
What force is required to produce a pressure of \[1000Pa\] on an area of \[250c{{m}^{2}}\]?
Answer
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Hint: In this question relation between force, pressure and area is used as these three terms are interrelate to each other and in these three terms value of two terms are given in the question so we can easily find out the third term and also remember to use proper system of unit to get the required result.
Complete step-by-step solution:
When perpendicular force or thrust acting on a unit area then the pressure on that surface is calculated. So pressure, force and area are interrelated to each other. This expression can be represented as :-
\[Pressure=\dfrac{Force}{Area}\].
The SI unit of pressure is \[\dfrac{N}{{{m}^{2}}}\] and it is also equivalent to\[Pascal(Pa)\].
Pressure is a scalar quantity while force is a vector quantity.
Since in the given question Pressure is given in Pascal. As Pascal is also the SI system of units , we do not need to convert it to any other system of units.
Let us assume pressure can be represented by \[P\].
So, in the question given pressure is
\[P=1000Pa\]
Let us assume the area can be reprinted by\[A\].
So, in the question given area is
\[A=250c{{m}^{2}}\]
Since Pascal is given in S.I system of unit and area is given in C.G.S system of unit, so area should also be converted in SI system so we will convert area in metre square by using the simple conversion factor between centimetre and metre.
Since we know that,
\[1cm={{10}^{-2}}m\]
So area becomes,
\[A=250\times {{10}^{-4}}{{m}^{2}}\]
Now we have to calculate the force , so we should apply the formula that relates force, pressure and area which can be represented mathematically as :-
\[F=PA\]
Where ,
\[F=\]Force.
So we will put the values of pressure and area in this above formula, we get
\[F=1000\times 250\times {{10}^{-4}}\]
on solving we get,
\[\therefore F=25N\]
The SI unit of force is Newton which is represented by the symbol\[N\].
So we can conclude that \[25N\]force is required to produce a pressure of \[1000Pa\]on an area of\[250c{{m}^{2}}\]
Note: As force is defined as simple push and pull, Force shows many effects when applied to some objects or bodies like it can change the shape and size of that object, it can change the state of rest or motion of object, it also changes the direction of object, it also can oscillate the body about mean position.
Complete step-by-step solution:
When perpendicular force or thrust acting on a unit area then the pressure on that surface is calculated. So pressure, force and area are interrelated to each other. This expression can be represented as :-
\[Pressure=\dfrac{Force}{Area}\].
The SI unit of pressure is \[\dfrac{N}{{{m}^{2}}}\] and it is also equivalent to\[Pascal(Pa)\].
Pressure is a scalar quantity while force is a vector quantity.
Since in the given question Pressure is given in Pascal. As Pascal is also the SI system of units , we do not need to convert it to any other system of units.
Let us assume pressure can be represented by \[P\].
So, in the question given pressure is
\[P=1000Pa\]
Let us assume the area can be reprinted by\[A\].
So, in the question given area is
\[A=250c{{m}^{2}}\]
Since Pascal is given in S.I system of unit and area is given in C.G.S system of unit, so area should also be converted in SI system so we will convert area in metre square by using the simple conversion factor between centimetre and metre.
Since we know that,
\[1cm={{10}^{-2}}m\]
So area becomes,
\[A=250\times {{10}^{-4}}{{m}^{2}}\]
Now we have to calculate the force , so we should apply the formula that relates force, pressure and area which can be represented mathematically as :-
\[F=PA\]
Where ,
\[F=\]Force.
So we will put the values of pressure and area in this above formula, we get
\[F=1000\times 250\times {{10}^{-4}}\]
on solving we get,
\[\therefore F=25N\]
The SI unit of force is Newton which is represented by the symbol\[N\].
So we can conclude that \[25N\]force is required to produce a pressure of \[1000Pa\]on an area of\[250c{{m}^{2}}\]
Note: As force is defined as simple push and pull, Force shows many effects when applied to some objects or bodies like it can change the shape and size of that object, it can change the state of rest or motion of object, it also changes the direction of object, it also can oscillate the body about mean position.
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