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Question

From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find : 
(i) the surface area of remaining solid, 
(ii) the volume of remaining solid, 
(iii) the weight of the material drilled out if it weighs 7 gm per cm3.


Solution

L = 42 cm
b = 30 cm
h= 20cm
assuming the solid is a cuboid

surface area = 2(lb +bh +lh) = 5400 cm2

Volume= lbh = 25200 cm2

For cone r = 7 cm
h = 24 cm
l = (h2+r2)=(576+49)=25 cm
then curved surface area = π×r×l=22×25=550cm2

volume =13πr2h=1232cm3

Area of bottom of cone (circle) =πr2=π×49=154cm2

Then surface area of remaining rectangle = Total surface area of rectangle - Cone's bottom part = 5400 - 154 = 5246 cm2

(i) Surface area of remaining solid = surface area of remaining rectangle + Curved surface area of cone = 5246 + 550 =5796 cm2

(ii) Volume of remaining solid = volume of solid - volume of cone = 25200 - 1232 = 23968 cm3

(iii) Volume of material drilled out is the same as volume of cone.
then weight = density×volume= 7×1232=8624 g.

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