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A hydrometer is a specific gravity indicator, the value being indicated by the level at which the free surface intersects the stem when floating in a liquid. The 1.0 mark is the level when in distilled water. For the unit shown, the immersed volume in distilled water is \(15 \mathrm{cm}^{3} .\) The stem is \(6 \mathrm{mm}\) in diameter. Find the distance, \(h,\) from the 1.0 mark to the surface when the hydrometer is placed in a nitric acid solution of specific gravity 1.5.

Short Answer

Expert verified
The distance, \( h \), from the 1.0 mark to the surface when the hydrometer is placed in a nitric acid solution of specific gravity 1.5 is 2.82 cm.

Step by step solution

01

Understand the hydrometer principle

A hydrometer works on the principle of buoyancy, where Archimedes' principle comes into picture which states that object submerged in a fluid experiences an upthrust or apparent loss in weight which is equal to the weight of the fluid displaced by it. Therefore, when it floats in a liquid, it displaces a volume of the liquid equivalent to its own weight.
02

Calculate the total volume of the hydrometer

The total volume of the hydrometer immersed in distilled water is given to be 15 cm³. This means that, in distilled water, the hydrometer displaces its volume, V = 15 cm³.
03

Calculate the new volume of hydrometer immersed

When the hydrometer is placed in nitric acid solution, it will float higher or lower depending on the specific gravity of the nitric acid solution. According to Archimedes' principle, the volume of nitric acid displaced is also equal to the weight of the hydrometer. If \( s_g \) is the specific gravity of nitric acid, then the volume \( V' \) of nitric acid displaced would be \( V' = V / s_g = 15 cm³ / 1.5 = 10 cm³ \).
04

Calculate the distance from the 1.0 mark to the surface

The change in volume is due to the change in height from the 1.0 mark. Let \( h \) be the distance. Change in volume \( \Delta V = \pi *(d/2)^2*h \), where \( d = 6mm \) is the diameter of the stem of hydrometer. Therefore, using the formula for cylinder volume, \( h = \Delta V / (\pi * (d/2)^2) = (15cm³ - 10cm³) / \pi * (0.6 cm )^2 = 2.82 cm \).

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