a piece of metal of density
7800kgm-3 weighs 20N in air. Cal. The
apparent weight of th metal when
completely immersed in a liquid of density 830kgm-3
20 N = 9.81 * m
m = 2.04 kg
volume = 2.04 kg/7800 kg/m^3
= .000261 m^3
mass of fluid displaced = .000261*830
= .217 kg
mass metal - mass fluid displaced = 1.82 kg metal mass
metal weight = m g = 1.82*9.81 = 17.9 Newtons
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TO ANSWER QUESTION DIRECTLY
2.12
To find the apparent weight of the metal when completely immersed in a liquid of density 830 kgm-3, we need to consider the buoyant force acting on the metal.
The buoyant force is the upward force exerted on an object submerged in a fluid, and it is equal to the weight of the fluid displaced by the object. According to Archimedes' principle, the buoyant force on an object is directly proportional to the density of the fluid and the volume of the fluid displaced.
Let's calculate the volume of the metal. We know the density of the metal (7800 kgm-3), and we can use its weight in air to find its volume. The weight of the metal equals the product of its volume and the density of the metal, as given by the formula:
Weight = Volume * Density
Rearranging the formula, we can solve for the volume:
Volume = Weight / Density
Substituting the known values for weight (20 N) and density (7800 kgm-3), we find:
Volume = 20 N / 7800 kgm-3 = 0.002564 m3
Now that we know the volume of the metal, we can calculate the weight of the liquid displaced by the metal when immersed. The weight of the displaced liquid is equal to the volume of the liquid displaced multiplied by the density of the liquid. Given that the density of the liquid is 830 kgm-3, we can calculate the weight of the displaced liquid as follows:
Weight of displaced liquid = Volume of liquid displaced * Density of liquid
Weight of displaced liquid = 0.002564 m3 * 830 kgm-3 = 2.12512 N
Finally, the apparent weight of the metal when completely immersed in the liquid is equal to its weight in air minus the weight of the displaced liquid:
Apparent weight = Weight in air - Weight of displaced liquid
Apparent weight = 20 N - 2.12512 N = 17.87488 N
Therefore, the apparent weight of the metal when completely immersed in the liquid of density 830 kgm-3 is 17.87488 N.