Animated Solution for Physics - Physics and Measurement: A quantity $x$ is given by $x=\frac{IF{v}^2}{W{L}^4}$ where, $I$ is moment of inertia, $F$ is force, $v$ is velocity, $W$ is work and $L$ is length. The dimensional formula for $x$ is same as that of

Welcome to another exciting and deeply illuminating journey into the world of dimensional analysis! Today, we are faced with a classic physics puzzle that will test our foundational understanding of physical quantities and their intrinsic nature. We have been handed a mysterious quantity, $x$, defined by a rather intimidating algebraic expression, and our mission is to uncover its true physical identity. This is not just a mathematical exercise; it is a profound exploration of how different physical concepts interlock and relate to one another.

Imagine you are a detective in the realm of physics, and the equation $x=\frac{IF{v}^2}{W{L}^4}$ is your primary clue. At first glance, it looks like a random jumble of variables thrown together to confuse you. We have moment of inertia, force, velocity, work, and length all mixed up in a single fraction. But in physics, there is a beautiful underlying order. Every valid equation, no matter how complex the combination of variables might appear, must be dimensionally consistent. This means that they must ultimately boil down to a fundamental dimensional structure composed of mass, length, and time. Our goal is to strip away the complexity and find that core structure.

To solve this mystery, we cannot tackle the entire equation at once. We need to interrogate each variable individually, understand its physical meaning, and extract its dimensional formula. Let's line up our suspects and break them down one by one:

- Moment of Inertia ($I$): Think of a rotating mass. Moment of inertia is the rotational analog of mass. It tells us how difficult it is to change the rotational state of an object. The fundamental formula for a point mass is $I=m{r}^2$. Mass has the dimension $[M]$, and radius (which is a distance) has the dimension $[L]$. Squaring the radius gives us $[L^2]$. Therefore, the dimensional formula for moment of inertia is $[M{L}^2]$.

- Force ($F$): This is the push or pull that causes an object to accelerate. We rely on good old Newton's second law of motion, $F=ma$. Mass gives us $[M]$. Acceleration is the rate of change of velocity, which is $[L{T}^{-2}]$. Multiplying them together, mass times acceleration gives us the familiar dimensional formula for force: $[ML{T}^{-2}]$.

- Velocity ($v$): This is simply the rate of change of displacement with respect to time. Displacement is a length $[L]$, and time is $[T]$. Therefore, velocity translates to $[L{T}^{-1}]$.

- Work ($W$): Work is done when a force is applied over a distance. The formula is $W=Fd$. We already know the dimension of force is $[ML{T}^{-2}]$, and distance is simply $[L]$. Multiplying these gives us the dimensional formula for work: $[M{L}^2{T}^{-2}]$. Notice that this is the exact same dimensional formula as energy, which makes perfect sense since work is the transfer of energy!

- Length ($L$): This is the most fundamental of them all. It is a base quantity, so its dimensional formula is simply $[L]$.

Now that we have successfully identified the dimensional formulas for all the individual components, it is time for the algebraic crunch. Let's substitute these fundamental dimensions back into our original equation for $x$. This is where we need to be incredibly careful and avoid any silly mistakes with exponents. A single misplaced negative sign can derail the entire investigation.

First, let's handle the square on the velocity term. According to the laws of exponents, when we raise a power to a power, we multiply the exponents.

Now, let's group all the terms in the numerator. We will multiply the dimensions of moment of inertia, force, and squared velocity together. To multiply these, we add the exponents of the corresponding base quantities. For Mass ($M$): $1+1=2$, so we have $M^2$. For Length ($L$): $2+1+2=5$, so we have $L^5$. For Time ($T$): $-2+(-2)=-4$, so we have $T^{-4}$. Putting it all together, the simplified numerator is:

Next, let's group the terms in the denominator. We will multiply the dimensions of work and length to the power of four. Again, we add the exponents of the corresponding base quantities. For Mass ($M$): $1$, so we have $M^1$. For Length ($L$): $2+4=6$, so we have $L^6$. For Time ($T$): $-2$, so we have $T^{-2}$. The simplified denominator is:

Bringing it all together, we divide the simplified numerator by the simplified denominator. Using the laws of exponents for division, we subtract the powers of the denominator from the powers of the numerator.

We have successfully distilled the complex, intimidating expression into its core dimensional signature! The mystery quantity $x$ has the dimensions of mass per unit length per squared time.

Now that we know the dimensional identity of $x$ is $[M{L}^{-1}{T}^{-2}]$, we need to check our options to see which physical quantity shares this exact signature. We have four suspects lined up. Let's interrogate them.

1. Planck's constant ($h$): This fundamental constant of quantum mechanics relates the energy of a photon to its frequency via the equation $E = h u$. Therefore, $h = E/ u$. Energy has dimensions $[M{L}^2{T}^{-2}]$ and frequency has dimensions $[T^{-1}]$. Dividing them gives $[h]=[M{L}^2{T}^{-1}]$. This is not a match for our quantity $x$.

2. Force constant ($K$): Also known as the spring constant, this relates the restoring force of a spring to its displacement via Hooke's Law, $F=Kx$ (where $x$ here is displacement, not our mystery quantity). Therefore, $K=F/x$. Force has dimensions $[ML{T}^{-2}]$ and displacement has dimensions $[L]$. Dividing them gives $[K]=[M{T}^{-2}]$. Close, but it is missing the $L^{-1}$ term. Not a match.

3. Coefficient of viscosity ($\eta$): This quantity describes a fluid's resistance to flow. We can find its dimensions using Stokes' law for the drag force on a sphere moving through a fluid: $F=6\pi \eta rv$. Rearranging for $\eta$, we get $\eta =F/(6\pi rv)$. Ignoring the dimensionless constant $6\pi$, we divide the dimensions of force $[ML{T}^{-2}]$ by the dimensions of radius $[L]$ and velocity $[L{T}^{-1}]$. This gives $[\eta ]=[M{L}^{-1}{T}^{-1}]$. Still not a match. We are looking for $T^{-2}$, not $T^{-1}$.

4. Energy density ($E_d$): This is defined as the amount of energy stored in a given system or region of space per unit volume. Let's calculate its dimensional formula from its definition: We know the dimensional formula for energy is $[M{L}^2{T}^{-2}]$, and the dimensional formula for volume is $[L^3]$. Subtracting the exponent of length in the denominator from the numerator ($2-3=-1$), we get:

Bingo! The dimensional formula for energy density perfectly matches the dimensional formula we calculated for our mystery quantity $x$.

It is truly fascinating how a bizarre combination of rotational inertia, linear force, velocity, mechanical work, and spatial length can perfectly mimic the dimensional footprint of energy packed into a three-dimensional volume. This is the profound beauty and utility of dimensional analysis. It allows us to verify equations, derive relationships, and, as we have done here, reveal the hidden connections between seemingly unrelated physical concepts.

By mastering dimensional analysis, you are not just learning a trick to solve exam problems; you are acquiring a powerful lens through which to view the entire physical universe. Keep practicing, keep questioning, and soon you will be reading these dimensional signatures like a seasoned physicist!