A solid of R.D. 4.2, is found to weigh 0.200kgf in air
Find its apparent weight in water
If the relative density is 4.2 compared to water, then in water it will weigh the same amount less the weight of the water displaced.
Example:
if the volume is 1m^3, with RD 4.2, then its weight is 4200N. In water, its apparent weigh is 4200N-1m^3*1E3kg*9.8= 4200-980=3400N
I wonder if you are learning kgf as force, instead of using the SI unit of Newtons.
To find the apparent weight of the solid in water, we need to use Archimedes' principle, which states that an object submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.
To calculate the apparent weight in water, we need to find the weight of the water displaced by the solid. Let's break down the steps:
Step 1: Calculate the actual weight of the solid in air.
Given that the solid has a relative density (R.D.) of 4.2 and weighs 0.200 kgf (kilogram-force) in air, we can calculate the actual weight using the formula:
Weight = Mass x Acceleration due to gravity.
The weight is measured in newtons (N), so we need to convert the kilogram-force to newtons by multiplying by the standard acceleration due to gravity (9.81 m/s²).
Weight in air = 0.200 kgf x 9.81 m/s².
Step 2: Calculate the weight of the water displaced.
The weight of the water displaced is equal to the weight of the solid in air.
Weight of water displaced = Weight in air.
Step 3: Calculate the apparent weight in water.
The apparent weight in water is the difference between the weight of the solid in air and the weight of the water displaced.
Apparent weight in water = Weight in air - Weight of water displaced.
Now we can substitute the values and calculate the apparent weight in water.