
Sneha weighing 50 kg is standing on pencil heels, each having an area of cross section of$1c{m^2}$. An elephant weighing 2000kg and foot area of cross section of $250c{m^2}$ standing on the floor. Which one of the following is correct about the force exerted by them:
A) Sneha exerts 12.5 times more pressure than an elephant.
B) Elephant exerts 2000 times more pressure than Sneha.
C) Both exerts the same pressure.
D) Sneha exerts 20 times more pressure than the elephant.
E) None of these.
Answer
573.9k+ views
Hint: The pressure is defined as the ratio of perpendicular force and cross sectional area. The S.I unit of the pressure is Pascal. The pressure is directly proportional to the force and is inversely proportional to the area of cross section.
Formula used:The formula of the pressure is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
Where P is the pressure the force is given by F and the area of the cross section is A.
Complete step by step solution:
It is given in the problem that Sneha weighing 50 kg is standing on pencil heels, each having an area of cross section of$1c{m^2}$, an elephant weighing 2000kg and foot area of cross section of $250c{m^2}$ standing on the floor and we need to tell which of the following option is correct.
The pressure on the ground by Sneha.
The formula of the pressure is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
Where P is the pressure the force is given by F and the area of the cross section is A.
The weight of the Sneha is 50 kg and area of cross section of the heel is$1c{m^2}$ the pressure on the ground is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow {P_S} = \dfrac{{50 \times 10}}{{2 \times {{10}^{ - 4}}}}$
$ \Rightarrow {P_S} = 25 \times {10^5}$
$ \Rightarrow {P_S} = 25 \times {10^5}Pa$………eq. (1)
The pressure due to elephant,
The formula of the pressure is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
Where P is the pressure the force is given by F and the area of the cross section is A.
$ \Rightarrow {P _E} = \dfrac{{2000 \times 10}}{{4 \times 250 \times {{10}^{ - 4}}}}$
$ \Rightarrow {P_E} = \dfrac{{500 \times 10}}{{250 \times {{10}^{ - 4}}}}$
$ \Rightarrow {P_E} = 2 \times {10^5}Pa$………eq. (2)
Dividing the pressure applied by Sneha to the pressure applied by the elephant.
$ \Rightarrow \dfrac{{{P_S}}}{{{P_E}}}$
$ \Rightarrow \dfrac{{{P_S}}}{{{P_E}}} = \dfrac{{25 \times {{10}^5}}}{{2 \times {{10}^5}}}$
$ \Rightarrow \dfrac{{{P_S}}}{{{P_E}}} = 12 \cdot 5$
The pressure applied by the Sneha is 12.5 times that of the pressure applied by the Elephant. The correct answer for this problem is option A.
Note: It is advisable to students to understand and remember the formula of the pressure. The pressure is inversely proportional to the cross sectional area which means the pressure increases with decrease in cross sectional area.
Formula used:The formula of the pressure is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
Where P is the pressure the force is given by F and the area of the cross section is A.
Complete step by step solution:
It is given in the problem that Sneha weighing 50 kg is standing on pencil heels, each having an area of cross section of$1c{m^2}$, an elephant weighing 2000kg and foot area of cross section of $250c{m^2}$ standing on the floor and we need to tell which of the following option is correct.
The pressure on the ground by Sneha.
The formula of the pressure is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
Where P is the pressure the force is given by F and the area of the cross section is A.
The weight of the Sneha is 50 kg and area of cross section of the heel is$1c{m^2}$ the pressure on the ground is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow {P_S} = \dfrac{{50 \times 10}}{{2 \times {{10}^{ - 4}}}}$
$ \Rightarrow {P_S} = 25 \times {10^5}$
$ \Rightarrow {P_S} = 25 \times {10^5}Pa$………eq. (1)
The pressure due to elephant,
The formula of the pressure is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
Where P is the pressure the force is given by F and the area of the cross section is A.
$ \Rightarrow {P _E} = \dfrac{{2000 \times 10}}{{4 \times 250 \times {{10}^{ - 4}}}}$
$ \Rightarrow {P_E} = \dfrac{{500 \times 10}}{{250 \times {{10}^{ - 4}}}}$
$ \Rightarrow {P_E} = 2 \times {10^5}Pa$………eq. (2)
Dividing the pressure applied by Sneha to the pressure applied by the elephant.
$ \Rightarrow \dfrac{{{P_S}}}{{{P_E}}}$
$ \Rightarrow \dfrac{{{P_S}}}{{{P_E}}} = \dfrac{{25 \times {{10}^5}}}{{2 \times {{10}^5}}}$
$ \Rightarrow \dfrac{{{P_S}}}{{{P_E}}} = 12 \cdot 5$
The pressure applied by the Sneha is 12.5 times that of the pressure applied by the Elephant. The correct answer for this problem is option A.
Note: It is advisable to students to understand and remember the formula of the pressure. The pressure is inversely proportional to the cross sectional area which means the pressure increases with decrease in cross sectional area.
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