What is the volume of a metal which weighs 28g less in kerosene and the densty is 800 kgm3 than it does in air
What do you mean by
" 800 k/gm^3 than it does in air " ?
The external fluid does not affect the density.
Did you mean to write:
"What is the volume of a metal which weighs 28g less in kerosene does in air, if the density of kerosene is 800 kg/m^3?
That question can be answered.
The metal will certainly sink to the bottom, and there will be an upward buoyancy force of (0.028 kg)*g = 0.274 N. This equals (kerosene density)*g*V, so
V = 0.274/(800*9.8) = 3.5*10^-5 m^3 = 35 cm^3
To find the volume of the metal, we can use the principle of buoyancy because the weight of an object in a fluid is equal to the weight of the displaced fluid. In this case, we are given that the metal weighs 28g less in kerosene than in air, and we know the density of kerosene is 800 kg/m³.
First, let's convert the weight difference from grams to kilograms:
28 g = 0.028 kg
Since we are comparing the weight in two different fluids, we need to use the concept of buoyancy. The weight difference is equal to the weight of the kerosene displaced by the metal:
Weight difference = Weight in air - Weight in kerosene
We know that the density of kerosene is 800 kg/m³, so we can calculate the volume of kerosene displaced by the metal using the formula:
Volume of kerosene = Weight difference / Density of kerosene
Volume of kerosene = 0.028 kg / 800 kg/m³
Now, we can find the volume of the metal using the principle of buoyancy:
Volume of metal = Volume of kerosene
Therefore, the volume of the metal is equal to the volume of kerosene, which is calculated to be 0.000035 m³ (or 35 cm³).