Express $1Pa$ in SI units.
$A)\text{ }1N/m$
$B)\text{ }1m/N$
$C)\text{ }1kg.{{m}^{-1}}{{s}^{-1}}$
$D)\text{ }1kg.{{m}^{-1}}{{s}^{-2}}$
Answer
653.4k+ views
Hint: We can solve this problem by writing the given unit in terms of its component units using the mathematical definition of pressure. Then we can convert these component units into the respective SI units to get the required expression of $1Pa$ in SI units.
Formula used:
$\text{Pressure = }\dfrac{\text{force}}{\text{area}}$
Complete step by step answer:
Pressure upon a surface is defined as the ratio of the normal force acting on the surface to the surface area.
Hence,
$\text{Pressure = }\dfrac{\text{force}}{\text{area}}$
Using this definition, $1Pa$ is also defined as
$1Pa=\dfrac{1N}{1{{m}^{2}}}$ --(1)
Where Newton $\left( N \right)$ is the SI unit of force and the SI unit for surface area is meter-squared $\left( {{m}^{2}} \right)$.
Now, the Newton $\left( N \right)$ is the SI unit of force and can also be written as
$1N=1kg.m.{{s}^{-2}}$ --(2)
Putting (2) in (1), we will get,
$1Pa=\dfrac{1N}{{{m}^{2}}}=\dfrac{\dfrac{1kg.m}{{{s}^{2}}}}{{{m}^{2}}}=\dfrac{1kg.m}{{{m}^{2}}{{s}^{2}}}=\dfrac{1kg}{m.{{s}^{2}}}=1kg.{{m}^{-1}}{{s}^{-2}}$
Hence, we have derived that
$1Pa=1kg.{{m}^{-1}}{{s}^{-2}}$.
Therefore, the correct option is $D)\text{ }1kg.{{m}^{-1}}{{s}^{-2}}$.
Additional information:
Pressure also has many other units such as the atmosphere $\left( atm \right)$, Torricelli $\left( torr \right)$ and bar $\left( bar \right)$. The unit Pascal $\left( Pa \right)$ is often too small for everyday applications and hence, these other units are used more.
The relation for atmosphere $\left( atm \right)$ in terms of Pascal is
$1atm=1.01325\times {{10}^{5}}Pa$.
The relation of bar $\left( bar \right)$ in terms of Pascal $\left( Pa \right)$ is
$1bar={{10}^{5}}Pa$.
Note: Students must know of some important units in systems other than the SI, that is, MKS system of units. There are many important units, especially in the CGS system of units such as the $dyne$, that are used quite commonly. Sometimes questions provide the information values in some other system of units and require the answer in SI units, just to check the knowledge of students and their ability to convert units between different systems. If the student is not aware of the conversion, he or she may not be able to proceed with the question at all.
Formula used:
$\text{Pressure = }\dfrac{\text{force}}{\text{area}}$
Complete step by step answer:
Pressure upon a surface is defined as the ratio of the normal force acting on the surface to the surface area.
Hence,
$\text{Pressure = }\dfrac{\text{force}}{\text{area}}$
Using this definition, $1Pa$ is also defined as
$1Pa=\dfrac{1N}{1{{m}^{2}}}$ --(1)
Where Newton $\left( N \right)$ is the SI unit of force and the SI unit for surface area is meter-squared $\left( {{m}^{2}} \right)$.
Now, the Newton $\left( N \right)$ is the SI unit of force and can also be written as
$1N=1kg.m.{{s}^{-2}}$ --(2)
Putting (2) in (1), we will get,
$1Pa=\dfrac{1N}{{{m}^{2}}}=\dfrac{\dfrac{1kg.m}{{{s}^{2}}}}{{{m}^{2}}}=\dfrac{1kg.m}{{{m}^{2}}{{s}^{2}}}=\dfrac{1kg}{m.{{s}^{2}}}=1kg.{{m}^{-1}}{{s}^{-2}}$
Hence, we have derived that
$1Pa=1kg.{{m}^{-1}}{{s}^{-2}}$.
Therefore, the correct option is $D)\text{ }1kg.{{m}^{-1}}{{s}^{-2}}$.
Additional information:
Pressure also has many other units such as the atmosphere $\left( atm \right)$, Torricelli $\left( torr \right)$ and bar $\left( bar \right)$. The unit Pascal $\left( Pa \right)$ is often too small for everyday applications and hence, these other units are used more.
The relation for atmosphere $\left( atm \right)$ in terms of Pascal is
$1atm=1.01325\times {{10}^{5}}Pa$.
The relation of bar $\left( bar \right)$ in terms of Pascal $\left( Pa \right)$ is
$1bar={{10}^{5}}Pa$.
Note: Students must know of some important units in systems other than the SI, that is, MKS system of units. There are many important units, especially in the CGS system of units such as the $dyne$, that are used quite commonly. Sometimes questions provide the information values in some other system of units and require the answer in SI units, just to check the knowledge of students and their ability to convert units between different systems. If the student is not aware of the conversion, he or she may not be able to proceed with the question at all.
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