The dimensions of energy are:
Understanding the Dimensions of Energy
The question asks for the dimensions of energy. Dimensions in physics represent the fundamental physical quantities, such as mass (M), length (L), and time (T), that make up a derived quantity. The dimension of a physical quantity indicates how it depends on the fundamental quantities.
Relating Energy to Work and Fundamental Quantities
Energy is defined as the capacity to do work. Therefore, the dimensions of energy are the same as the dimensions of work.
Work is done when a force causes a displacement. The formula for work (W) is:
\( W = F \times d \)
Where:
- \( F \) is the force applied
- \( d \) is the displacement
Finding the Dimensions of Force
To find the dimensions of work, we first need the dimensions of force. Force is defined by Newton's second law of motion:
\( F = m \times a \)
Where:
- \( m \) is mass
- \( a \) is acceleration
The fundamental dimensions are:
- Mass (\( m \)): [\( M^1 \)]
- Length (\( d \), related to acceleration's units): [\( L^1 \)]
- Time (related to acceleration's units): [\( T^1 \)]
Now let's find the dimensions of acceleration (\( a \)). Acceleration is the rate of change of velocity with respect to time. Velocity is the rate of change of displacement with respect to time.
- Dimensions of velocity: Displacement / Time = [\( L^1 \)] / [\( T^1 \)] = [\( L^1 T^{-1} \)]
- Dimensions of acceleration: Velocity / Time = [\( L^1 T^{-1} \)] / [\( T^1 \)] = [\( L^1 T^{-2} \)]
Now we can find the dimensions of force:
Dimensions of Force = Dimensions of Mass \(\times\) Dimensions of Acceleration
Dimensions of Force = [\( M^1 \)] \(\times\) [\( L^1 T^{-2} \)]
Dimensions of Force = [\( M^1 L^1 T^{-2} \)]
Calculating the Dimensions of Energy (Work)
Now we can find the dimensions of work (and thus energy) using the formula \( W = F \times d \):
Dimensions of Work = Dimensions of Force \(\times\) Dimensions of Displacement
Dimensions of Work = [\( M^1 L^1 T^{-2} \)] \(\times\) [\( L^1 \)]
Dimensions of Work = [\( M^1 L^{1+1} T^{-2} \)]
Dimensions of Work = [\( M^1 L^2 T^{-2} \)]
Since energy has the same dimensions as work, the dimensions of energy are [\( M^1 L^2 T^{-2} \)].
Comparing with the Options
Let's compare the derived dimensions with the given options:
- [\( M^1 L^{-1} T^{-1} \)]
- [\( M^2 L^2 T^2 \)]
- [\( M^1 L^{-2} T^{-1} \)]
- [\( M^1 L^2 T^{-2} \)]
Our derived dimension [\( M^1 L^2 T^{-2} \)] matches option 4.
Revision Table: Key Physical Quantities and Their Dimensions
| Physical Quantity | Formula | Dimensions |
|---|---|---|
| Mass | - | [\( M^1 \)] |
| Length | - | [\( L^1 \)] |
| Time | - | [\( T^1 \)] |
| Velocity | Displacement / Time | [\( L^1 T^{-1} \)] |
| Acceleration | Velocity / Time | [\( L^1 T^{-2} \)] |
| Force | Mass \(\times\) Acceleration | [\( M^1 L^1 T^{-2} \)] |
| Work / Energy | Force \(\times\) Displacement | [\( M^1 L^2 T^{-2} \)] |
| Power | Work / Time | [\( M^1 L^2 T^{-3} \)] |
Additional Information: Importance of Dimensional Analysis
Dimensional analysis is a powerful tool in physics. It is used for:
- Checking the consistency of equations: Both sides of an equation must have the same dimensions. If they don't, the equation is incorrect.
- Deriving relationships between physical quantities: Based on the factors a quantity depends on, dimensional analysis can help propose possible formulas.
- Unit conversion: Dimensions help ensure that units are converted correctly between different systems (e.g., SI to CGS).
Understanding the dimensions of energy, force, and other quantities is fundamental to solving many physics problems and verifying results.
Important Questions from Dimensional formulae and dimensional equations
- Considering the Lorentz force $\vec{F} = q(\vec{v} \times \vec{B})$, where $F$ is force, $q$ is electric charge, and $v$ is velocity, what is the dimensional formula for magnetic flux density $B$?
Given that the energy stored in an inductor is expressed as $U_L = \frac{1}{2}LI^2$ and the power dissipated in a resistor is $P = I^2R$, where $L$ is inductance, $R$ is resistance, and $I$ is current, determine the dimension of the ratio $\frac{L}{R}$.
If force $[F]$, acceleration $[A]$ and time $[T]$ are chosen as the fundamental physical quantities. Find the dimensions of pressure.
- The characteristic impedance of free space, $Z_0$, is given by the expression $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$. If $\mu_0$ represents the magnetic permeability and $\epsilon_0$ represents the electric permittivity, what are the dimensions of $Z_0$?
- Determine the dimensional formula for the quantity represented by the product of pressure and volume.
