If speed (v), acceleration (A) and force (F) are considered as fundamental units, the dimension of Y

Q: Why do we assume $[Y] = [v]^\alpha [A]^\beta [F]^\gamma$?

This is the standard method of dimensional analysis. If a physical quantity can be expressed in terms of other fundamental quantities, its dimensions must be a product of some powers of those fundamental quantities.

Q: How do I quickly recall the dimension of Young's modulus?

Remember its definition: $Y = \text{Stress} / \text{Strain}$. Since strain ($\Delta L / L$) is dimensionless, the dimension of $Y$ is exactly the same as stress, which is Force / Area, giving $[ML^{-1}T^{-2}]$.

Q: What if I get a fractional power when solving the equations?

Fractional powers are completely fine in dimensional analysis! However, in this specific problem, the equations resolve cleanly into integers. If you get fractions here, double-check your algebraic steps and signs.

Q: Can we use this method to find exact physical formulas?

No, dimensional analysis only gives the proportional dependence (the powers). It cannot determine dimensionless constants (like $2\pi$ or $1/2$) that might be part of the actual physical formula.

Q: Why is the power of M only dependent on Force?

If you look at the standard dimensions of speed ($[LT^{-1}]$) and acceleration ($[LT^{-2}]$), neither contains mass ($M$). Therefore, the entire mass dimension in Young's modulus must come exclusively from the Force term ($[MLT^{-2}]$).

Question

If speed ($v$), acceleration ($A$) and force ($F$) are considered as fundamental units, the dimension of Young's modulus will be

Options:

(A) $[v^{-4}A^{-2}F]$

(B) $[v^{-2}A^{2}F^{2}]$

(C) $[v^{-2}A^{2}F^{-2}]$

(D) $[v^{-4}A^{2}F]$

Sigma Percentile · Accepted Answer

### The Power of Dimensional Analysis

Imagine you are building a completely new system of physics. Instead of using the traditional fundamental units of Mass ($M$), Length ($L$), and Time ($T$), you decide to use Speed ($v$), Acceleration ($A$), and Force ($F$) as your foundational building blocks. This is exactly what this problem asks us to do! It's a classic exercise in dimensional analysis that tests your understanding of how physical quantities relate to one another.

To translate any quantity into this new system, we must first understand the standard dimensions of our new 'fundamental' units.

Speed ($v$) is distance over time, giving us $[LT^{-1}]$.

Acceleration ($A$) is velocity over time, which gives $[LT^{-2}]$.

Force ($F$), derived from Newton's Second Law ($F = ma$), has the dimension $[MLT^{-2}]$.

### Decoding the Target Quantity

Our target is Young's Modulus ($Y$). Before we can translate it, we need its standard dimension.

Recall that Young's Modulus is defined as the ratio of stress to strain.

$$Y = \frac{	ext{Stress}}{	ext{Strain}}$$

Since strain is simply a ratio of lengths ($\Delta L / L$), it is dimensionless. Therefore, the dimension of Young's Modulus is identical to that of stress, which is force per unit area ($F/A$).

$$[Y] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$$

### Constructing the Master Equation

Now comes the core logic. If Young's Modulus can be expressed in our new system, it must be a product of some unknown powers of $v$, $A$, and $F$. Let's call these powers $\alpha$, $\beta$, and $\gamma$.

$$[Y] = [v]^\alpha [A]^\beta [F]^\gamma$$

By substituting the standard $MLT$ dimensions into this master equation, we set the stage for an algebraic resolution.

$$[ML^{-1}T^{-2}] = [LT^{-1}]^\alpha [LT^{-2}]^\beta [MLT^{-2}]^\gamma$$

### The Algebraic Resolution

To solve this, we must group the powers of $M$, $L$, and $T$ on the right side of the equation.

$$[ML^{-1}T^{-2}] = [M^\gamma L^{\alpha+\beta+\gamma} T^{-\alpha-2\beta-2\gamma}]$$

According to the **Principle of Dimensional Homogeneity**, the powers of each fundamental dimension must be identical on both sides of the equation. This gives us a neat system of three linear equations:

For $M$: $\gamma = 1$

For $L$: $\alpha + \beta + \gamma = -1$

For $T$: $-\alpha - 2\beta - 2\gamma = -2$

We immediately know that $\gamma = 1$. Substituting this into the $L$ equation yields:

$$\alpha + \beta + 1 = -1 \implies \alpha + \beta = -2$$

Substituting $\gamma = 1$ into the $T$ equation yields:

$$-\alpha - 2\beta - 2(1) = -2 \implies -\alpha - 2\beta = 0 \implies \alpha = -2\beta$$

Now, we substitute $\alpha = -2\beta$ into our simplified $L$ equation:

$$-2\beta + \beta = -2 \implies -\beta = -2 \implies \beta = 2$$

With $\beta = 2$, we easily find $\alpha$:

$$\alpha = -2(2) = -4$$

### The Final Result

We have successfully cracked the code! The powers are $\alpha = -4$, $\beta = 2$, and $\gamma = 1$.

Plugging these back into our initial assumption, the dimension of Young's Modulus in terms of speed, acceleration, and force is:

$$[Y] = [v^{-4}A^2F^1]$$

This perfectly matches option (d). Dimensional analysis is a powerful tool, not just for checking equations, but for translating between entirely different physical frameworks!

Animated Solution for Physics - Physics and Measurement: If speed ( $v$ ), acceleration ( $A$ ) and force ( $F$ ) are considered as fundamental units, the dimension of Young's modulus will be

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Visualized Solution

\text{Dimensions of } v, A, \text{ and } F

\text{Dimension of Young's Modulus } (Y)

\text{The Dimensional Equation}

\text{Substituting MLT Dimensions}

\text{Grouping Powers of M, L, and T}

\text{Equating the Powers}

\text{Solving for } \alpha \text{ and } \beta

\text{Final Dimensional Formula}

The Sigma Insight: Dimensional Analysis

The Power of Dimensional Analysis

Decoding the Target Quantity

Constructing the Master Equation

The Algebraic Resolution

The Final Result

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