Animated Solution for Physics - Physics and Measurement: Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of L, which of the following statement (s) is/are correct ?

The beauty of dimensional analysis lies in its flexibility. We are so used to the standard SI system where Mass ($M$), Length ($L$), and Time ($T$) are the absolute rulers of the physical world. But what if we change the rules of the game? What if we create a completely new universe where mass and angular momentum are just pure numbers, completely dimensionless? Let's dive into this fascinating problem and see how the dimensions of other physical quantities adapt to this new reality.

The problem introduces a hypothetical system of units with two critical conditions: 1. Mass is dimensionless: This means the dimensional formula for mass is simply $[M]=M^0{L}^0{T}^0=1$. 2. Angular momentum is dimensionless: Similarly, the dimensional formula for angular momentum is $[J]=M^0{L}^0{T}^0=1$. 3. Length retains its dimension: The dimension of length remains $[L]=L$.

Our first goal is to understand how Time ($T$) behaves in this new system. To do this, we need to express angular momentum in terms of the fundamental quantities $M$, $L$, and $T$.

Recall the formula for angular momentum ($J$). A simple way to remember it is the product of mass, velocity, and radius:

Let's write down the standard dimensions for each of these components: - Mass ($m$) $\to [M]$ - Velocity ($v$) $\to [L{T}^{-1}]$ - Radius ($r$) $\to [L]$

Multiplying these together, we get the standard dimensional formula for angular momentum:

Now, we apply the rules of our new universe. We know that angular momentum is dimensionless, so we equate its dimensional formula to 1:

We also know that mass is dimensionless, so we can substitute $M=1$ into the equation:

Rearranging this equation to solve for $T$, we find our master relation:

This is a profound result! In this new system, time is dimensionally equivalent to the square of length. Now, we can use this master relation, along with $M=1$, to evaluate the dimensions of the quantities given in the options.

Let's systematically check each option by substituting $M=1$ and $T=L^2$ into their standard dimensional formulas.

Checking Option A: Force The standard dimension of force is: Substituting our new rules: This matches Option A perfectly. Option A is correct.

Checking Option B: Energy The standard dimension of energy (work done) is: Substituting our new rules: This matches Option B perfectly. Option B is correct.

Checking Option C: Power Power is the rate of doing work, so its standard dimension is: Substituting our new rules: Option C claims the dimension is $L^{-5}$, which contradicts our result. Option C is incorrect.

Checking Option D: Linear Momentum The standard dimension of linear momentum is: Substituting our new rules: This matches Option D perfectly. Option D is correct.

By carefully applying the given conditions to the standard dimensional formulas, we successfully translated the dimensions of force, energy, power, and linear momentum into the new system. The correct statements are indeed (A), (B), and (D). This problem is a fantastic exercise in understanding the fundamental relationships between physical quantities and how they scale when the base units are redefined.