Problem 4 - A Complete Resource Book in Chemistry for JEE Main

Question

The dimensions of pressure are same as that of (a) energy (b) energy per unit volume (c) force per unit area (d) force per unit volume

Step-by-Step Solution

Answer

Pressure has the same dimensions as energy per unit volume (b) and force per unit area (c).

Step 1: Understand the Dimension of Pressure

Pressure is defined as force per unit area. The formula for pressure is given by \( P = \frac{F}{A} \), where \( P \) is pressure, \( F \) is force, and \( A \) is area. The dimensional formula for force \( F = ma = MLT^{-2} \), and for area \( A = L^2 \), leading to the dimensional formula for pressure \( P = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \).

Step 2: Analyze Dimensions of Given Options

Let's consider each option individually: (a) Energy: The dimensional formula for energy (work done) is \( [ML^2T^{-2}] \).(b) Energy per unit volume: Energy per unit volume is given by \( \frac{energy}{volume} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2} \).(c) Force per unit area: We have already seen this is the definition of pressure, \( [ML^{-1}T^{-2}] \).(d) Force per unit volume: The force per unit volume is given by \( \frac{MLT^{-2}}{L^3} = ML^{-2}T^{-2} \).

Step 3: Match Dimensions with Pressure

From Step 2, the dimensions of pressure \( [ML^{-1}T^{-2}] \) match with both option (b) Energy per unit volume \( [ML^{-1}T^{-2}] \) and option (c) Force per unit area \( [ML^{-1}T^{-2}] \). This means that pressure has the same dimensions as both energy per unit volume and force per unit area.

Key Concepts

Pressure and Its Dimensions

Pressure is a concept that helps us understand the force applied over a certain area. To define it in simple terms, pressure is the amount of force that is exerted per unit area. The formula for calculating pressure is given by:
\[ P = \frac{F}{A} \]where \( P \) represents pressure, \( F \) stands for force, and \( A \) refers to area. The dimensions of pressure can be extracted from this formula and are expressed as \( [ML^{-1}T^{-2}] \). Here's a more in-depth look:

• Force \( F \) is calculated as mass times acceleration, \( ma \), with the dimensional formula \( [MLT^{-2}] \).
• Area \( A \) is the space an object covers, with dimensions \( [L^2] \).

When you divide the force by area, as in the pressure formula, the area \( L^2 \) affects the dimensions of force to result in pressure with \([ML^{-1}T^{-2}]\). This understanding helps in visualizing how intensely a force acts on a specific surface.

Energy per Unit Volume

Energy per unit volume is a concept closely tied to the idea of density. It involves the energy contained in a certain volume of space. This becomes particularly relevant when considering phenomena like fluid pressure or energy distribution in space.
The dimensional formula for energy is \( [ML^2T^{-2}] \). To find the energy per unit volume, you divide the energy by volume which has the dimensional formula \( [L^3] \). This division leads to:\[\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]\]These dimensions, \( [ML^{-1}T^{-2}] \), are identical to the dimensions of pressure. Understanding energy per unit volume is vital in physics because it allows us to calculate aspects like the intensity of energy fields and pressures arising from accumulated energies.

Force per Unit Area

We often encounter force being distributed in various ways, and understanding how force acts over an area is key to many physics applications. Force per unit area is essentially what defines pressure. The common formula given by:\[ P = \frac{F}{A} \]is a direct representation of the concept of pressure, the concentration of force applied upon a surface.

• Force has dimensions \( [MLT^{-2}] \), a product of mass and acceleration.
• Area dimensions are \( [L^2] \).

When you divide force by area to ascertain pressure, the dimensions become \([ML^{-1}T^{-2}]\). This formula is the foundation for numerous applications, such as calculating the stress on structural beams, the behavior of gases, and the pressure exerted by fluids.

Approaching Physics Problems

Physics problem solving requires a structured approach, breaking down complex concepts into understandable and manageable parts. Dimensional analysis plays a significant role in this by ensuring consistency across measurements, providing a check against errors in derivations or calculations.
When faced with problems in physics, follow these steps:

• Identify the pertinent equations and concepts, like understanding force, area, energy, and volume.
• Assign and ensure the correct dimensions to various quantities using the fundamental dimensions of mass \( M \), length \( L \), and time \( T \).
• Perform dimensional analysis to verify the correctness of physical equations, ensuring the dimensions align with expected units.
• Approach problems methodically, forming a clearer understanding of interaction between different physical forces and principles involved.

Dimensional analysis not only aids in confirming derived equations but also enriches one's understanding of how various physical quantities relate to each other, making problem solving in physics a logical and streamlined process.