Question

Equation of physical quantity \(\mathrm{v}=\mathrm{at}+\mathrm{bt}^{2}\) where \(\mathrm{v}=\) velocity \(\mathrm{t}=\) time so write the dimensional formula of a in this equation (a) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{0}\)

Short answer

The dimensional formula of 'a' in the equation is (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\).

Step-by-step solution

Write the given equation

Rewrite the given equation as it is: \(v = at + bt^2\)

Determine the dimensions of the given quantities

We already know the dimensions of the following physical quantities: 1. Velocity (v): \([v] = [LT^{-1}]\) or \(\mathrm{M}^{0}\mathrm{L}^{1}\mathrm{T}^{-1}\) 2. Time (t): \([t] = [T]\) or \(\mathrm{M}^{0}\mathrm{L}^{0}\mathrm{T}^{1}\) For the second part of the equation involving 'b', we can see it is multiplied by the square of time (t^2). Thus, the dimensions of 'bt^2' should also match the dimensions of velocity.

Determine the dimensions of 'bt^2'

Since the dimensions of 'bt^2' should match the dimensions of velocity, we can write: \([b]\times [t]^2 = [LT^{-1}]\) Plugging in the dimensions of time, we get: \([b]\times (T^2) = LT^{-1}\) Solving for the dimensions of 'b', we find: \([b] = L T^{-3}\) or \(\mathrm{M}^{0}\mathrm{L}^{1}\mathrm{T}^{-3}\)

Determine the dimensions of 'at'

Since the lefthand side of the equation has dimensions of velocity, the dimensions of 'at' should also match those of velocity. So, we can write: \([a]\times [t] = [LT^{-1}]\) Plugging in the dimensions of time, we get: \([a] \times T = LT^{-1}\) Solving for the dimensions of 'a', we find: \([a] = LT^{-2}\) or \(\mathrm{M}^{0}\mathrm{L}^{1}\mathrm{T}^{-2}\) Answer: The dimensional formula of 'a' in the equation is (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\).

Key concepts

Velocity

Velocity is a fundamental concept in physics that describes how fast something is moving in a specific direction. It is a vector quantity, which means it has both a magnitude (how fast it's moving) and a direction (which way it's going).

Unlike speed, which only tells you how fast something is moving regardless of the direction, velocity gives a more complete picture by including the direction. For example, if you're driving at 60 km/h heading north, your speed is 60 km/h, but your velocity is 60 km/h to the north.

In terms of dimensional analysis, velocity has the dimensions of \([LT^{-1}]\).

• Here, \([L]\) stands for length (distance), which is typically measured in meters (m).
• \([T^{-1}]\) denotes the inverse of time, measured in seconds (s). This tells us how length changes over time.

Whenever solving problems involving velocity, it's crucial to remember that both speed and direction are important. This makes it easier to understand the motion of objects.

Time

Time is one of the most fundamental aspects of physics and everyday life. It's a scalar quantity, meaning it only has magnitude and no direction. Often measured in seconds, time helps us understand and describe events sequentially.

In dimensional analysis, time is denoted as \([T]\). This is all you need to know about time for basic physics problems, making it straightforward.

• In the context of motion, time allows us to measure the duration of events.
• It helps calculate other essential quantities like velocity and acceleration.
• Since time is a key element in the equations of motion, understanding its dimensional role can simplify complex calculations.

Time serves as a backbone for other calculations, making tasks like predicting future positions of moving objects much more manageable.

Dimensional Formula

The dimensional formula is a powerful tool in physics that helps us describe physical quantities in terms of their fundamental dimensions. By using specific symbols, it ensures consistency when solving problems involving different physical units.

The basic dimensions are:

• Mass \([M]\),
• Length \([L]\),
• and Time \([T]\).

The purpose of a dimensional formula is to express a physical quantity in terms of these basic dimensions. For instance, velocity has the formula \([LT^{-1}]\), indicating it involves both length and time.

In the exercise we looked at, dimensional analysis ensures that the dimensional formula of constants (like \([a]\) in the equation) balances the units on both sides of the equation. This helps verify that equations are dimensionally consistent, which is crucial for correctness. By breaking down complex equations into their dimensional components, dimensional analysis offers a clear pathway to understanding and solving them.