A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion is floating vertically in water at 0$$^\circ$$C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4$$^\circ$$C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4$$^\circ$$C, relative to the density at T = 0$$^\circ$$C is close to:

We begin by noting the basic principle of flotation: a body floats when the buoyant force exerted by the liquid equals the weight of the body. Mathematically, for a floating object, we have the equality $$\text{Buoyant force}= \text{Weight of the body}.$$

The buoyant force is given by Archimedes’ principle, which states that $$\text{Buoyant force}= \rho g V_{\text{sub}},$$ where $$\rho$$ is the density of the liquid, $$g$$ is the acceleration due to gravity, and $$V_{\text{sub}}$$ is the volume of the liquid displaced, i.e. the submerged volume of the object.

Let us denote:

$$\rho_0=$$ density of water at $$0^\circ\text{C},$$

$$\rho_4=$$ density of water at $$4^\circ\text{C},$$

$$A=$$ cross‑sectional area of the cylinder (constant),

$$L=1\ \text{m}=100\ \text{cm},$$ the length of the cylinder (constant, since the metal has negligible expansion).

The cylinder stands vertically and floats. Therefore the submerged length at each temperature can be found by subtracting the height that projects above the water level from the total length.

At $$0^\circ\text{C}:$$ the height above water is $$20\ \text{cm}=0.20\ \text{m}.$$ Hence the submerged length is

$$L_{\text{sub,0}}=L-0.20\ \text{m}=1.00\ \text{m}-0.20\ \text{m}=0.80\ \text{m}.$$

The submerged volume at $$0^\circ\text{C}$$ is therefore

$$V_{\text{sub,0}}=A\,L_{\text{sub,0}}=A(0.80\ \text{m}).$$

Applying Archimedes’ principle, the buoyant force equals the weight of the cylinder, so

$$\rho_0 g V_{\text{sub,0}}=mg,$$

which simplifies to

$$\rho_0 g \,(0.80A)=mg.$$

From this we can express the mass of the cylinder as

$$m=\rho_0(0.80A).$$

At $$4^\circ\text{C}:$$ the height above water becomes $$21\ \text{cm}=0.21\ \text{m}.$$ The submerged length is now

$$L_{\text{sub,4}}=L-0.21\ \text{m}=1.00\ \text{m}-0.21\ \text{m}=0.79\ \text{m}.$$

The submerged volume at $$4^\circ\text{C}$$ is correspondingly

$$V_{\text{sub,4}}=A\,L_{\text{sub,4}}=A(0.79\ \text{m}).$$

Again using Archimedes’ principle at $$4^\circ\text{C},$$ we have

$$\rho_4 g V_{\text{sub,4}}=mg.$$

Substituting the expressions we have just obtained gives

$$\rho_4 g\,(0.79A)=mg.$$

But we already expressed $$mg$$ from the $$0^\circ\text{C}$$ condition as $$\rho_0 g(0.80A).$$ Therefore we can write

$$\rho_4 g\,(0.79A)=\rho_0 g\,(0.80A).$$

We may now cancel the common factors $$g$$ and $$A,$$ yielding the simple ratio

$$\rho_4\,\bigl(0.79\bigr)=\rho_0\,\bigl(0.80\bigr).$$

Hence,

$$\frac{\rho_4}{\rho_0}=\frac{0.80}{0.79}.$$

Evaluating this fraction gives

$$\frac{\rho_4}{\rho_0}=\frac{80}{79}\approx1.012658\ldots\approx1.01.$$

This value of approximately $$1.01$$ corresponds to a relative increase of about $$1\%$$ in the density of water from $$0^\circ\text{C}$$ to $$4^\circ\text{C}.$$

Hence, the correct answer is Option C.