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An iron pillar consists of a cylindrical portion 2.8m high and 20cm in diameter and a cone 42cm high is surmounted on it. Find the weight of the pillar, given that 1cm3 of iron weights 7.5gm.

Answer
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Hint: Find the volume of the pillar by finding the volumes of cylindrical and conical portions. And then calculate its weight as per the given information.
Given, the height of the cylindrical portion of the pillar, hcylinder=2.8m=280cm.
And the height of the conical portion hcone=42cm.
Since, the cone is surmounted on the cylinder, the radius of the base will be the same for cylinder and cone. And diameter is given as 20cm in the question. So, radius will be:
r=d2=202,r=10cm
And we know formulae of volume of cylinder and cone, then:
Vcylinder=πr2hcylinder,Vcone=13πr2hcone
Thus, the total volume of the pillar will be:
V=Vcylinder+Vcone,V=πr2hcylinder+13πr2hcone,V=πr2[hcylinder+hcone3]
Putting values of respective heights and radius, we’ll get:
V=227×(10)2[280+423],V=227×100×294,V=92400
Thus the volume of the pillar is 92400cm3. Now, we have to determine the weight of the pillar. So, according to question:
Weight of 1cm3of iron =7.5gm,
Therefore, the weight of 92400cm3of iron will be:
w=92400×7.5gm,w=693000gm,w=693kg.
Thus, the total weight of the pillar is 693kg.
Note: If we have to calculate the weight of a solid when its density is given, we always have to calculate the volume of the solid first because weight is directly related to volume and density as:
Density=weightVolume,weight=Density×Volume.
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