m}}^2}\].**

(b) In the second case, the girl is standing on the ground with her one foot.Hence, the area \[A\] of the shoe in contact with the ground is the area of contact of the one shoe with the ground.The force acting on the one feet of the girl is her own weight.Let us calculate the pressure \[{P_2}\] exerted by the girl on the ground when she is standing on her one foot.Substitute \[{P_2}\] for \[P\] and \[mg\] for \[F\] in equation (1).

\[{P_2} = \dfrac{{mg}}{A}\]

Substitute \[550\,{\text{N}}\] for \[mg\] and \[160 \times {10^{ - 4}}\,{{\text{m}}^2}\] for \[A\] in the above equation.

\[{P_2} = \dfrac{{550\,{\text{N}}}}{{160 \times {{10}^{ - 4}}\,{{\text{m}}^2}}}\]

\[ \Rightarrow {P_2} = 3.4375 \times {10^4}\,{\text{N/}}{{\text{m}}^2}\]

**Hence, the pressure exerted by the girl on the ground when she stands on her one foot is \[3.4375 \times {10^4}\,{\text{N/}}{{\text{m}}^2}\].**

**Note:**The students may think that when the girl is standing on her one foot then the area of contact of the shoe with the ground reduces to half in the same way like the force of weight of the girl acting on the ground also reduces to half. But the students should keep in mind that the total weight of the girl acts on the ground of surface area equal to are of contact of one shoe with the ground and not the half weight.

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