If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the

Q: Why do we assume $[D] = [F]^a [L]^b [T]^c$ in the first place?

Because we are told that $F$, $L$, and $T$ are the new fundamental quantities. Any derived quantity (like density) must be expressible as some combination of these fundamental quantities raised to certain powers. This is the core definition of a dimensional formula.

Q: What exactly is the Principle of Dimensional Homogeneity?

It states that you can only equate, add, or subtract physical quantities that have the exact same dimensions. In an equation like $A = B$, the fundamental dimensions (like mass, length, time) must have the exact same exponents on both sides.

Q: Can we choose literally any physical quantities to be our fundamental base?

Almost! You can choose any set of quantities as long as they are independent of each other. For example, you cannot choose Velocity, Acceleration, and Time as your three fundamental quantities because Velocity is just Acceleration multiplied by Time. They must form an independent basis.

Q: I keep messing up the powers when grouping terms. Any tips?

Always write out the intermediate step explicitly. When you have $[MLT^{-2}]^a$, distribute the $a$ to every term inside the bracket first: $[M^a L^a T^{-2a}]$. Then, when multiplying by $[L]^b$, remember that $x^m \cdot x^n = x^{m+n}$, so it becomes $L^{a+b}$.

Q: What if the equations for $a$, $b$, and $c$ give contradictory results?

If you get contradictory results (e.g., $a=1$ from one equation and $a=2$ from another), it means the quantity you are trying to express cannot be formed using the chosen fundamental quantities. However, in JEE problems, the sets provided are always valid bases.

Question

If force ($F$), length ($L$) and time ($T$) are taken as the fundamental quantities. Then what will be the dimension of density?

Options:

(A) $[FL^{-4}T^2]$

(B) $[FL^{-3}T^2]$

(C) $[FL^{-5}T^2]$

(D) $[FL^{-3}T^3]$

Sigma Percentile · Accepted Answer

### The Power of Dimensional Analysis

Imagine you are an architect of the universe, and you decide that instead of using Mass, Length, and Time as your fundamental building blocks, you want to use **Force**, **Length**, and **Time**. How would you describe everything else? This is the core of dimensional analysis, a favorite concept in JEE physics.

In this problem, we are asked to find the dimensional formula for **density** in this new system. Density, as we know, is mass per unit volume. But how do we express it without explicitly using mass? Let's dive into the thought process.

### Setting Up the Master Equation

We start by assuming that density ($[D]$) depends on some unknown powers of force ($[F]$), length ($[L]$), and time ($[T]$). We can write this mathematically as:

$$[D] = [F]^a [L]^b [T]^c$$

Our goal is to find the exact values of these exponents: $a$, $b$, and $c$. To do this, we need a common ground. We will translate everything back into the standard SI fundamental quantities: Mass ($M$), Length ($L$), and Time ($T$).

### The Translation Phase

Let's write down the standard dimensional formulas for each quantity involved:

- Density is mass per unit volume: $[D] = [ML^{-3}T^0]$
- Force is mass times acceleration: $[F] = [MLT^{-2}]$
- Length is simply: $[L] = [L]$
- Time is simply: $[T] = [T]$

Now, we substitute these standard dimensions back into our assumed master equation. This is where we must be careful not to make any silly algebraic mistakes.

$$[ML^{-3}T^0] = [MLT^{-2}]^a [L]^b [T]^c$$

Next, we group the terms with the same base on the right-hand side. By applying the laws of exponents, we collect all the $M$'s, $L$'s, and $T$'s together:

$$[ML^{-3}T^0] = [M^a L^{a+b} T^{-2a+c}]$$

### The Principle of Dimensional Homogeneity

Here comes the magic trick. According to the **Principle of Dimensional Homogeneity**, for any physical equation to be valid, the dimensions on both sides of the equals sign must be identical. This means the power of $M$ on the left must equal the power of $M$ on the right, and similarly for $L$ and $T$.

Let's compare the powers one by one.

**Comparing powers of $M$:**
On the left, the power is $1$. On the right, it is $a$. Therefore, we immediately get:

$$a = 1$$

**Comparing powers of $L$:**
On the left, the power is $-3$. On the right, it is $a + b$. Since we already know $a = 1$, we can substitute it in:

$$-3 = 1 + b \implies b = -4$$

**Comparing powers of $T$:**
On the left, the power is $0$. On the right, it is $-2a + c$. Substituting $a = 1$ again:

$$0 = -2(1) + c \implies c = 2$$

### The Final Reveal

We have successfully cracked the code! The exponents are $a = 1$, $b = -4$, and $c = 2$. Now, we simply substitute these values back into our original assumed equation:

$$[D] = [F]^1 [L]^{-4} [T]^2$$

So, the dimensional formula for density in this new system is **$[FL^{-4}T^2]$**. This elegant method can be used to derive relationships between any set of physical quantities, proving that the language of physics is universally adaptable.

Animated Solution for Physics - Physics and Measurement: If force ( $F$ ), length ( $L$ ) and time ( $T$ ) are taken as the fundamental quantities. Then what will be the dimension of density?

Select Answer:

Visualized Solution

$[D] = [F]^{a} [L]^{b} [T]^{c}$

\text{Standard Dimensions}

\text{Substitution}

\text{Grouping Terms}

\text{Comparing } M

\text{Comparing } L

\text{Comparing } T

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Dimensional Analysis

The Power of Dimensional Analysis

Setting Up the Master Equation

The Translation Phase

The Principle of Dimensional Homogeneity

The Final Reveal

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