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 the remaining solid
(ii) the volume of remaining solid
(iii) the weight of the material drilled out if it weighs 7 gm per cm³.

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 the remaining solid
(ii) the volume of remaining solid
(iii) the weight of the material drilled out if it weighs 7 gm per cm³.

KnowledgeBoat · Accepted Answer

Given, Dimensions of rectangular solid are: l = 42 cm, b = 30 cm and h = 20 cm. Conical cavity’s diameter = 14 cm So, its radius (r) = 7 cm Depth (h) = 24 cm (i) Total surface area of cuboid = 2(lb + bh + lh) = 2 (42 x 30 + 30 x 20 + 20 x 42) = 2 (1260 + 600 + 840) = 2 (2700) = 5400 cm² Area of circular base of conical cavity = πr² = x 7 x 7 = 154 cm². By formula, ⇒ l² = r² + h² ⇒ l² = (7)² + (24)² ⇒ l² = 49 + 576 ⇒ l² = 625 ⇒ l = ⇒ l = 25 cm. Area of curved surface area of cone = πrl = = 22 x 25 = 550 cm² Surface area of remaining part = Surface area of rectangular solid + Surface area of cone - Area of base of conical cavity = 5400 + 550 - 154 = 5796 cm². Hence, surface area of remaining part = 5796 cm². (ii) Volume of the rectangular solid = lbh = (42 x 30 x 20) cm³ = 25200 cm³ Radius of conical cavity (r) = 7 cm Depth (h) = 24 cm Volume of cone = = = 22 × 7 × 8 = 1232 cm³. Volume of remaining solid = Volume of rectangular solid - Volume of cone = 25200 - 1232 = 23968 cm³. Hence, volume of remaining solid = 23968 cm³. (iii) Volume of material drilled out = Volume of cone = 1232 cm³. Weight of material drilled out = 1232 × 7 = 8624 g = 8.624 kg. Hence, weight of material drilled out = 8.624 kg.

Mathematics

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 the remaining solid
(ii) the volume of remaining solid
(iii) the weight of the material drilled out if it weighs 7 gm per cm³.

Mensuration

Related Questions

A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.

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Mathematics

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 the remaining solid(ii) the volume of remaining solid(iii) the weight of the material drilled out if it weighs 7 gm per cm3.

Mensuration

Related Questions

A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.

The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.

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 the remaining solid
(ii) the volume of remaining solid
(iii) the weight of the material drilled out if it weighs 7 gm per cm³.