A hydrometer has a weighed bulb of volume 30cm3 and a uniform cross sectional area of 1cm3. When floating in water 3cm of the stem is immersed. What is the length of the stem when immersed in water when the hydrometer floats in the liquid of density 0.89gcm-3?
To solve this problem, we can use the principle of flotation. According to this principle, a floating object displaces its own weight of the liquid in which it floats.
1. Let's first calculate the weight of the displaced water when the hydrometer is floating in water. The weight of the displaced water is equal to the weight of the hydrometer itself, which can be calculated using its volume and density of water.
Weight of displaced water = Volume of displaced water × Density of water
The volume of displaced water can be calculated using the following formula:
Volume of displaced water = Volume of the hydrometer - Volume of the submerged part of the hydrometer
In this case, the volume of the hydrometer is 30 cm^3, and 3 cm of the stem is submerged. Therefore,
Volume of displaced water = 30 cm^3 - 3 cm^3 = 27 cm^3
The density of water is 1 g/cm^3.
So, the weight of the displaced water = 27 cm^3 × 1 g/cm^3 = 27 g
Therefore, when the hydrometer is floating in water, it displaces 27 g of water.
2. Now, let's calculate the weight of the displaced liquid when the hydrometer floats in the liquid of density 0.89 g/cm^3. Using the principle of flotation, the weight of the displaced liquid will be equal to the weight of the hydrometer itself.
Weight of displaced liquid = Weight of the hydrometer
The weight of the hydrometer can be calculated using its volume and the density of the liquid. The volume of the hydrometer is 30 cm^3, and the density of the liquid is 0.89 g/cm^3.
Weight of the hydrometer = 30 cm^3 × 0.89 g/cm^3 = 26.7 g
Therefore, when the hydrometer is floating in the liquid, it displaces 26.7 g of the liquid.
3. Finally, let's calculate the length of the stem submerged in the liquid. Since the hydrometer displaces its own weight of the liquid, the length of the stem submerged in the liquid is proportional to the weight of the displaced liquid.
Length of stem submerged in the liquid = (Weight of displaced liquid / Density of liquid) × Cross-sectional area of the stem
Plugging in the values, we get:
Length of stem submerged in the liquid = (26.7 g / 0.89 g/cm^3) × 1 cm^3 = 30 cm
Therefore, the length of the stem submerged in the liquid when the hydrometer floats in the liquid of density 0.89 g/cm^3 is 30 cm.