Which among the following is (are) the unit of pressure? (1) \(\mathrm{N} \mathrm{m}^{-2}\) (2) \(\mathrm{gwt} \mathrm{cm}^{-2}\) (3) kgwt \(\mathrm{m}^{-2}\) (4) All the above

Short Answer

Expert verified
(1) N m^(-2) (2) gwt cm^(-2) (3) kgwt m^(-2) (4) All the above Answer: (4) All the above

Step by step solution

01

Understand the definition of pressure

Pressure is defined as the force applied per unit area. In SI units, force is measured in Newton (N) and area in square meters (\(\mathrm{m}^2\)), so the unit of pressure will have N in the numerator and \(\mathrm{m}^2\) in the denominator.
02

Analyze option (1) N \(\mathrm{m}^{-2}\)

Option (1) has Newton (N) in the numerator and square meters (\(\mathrm{m}^2\)) in the denominator. This is consistent with the definition of pressure (force per unit area). So, option (1) is a unit of pressure.
03

Analyze option (2) gwt \(\mathrm{cm}^{-2}\)

In option (2), we have gram weight (gwt) in the numerator and square centimeters (\(\mathrm{cm}^2\)) in the denominator. This is also consistent with the definition of pressure as it represents the force (in terms of gram weight) per unit area (in square centimeters). So, option (2) is a unit of pressure.
04

Analyze option (3) kgwt \(\mathrm{m}^{-2}\)

In option (3), we have kilogram weight (kgwt) in the numerator and square meters (\(\mathrm{m}^2\)) in the denominator. This is consistent with the definition of pressure as it represents the force (in terms of kilogram weight) per unit area (in square meters). So, option (3) is a unit of pressure.
05

Conclude the final answer

From our analysis, we found that all options represent a unit of pressure consistent with the force per unit area definition. Therefore, the correct answer is (4) All the above.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Definition
Understanding what pressure means in physics is crucial for many applications, from checking your car's tires to measuring atmospheric conditions. Pressure is simply the amount of force exerted over a specific area. When you press your hand against a wall, you're applying force to the wall, and this force spread over the area of your hand that touches the wall represents pressure.

Imagine you're using your finger to push a pushpin into a notice board. Even though the force you exert is relatively small, the area over which it is applied is tiny (the tip of the pin), so the resulting pressure is high enough to push the pin into the board. On the other hand, if you were to press the flat side of your hand against the pin, even with the same force, you wouldn't be able to push it in because the area is much larger, resulting in lower pressure.
Force per Unit Area
Let's delve deeper into the fundamental principle of pressure—force per unit area. It's a straightforward concept: pressure increases if you apply the same force over a smaller area, and pressure decreases when you spread that force out over a larger area. The formula to calculate pressure is thus given by:

\[\begin{equation}Pressure = \frac{Force}{Area}\end{equation}\]

In this equation, 'Force' stands for the push or pull you're exerting on an object, measured in Newtons (N) for SI units. The 'Area' is the surface area over that force is distributed, measured in square meters (\[\begin{equation}\mathrm{m}^2\end{equation}\]).

Real-World Application

Consider a snowshoe: its wide base spreads the force exerted by your weight over a larger area, reducing the pressure on the snow and preventing you from sinking. That's the force per unit area concept in action.
SI Units of Pressure
The SI (International System of Units) unit of pressure is the pascal (Pa). This unit is named after Blaise Pascal, a French mathematician, physicist, and inventor, who did significant work in the field of hydrodynamics and hydrostatics and the study of fluids. One pascal is defined as one newton per square meter (\[\begin{equation}1\;Pa = 1\;N/m^2\end{equation}\]).

While the pascal is the standard unit of pressure in the SI system, various other units are used in different contexts, such as the bar (used in meteorology) or the atmosphere (atm), which is used to describe the pressure exerted by the Earth's atmosphere at sea level. Additionally, pressure can also be measured in terms of the weight of an object. For example, the gram weight per square centimeter (\[\begin{equation}\mathrm{gwt} \; \mathrm{cm}^{-2}\end{equation}\]) or the kilogram weight per square meter (\[\begin{equation}\mathrm{kgwt} \; \mathrm{m}^{-2}\end{equation}\]) are both valid units that express pressure.

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Most popular questions from this chapter

The distance between two cities \(A\) and \(B\) in a map is \(7.5 \mathrm{~cm}\). The scale taken for drawing this map is \(1 \mathrm{~cm}=1,50,000 \mathrm{~m}\). The actual distance between \(\mathrm{A}\) and \(\mathrm{B}\) is _______ \(\mathrm{km}\). (1) 1125000 (2) \(\quad 20000\) (3) \(\quad 200\) (4) 1125

Density of the material of a paper is given as \(0.5 \mathrm{~g} \mathrm{~cm}^{-3}\). The mass of the paper is \(1 \mathrm{~g}\) and its length and breadth are \(10 \mathrm{~cm}\) and \(5 \mathrm{~cm}\) respectively. Arrange the following steps in a sequence to find the thickness of the paper. (a) The thickness of the paper is \(=\frac{\text { volume }(\mathrm{V}) \text { of the paper }}{\text { length } \times \text { breadth }}\) (b) The density (d) of the material of the paper is \(=\frac{\text { mass of the paper(m) }}{\text { volume of the paper(V) }}\) (c) Then the volume \((\mathrm{V})\) of the paper \(=\frac{\text { mass of the paper }(\mathrm{m})}{\text { density }(\mathrm{d}) \text { of the paper }}\) (d) The volume \((\mathrm{V})\) of the paper is \(=\) length \(\times\) breadth \(\times\) thickness of the paper. (1) abcd (2) badc (3) \(\mathrm{abdc}\) (4) \(\mathrm{bcda}\)

What are the common multiple and submultiple units of mass? Express them in terms of \(\mathrm{kg}\).

Name the method used to measure the distance between stars.

Match the entries given in column A with the appropriate ones in column \(B\). $$ \begin{array}{lllll} \hline & {\text { Column A }} & {\text { Column B }} \\ \hline \text { A. } & 1 \mathrm{~g} \mathrm{~cm}^{-3} & (\quad) & \text { a. } 1000 \mathrm{~kg} \mathrm{~m}^{-3} \\ \text { B. } & \text { Convection current } & ( \quad ) & \text { b. } \mathrm{cm}^{3} \\ \text { C. } & \text { Volume } & ( \quad ) & \text { c. } \text { Measurement of large distances } \\ \text { D. } & \text { Triangulation method } & (\quad ) & \text { d. } \text { Mass/volume } \\ \text { E. } & \text { Mass } & (\quad ) & \text { e. } \text { Change in density } \\ \text { F. } & \text { Density } & \text { ( ) } & \text { f. } \text { kg } \\\ \hline \end{array} $$

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