The density of a material in SI units is 128 kg m^-3. In certain units in which the unit of length i

Q: Why do we cube the length unit in the denominator?

Density is defined as mass per unit volume. Since volume is derived from length cubed ($V = L^3$), whatever unit of length you use must be cubed to represent volume correctly.

Q: Can I convert the new units into SI units instead of the other way around?

Absolutely! You could convert $50 \text{ g}$ to $0.05 \text{ kg}$ and $25 \text{ cm}$ to $0.25 \text{ m}$. The math will work out exactly the same, though dealing with decimals might be slightly more prone to calculation errors than using whole numbers.

Q: What does $N_1 U_1 = N_2 U_2$ actually mean physically?

It means that the actual 'amount' of stuff doesn't change just because you measure it differently. If you measure a table in inches, the number is bigger than if you measure it in meters, but the table's length is identical. The product of the number and the size of the unit is constant.

Q: Why did we expand the cubes instead of calculating $100^3$ as $1000000$?

Expanding the terms as $100 \times 100 \times 100$ allows us to spot direct cancellations with the $25 \times 25 \times 25$ on the other side. It's a strategic algebraic move to avoid multiplying large numbers only to divide them again.

Q: Would this method work for quantities with time, like velocity or force?

Yes! The method is universal. You just need the correct dimensional formula. For force ($[MLT^{-2}]$), you would substitute the respective units for mass, length, and time in both systems and equate them.

Question

The density of a material in SI units is $128 	ext{ kg m}^{-3}$. In certain units in which the unit of length is $25 	ext{ cm}$ and the unit of mass is $50 	ext{ g}$, the numerical value of density of the material is

Options:

(A) 40

(B) 16

(C) 640

(D) 410

Sigma Percentile · Accepted Answer

### The Philosophy of Measurement

Imagine you are an explorer traveling between two completely different worlds. In our world, we measure mass in kilograms and length in meters.

But in this new, alien world, the inhabitants have decided that their standard unit of mass is $50 	ext{ g}$, and their standard unit of length is $25 	ext{ cm}$.

Despite these different languages of measurement, the actual physical reality—the "heaviness" or "compactness" of a material—does not change. A block of iron is just as dense here as it is there.

This brings us to the most beautiful and fundamental rule of dimensional analysis: **the physical quantity remains constant regardless of the system of units used to measure it**.

### The Master Equation

Because the physical quantity is constant, we can mathematically state that the numerical value in the first system multiplied by its unit must equal the numerical value in the second system multiplied by its unit.

This gives us our master equation:

$$N_1 U_1 = N_2 U_2$$

Here, $N_1$ and $N_2$ are the numerical values, and $U_1$ and $U_2$ are the respective units.

In our problem, we are dealing with density. The dimensional formula for density is mass divided by volume, or $[ML^{-3}]$.

Therefore, the unit $U$ can be written as $\frac{M}{L^3}$.

### Setting Up the Two Worlds

Let's carefully substitute the given values into our master equation.

For the SI system (System 1), the numerical value $N_1$ is $128$, the mass unit is $	ext{kg}$, and the length unit is $	ext{m}$.

For the new system (System 2), we need to find $N_2$. The mass unit is $50 	ext{ g}$, and the length unit is $25 	ext{ cm}$.

Plugging these into our equation, we get:

$$128 \left(\frac{	ext{kg}}{	ext{m}^3}ight) = N_2 \left(\frac{50 	ext{ g}}{(25 	ext{ cm})^3}ight)$$

**A critical trap to avoid here is forgetting to cube the entire length unit in the denominator.** The volume is derived from length cubed, so the $25 	ext{ cm}$ must be fully enclosed in parentheses and cubed.

### The Art of Conversion

To solve for $N_2$, we need to speak a common language. We cannot directly compare kilograms to grams or meters to centimeters.

Let's convert the SI units on the left side into the CGS units (grams and centimeters) used on the right side.

We know that $1 	ext{ kg} = 1000 	ext{ g}$ and $1 	ext{ m} = 100 	ext{ cm}$. Substituting these conversions, our equation transforms into:

$$128 \left(\frac{1000 	ext{ g}}{(100 	ext{ cm})^3}ight) = N_2 \left(\frac{50 	ext{ g}}{(25 	ext{ cm})^3}ight)$$

Now, let's expand the cubes in the denominators. Writing them out explicitly helps prevent calculation errors and makes cancellations obvious.

$$128 	imes \frac{1000}{100 	imes 100 	imes 100} = N_2 	imes \frac{50}{25 	imes 25 	imes 25}$$

Notice how the units of grams ($	ext{g}$) and cubic centimeters ($	ext{cm}^3$) have perfectly cancelled out from both sides, leaving us with a pure numerical equation.

### The Final Calculation

Now, we simply need to isolate $N_2$. By cross-multiplying, we move the three $25$s to the numerator on the left side, and the $50$ to the denominator.

$$N_2 = \frac{128 	imes 1000 	imes 25 	imes 25 	imes 25}{50 	imes 100 	imes 100 	imes 100}$$

This might look like a terrifying calculation, but let's take a breath and look for elegant cancellations.

First, notice that $25$ goes into $100$ exactly $4$ times. Since we have three pairs of these, we get three $4$s in the denominator.

Next, $50$ goes into $1000$ exactly $20$ times.

$$N_2 = \frac{128 	imes 20}{4 	imes 4 	imes 4}$$

Since $4 	imes 4 	imes 4 = 64$, the equation simplifies beautifully:

$$N_2 = \frac{128 	imes 20}{64}$$

Finally, $64$ goes into $128$ exactly $2$ times.

$$N_2 = 2 	imes 20$$

**$$N_2 = 40$$**

The numerical value of the density in this new, alien system of units is exactly $40$.

By trusting the principle of dimensional homogeneity and carefully executing our algebraic cancellations, we turned a complex-looking conversion into a smooth and satisfying sequence of logical steps.

Animated Solution for Physics - Physics and Measurement: The density of a material in SI units is $128 kg m^{- 3}$ . In certain units in which the unit of length is $25 cm$ and the unit of mass is $50 g$ , the numerical value of density of the material is

Select Answer:

Visualized Solution

$Visualizing the Systems$

$The Principle of N_{1} U_{1} = N_{2} U_{2}$

$Setting up Density U = \frac{M}{L ^{3}}$

$Converting to CGS Units$

$Expanding the Denominators$

$Isolating the Unknown N_{2}$

$Algebraic Simplification$

$Calculating the Final Value$

$Extending to Other Dimensions$

The Sigma Insight: Dimensional Analysis

The Philosophy of Measurement

The Master Equation

Setting Up the Two Worlds

The Art of Conversion

The Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Principle of N1​U1​=N2​U2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up Density U=L3M​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Converting to CGS Units

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating the Unknown N2​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extending to Other Dimensions