A metal weighs 20N in air, 12N in water and 14N in kerosene. Calculate the relative density of :a. The metal b. The kerosene
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I would think of it slightly differently to the gents below. In a manner more intuitively to me.
The density is the mass / volume
And that the volume is the mass of water displaced.
So, given that we know that the acceleration due to gravity is something like 9.801m/seconds^2, the masses (or mass equivalents) are
7.54N / 9.801m/second^2 = 0.7693kg and
6.56N / 9.801 m/second^2 = 0.6693kg respectively
The volume of water displaced is the difference between those two masses (if the temperature is such that the water density is 1.000kg/litre). Ie the volume of the object is 0.7693kg - 0.6693kg = 0.100kg = 0.100 litres
The density is then 0.7693kg/0.100 litres = 7.693kg/litre or 7693kg/m^3. Which might do with being rounded to (say) 7690kg/m^3.
Interestingly, we did not use the dimensions provided, and that the acceleration due to gravity was not necessary (but most of us "weigh" things on a balance calibrated in kg, not N).
But, just as a check, 4.5cm x 4.5cm x 1.2cm is 24.3 cubic centimetres, or 0.0243 litres. *Ooops*! Something does not add up.
Supposing that the clue to this lies in the fact that the question states that the "dimension of the *metal* is..." (emphasis mine), maybe the whole item is not just a simple chunk of metal. Ie it is a chunk of metal, surrounded by something of negligible density. So, the density of the metal itself is 0.7693kg/0.0243litres or 31.7kg/litre or 31700 kg/m^3. Impressive density. 1.5 times the density of the most dense element (osmium).
Or, have I missed something?
Plz Brain mark
The density is the mass / volume
And that the volume is the mass of water displaced.
So, given that we know that the acceleration due to gravity is something like 9.801m/seconds^2, the masses (or mass equivalents) are
7.54N / 9.801m/second^2 = 0.7693kg and
6.56N / 9.801 m/second^2 = 0.6693kg respectively
The volume of water displaced is the difference between those two masses (if the temperature is such that the water density is 1.000kg/litre). Ie the volume of the object is 0.7693kg - 0.6693kg = 0.100kg = 0.100 litres
The density is then 0.7693kg/0.100 litres = 7.693kg/litre or 7693kg/m^3. Which might do with being rounded to (say) 7690kg/m^3.
Interestingly, we did not use the dimensions provided, and that the acceleration due to gravity was not necessary (but most of us "weigh" things on a balance calibrated in kg, not N).
But, just as a check, 4.5cm x 4.5cm x 1.2cm is 24.3 cubic centimetres, or 0.0243 litres. *Ooops*! Something does not add up.
Supposing that the clue to this lies in the fact that the question states that the "dimension of the *metal* is..." (emphasis mine), maybe the whole item is not just a simple chunk of metal. Ie it is a chunk of metal, surrounded by something of negligible density. So, the density of the metal itself is 0.7693kg/0.0243litres or 31.7kg/litre or 31700 kg/m^3. Impressive density. 1.5 times the density of the most dense element (osmium).
Or, have I missed something?
Plz Brain mark
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Answer:
Explanation:
When something is dipped in a liquid, a force pushes it up (upthrust). That force is equal to the volume displaced by the solid times the density of the liquid. As you want a density (here in kg/lit), let’s convert the N into kg : 20 / 9.80665 = 2.0364 kg. Similarly 12 N = 1.2236 kg.
Hence:
2.0364 - 1 x V = 1.2236 (density of water = 1 and V is the volume of the metal). Therefore V = 0.81577 and the density will be 2.0364/0.81577 = 2.5 kg/lit (or for the fanatical people 2500 kg/m3).
for the kerosene to reach the same figure, its density should be 0.75 which is way too low. Actually 0.78 to 0.81.However we did not need this figure to answer the question.
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