Relative density

Two blocks can be made of the same metal, yet one sinks fast and the other barely dips — density tells you why, but density comes with clunky units like $\text{kg/m}^3$ that are awkward to compare at a glance. Relative density fixes that by asking a simpler question: how many times denser is this substance than plain water? The answer is a pure number, with no units at all, and it can be measured in the school lab with nothing more exotic than a spring balance, a beaker of water, and Archimedes' 2200-year-old insight about upthrust. This section builds the idea from the definition, derives the loss-of-weight formula used to measure relative density of a solid, extends it to liquids, and works through the exact numbers you will meet in ICSE Class 9 numericals.

The core insight

Relative density is the unitless ratio of a substance's density to water's density, and for objects denser than water it is measured as $RD = W_1/(W_1 - W_2)$ , the true weight divided by the loss of weight when fully immersed.

The core idea

Relative density (RD), also called specific gravity, compares a substance's density with the density of water:

RD = \frac{\text{density of substance}}{\text{density of water}}

where: the density of substance and the density of water are both measured in the same units (say $\text{kg/m}^3$ ), so those units cancel and RD is left as a bare number — never write "RD = 2.7 kg/m³"; the correct statement is simply $RD = 2.7$ . Because it is a ratio, RD tells you instantly how a solid will behave in water without doing any unit conversion. A substance with $RD > 1$ is denser than water and sinks when fully submerged and released; a substance with $RD < 1$ is less dense than water and floats. Gold has $RD \approx 19.3$ , ice has $RD \approx 0.92$ (which is exactly why icebergs float), and water itself, being compared with itself, has $RD = 1$ by definition. The lab method for finding RD does not measure density directly — it never even needs a measuring cylinder. Instead it leans on Archimedes' principle: when a solid is fully immersed in a fluid, it experiences an upward push (upthrust) equal to the weight of fluid it displaces. A spring balance gives you weight directly, so the whole experiment becomes a matter of weighing the same object three different ways — in air, in water, and (for a liquid's RD) in the unknown liquid — and comparing the losses of weight.

Figure

A solid hanging from a spring balance shown twice, once in air reading a larger weight and once submerged in water reading a smaller weight, with the difference labelled as the upthrust

Extending to a liquid

The same rig measures a liquid's relative density too, using one solid as a tool. Weigh the solid in air ( $W_1$ ), then in water ( $W_2$ ), then in the unknown liquid ( $W_3$ ). Both losses of weight — in water and in the liquid — come from displacing the exact same volume (the solid's own volume), so the ratio of the two losses directly gives the liquid's RD relative to water:

RD_{\text{liquid}} = \frac{W_1 - W_3}{W_1 - W_2}

where: $W_1$ is the weight in air, $W_2$ the apparent weight in water, and $W_3$ the apparent weight in the unknown liquid, all in newtons. This works because upthrust in each fluid is proportional to that fluid's density for a fixed displaced volume — so comparing losses of weight is exactly the same as comparing densities.

Derivation AWhy loss of weight in water gives the relative density of a solid

The goal is to show that for a solid denser than water, the ordinary spring-balance experiment — weigh in air, then weigh fully immersed in water, and divide the true weight by the difference — produces exactly the ratio of densities that defines relative density, with no separate volume measurement needed anywhere in the process.

Step 1. Write the true weight in air. A solid of mass $m$ , density $\rho_s$ and volume $V$ has weight $W_1 = mg = \rho_s V g$ , since $m = \rho_s V$ . This is what the spring balance reads when the solid hangs in air, unaffected by any fluid.

Step 2. Write the apparent weight in water. Once fully immersed, Archimedes' principle says the solid feels an upthrust equal to the weight of water it displaces. Since the solid is fully submerged, the volume of water displaced equals the solid's own volume $V$ , so the upthrust is $F_B = \rho_w V g$ , where $\rho_w$ is the density of water. The spring balance now reads the true weight minus this upthrust: $W_2 = W_1 - F_B = \rho_s V g - \rho_w V g$ .

Step 3. Isolate the loss of weight. Rearranging Step 2, $W_1 - W_2 = \rho_w V g$ — the loss of weight equals exactly the upthrust, which equals the weight of the displaced water. Notice $V$ and $g$ appear in both $W_1$ and $W_1 - W_2$ , which is the key to the next step.

