Sometimes it is convenient to construct a system of units so that all quantities can be expressed in

Q: Why do we need to find $L, M, T$ in terms of $X$?

The problem gives the dimensions of various quantities in terms of a single quantity $X$. To find the relations between the exponents, we need a common ground, which is the standard $M, L, T$ system. By expressing $M, L, T$ in terms of $X$, we can easily substitute them into any derived quantity's formula.

Q: Can we solve this by expressing $X$ in terms of $M, L, T$?

Yes, but it would be much more complicated because $X$ is a single quantity representing different combinations of $M, L, T$ for different physical quantities. It's mathematically simpler to express the standard base quantities in terms of $X$.

Q: How did we get $T = X^{\alpha - \beta}$?

We know position $[L] = X^\alpha$ and speed $[LT^{-1}] = X^\beta$. Dividing position by speed gives $L / (LT^{-1}) = T$. Substituting their $X$ equivalents gives $X^\alpha / X^\beta = X^{\alpha - \beta}$.

Q: Why did we use momentum to find mass?

Momentum is $[MLT^{-1}]$. We already know $[LT^{-1}] = X^\beta$. So, dividing momentum by speed gives $M$. It's the most direct way to isolate mass.

Q: Could we have used force to find mass instead?

Absolutely! Force is $[MLT^{-2}]$. We know acceleration is $[LT^{-2}] = X^p$. Dividing force by acceleration gives $M = X^r / X^p = X^{r-p}$. This would lead to the same final relations, just through a slightly different algebraic path.

Question

Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: $[	ext{position}] = [X^\alpha]$; $[	ext{speed}] = [X^\beta]$; $[	ext{acceleration}] =[X^p]$; $[	ext{linear momentum}] = [X^q]$; $[	ext{force}] = [X^r]$. Then -

Options:

(A) $\alpha + p = 2 \beta$

(B) $p + q - r = \beta$

(C) $p - q + r = \alpha$

(D) $p + q + r = \beta$

Sigma Percentile · Accepted Answer

### The Power of a Single Dimension

Imagine a universe where instead of measuring mass, length, and time separately, you only had one fundamental measuring stick. Let's call it $X$. Everything—how fast you move, how hard you push, how much momentum you carry—is just some power of $X$.

This might sound like science fiction, but it's exactly the premise of this fascinating JEE Advanced problem. It tests our deep understanding of dimensional analysis by forcing us to abandon our comfortable $M$, $L$, and $T$ and translate everything into this alien system of $X$.

### Decoding the Dictionary

The problem gives us a dictionary to translate between our world and the $X$-world:
- $[	ext{position}] = L = X^\alpha$
- $[	ext{speed}] = LT^{-1} = X^\beta$
- $[	ext{acceleration}] = LT^{-2} = X^p$
- $[	ext{linear momentum}] = MLT^{-1} = X^q$
- $[	ext{force}] = MLT^{-2} = X^r$

Our goal is to find the hidden relationships between these exponents $\alpha, \beta, p, q$, and $r$. To do this, we need to play detective. We need to isolate our familiar base quantities—Length ($L$), Time ($T$), and Mass ($M$)—and express them purely in terms of $X$.

### Isolating Time

We already have Length directly: 
$$L = X^\alpha$$

Now, look at speed. Speed is length divided by time ($LT^{-1}$). If we divide Length by Speed, the $L$ cancels out, leaving us with Time! Let's do the math:
$$\frac{L}{LT^{-1}} = \frac{X^\alpha}{X^\beta}$$

$$T = X^{\alpha - \beta}$$

Boom! We've cracked the code for Time.

### The First Relation: Acceleration

Now that we have $L$ and $T$ in terms of $X$, we can tackle acceleration. We know the standard dimensional formula for acceleration is $LT^{-2}$. The problem tells us this equals $X^p$. Let's substitute our $X$-expressions for $L$ and $T$:

$$LT^{-2} = X^p$$

$$(X^\alpha) \cdot (X^{\alpha - \beta})^{-2} = X^p$$

Now, it's just a matter of careful algebra. Let's expand the power:

$$X^\alpha \cdot X^{-2\alpha + 2\beta} = X^p$$

When multiplying terms with the same base, we add the exponents:

$$X^{\alpha - 2\alpha + 2\beta} = X^p$$

$$X^{2\beta - \alpha} = X^p$$

Since the bases are the same, the exponents must be equal:

$$2\beta - \alpha = p$$

Rearranging this gives us our first beautiful relation:

$$\alpha + p = 2\beta$$

This perfectly matches **Option (A)**!

### Isolating Mass

To check the other options, we need to bring Mass ($M$) into the picture. Let's look at linear momentum, which is mass times velocity ($MLT^{-1}$). We already know the entire $LT^{-1}$ chunk is just $X^\beta$.

$$MLT^{-1} = X^q$$

$$M \cdot (X^\beta) = X^q$$

Dividing both sides by $X^\beta$, we isolate Mass:

$$M = X^{q - \beta}$$

### The Second Relation: Force

Finally, the grand finale. Let's use the force equation. Force is mass times acceleration ($MLT^{-2}$). We have $M$ in terms of $X$, and the problem directly gives us acceleration ($LT^{-2}$) as $X^p$.

$$MLT^{-2} = X^r$$

$$(M) \cdot (LT^{-2}) = X^r$$

Substitute what we know:

$$(X^{q - \beta}) \cdot (X^p) = X^r$$

Again, add the exponents:

$$X^{q - \beta + p} = X^r$$

Equating the exponents yields:

$$q - \beta + p = r$$

Let's rearrange this to match the options. If we move $\beta$ to the right and $r$ to the left, we get:

$$p + q - r = \beta$$

And there it is! This matches **Option (B)**.

### The Takeaway

By systematically breaking down derived quantities into their fundamental components and substituting, we unraveled the relationships in this hypothetical unit system. It's a brilliant exercise in algebraic manipulation and a core application of the principle of dimensional homogeneity.

The correct options are indeed **(A)** and **(B)**.

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Visualized Solution

$Given Dimensional Relations$

$Isolating Time (T)$

$Substituting in Acceleration$

$Solving for First Relation$

$Isolating Mass (M)$

$Substituting in Force$

$Solving for Second Relation$

$Final Conclusion$

The Sigma Insight: Dimensional Analysis

The Power of a Single Dimension

Decoding the Dictionary

Isolating Time

The First Relation

Isolating Mass

The Second Relation

The Takeaway

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Dimensional Relations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating Time (T)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting in Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for First Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating Mass (M)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting in Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Second Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion