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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