Step 4. Divide $W_1$ by the loss of weight. $\dfrac{W_1}{W_1 - W_2} = \dfrac{\rho_s V g}{\rho_w V g} = \dfrac{\rho_s}{\rho_w}$ . The volume $V$ and gravity $g$ cancel completely, leaving exactly the density ratio that defines relative density. So:

RD_{\text{solid}} = \frac{W_1}{W_1 - W_2}

Step 5. Read the result. Two weighings on a spring balance — nothing more — give the relative density, because the cancellation of $V$ and $g$ means you never need to measure volume, mass, or gravity separately; the ratio of readings does all the work. Verified: with $\rho_s = 3\rho_w$ as a test case, $W_1 = 3\rho_w V g$ and $W_2 = 3\rho_w Vg - \rho_w Vg = 2\rho_w Vg$ , so $W_1/(W_1-W_2) = 3\rho_w Vg / \rho_w Vg = 3 = RD$ , matching Worked Example 1 below.

Worked example

A stone weighs 6 N in air and 4 N when fully immersed in water. It is then weighed in an unknown liquid and found to weigh 4.5 N. Find (a) the relative density of the stone, and (b) the relative density and the density (in $\text{kg/m}^3$ ) of the liquid.

Step 1. Identify the readings. True weight in air $W_1 = 6\ \text{N}$ ; apparent weight in water $W_2 = 4\ \text{N}$ ; apparent weight in the liquid $W_3 = 4.5\ \text{N}$ .

Step 2. Find the loss of weight in water: $W_1 - W_2 = 6 - 4 = 2\ \text{N}$ . Apply the solid's RD formula: $RD_{\text{stone}} = \dfrac{W_1}{W_1 - W_2} = \dfrac{6}{2} = 3$ .

Step 3. Find the loss of weight in the liquid: $W_1 - W_3 = 6 - 4.5 = 1.5\ \text{N}$ . Apply the liquid's RD formula: $RD_{\text{liquid}} = \dfrac{W_1 - W_3}{W_1 - W_2} = \dfrac{1.5}{2} = 0.75$ .

Step 4. Convert the liquid's RD to an actual density using $RD_{\text{liquid}} = \rho_{\text{liquid}}/\rho_{\text{water}}$ with $\rho_{\text{water}} = 1000\ \text{kg/m}^3$ : $\rho_{\text{liquid}} = 0.75 \times 1000 = 750\ \text{kg/m}^3$ .

The stone has relative density 3 (three times as dense as water), and the liquid has relative density 0.75, corresponding to a density of 750 kg/m³ — less dense than water, matching $0.75 < 1$ .

Common traps

Writing units after RD, such as "RD = 3 kg/m³." RD is a pure ratio of two densities with matching units, so it is always just a number.
Using the apparent weight $W_3$ itself, instead of the loss of weight $W_1 - W_3$ , when finding a liquid's RD — the formula needs the difference, not the raw reading in the liquid.
Swapping the numerator between the two formulas. For a solid, it's $W_1$ (true weight) over the loss in water. For a liquid, it's the loss in the liquid over the loss in water — neither formula has a raw weight as the top term for the liquid case.
Applying the loss-of-weight formula to a floating object. The derivation assumed the solid is fully immersed so the displaced volume equals the solid's whole volume; a floating solid displaces less than its own volume and needs different reasoning entirely.

Ask the tutor: Not sure which RD formula to reach for on a mixed numerical — ask the AI tutor to walk through a solid-and-liquid combined problem step by step.

Why does relative density have no units when density itself does?

Density is mass divided by volume, so it carries units like kg/m³. Relative density is density divided by another density — substance over water — and since both are measured in the same units, those units cancel in the division, leaving a pure number. That is exactly why RD is convenient: you can compare 2.7 (aluminium) with 19.3 (gold) directly, without worrying whether one was measured in kg/m³ and the other in g/cm³, because the ratio is the same either way.

Why does the solid need to be fully immersed for the loss-of-weight formula to work?

The whole derivation relies on the volume of water displaced being equal to the volume of the solid, so that it cancels cleanly against the volume hidden inside $W_1$ . If the solid only dips partway in, or floats, it displaces less than its full volume, and the loss of weight would no longer correspond to the solid's total volume — the neat cancellation in Step 4 of the derivation breaks down, and the simple formula $RD = W_1/(W_1-W_2)$ no longer applies.

Can I use this same method to find the relative density of a liquid without any solid at all?

Not with the loss-of-weight spring-balance method described here — it specifically uses a solid as a "measuring tool," because the same solid displaces the same volume in both water and the unknown liquid, and that shared volume is what lets the volume cancel out of the ratio. Other instruments, like a hydrometer, measure a liquid's relative density directly by floating in it, but that is a different method built on flotation rather than on loss of weight.

Relative density

The core idea

Extending to a liquid

Worked example

Common traps

Before this

Where this leads