a solid of density 5000 kgf weights
0.5 kgf in air . it is completely immersed in water of density 1000 kg m-3.calculate the apparent weight of the solid in water.
To calculate the apparent weight of the solid in water, we need to consider the buoyant force acting on the solid when it is immersed in water.
The buoyant force is equal to the weight of the water displaced by the solid. Therefore, we can use Archimedes' principle to find the buoyant force:
Buoyant force = Weight of water displaced = Density of water x Volume of water displaced x acceleration due to gravity
First, we need to find the volume of water displaced by the solid. Since the solid is completely immersed in water, the volume of water displaced is equal to the volume of the solid. We can find this using the following formula:
Volume of solid = Mass of solid / Density of solid
In this case, the mass of the solid is given as 0.5 kgf, which can be converted to kg by dividing by the acceleration due to gravity (9.8 m/s^2):
Mass of solid = 0.5 kgf / 9.8 m/s^2 = 0.051 kg
The density of the solid is given as 5000 kgf, which can also be converted to kg by dividing by the acceleration due to gravity:
Density of solid = 5000 kgf / 9.8 m/s^2 = 510.2 kg/m^3
Now, we can calculate the volume of the solid:
Volume of solid = Mass of solid / Density of solid = 0.051 kg / 510.2 kg/m^3 = 0.0001 m^3
Next, we calculate the buoyant force:
Buoyant force = Density of water x Volume of water displaced x acceleration due to gravity
= 1000 kg/m^3 x 0.0001 m^3 x 9.8 m/s^2
= 0.98 N
Finally, the apparent weight of the solid in water is the difference between its weight in air and the buoyant force:
Apparent weight = Weight in air - Buoyant force
= 0.5 kgf - 0.98 N
Note: It is important to convert the weight from kgf to Newtons (N) as kgf is not part of the SI unit system.
Therefore, the apparent weight of the solid in water is 0.5 kgf - 0.98 N.