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 :

1. the surface area of remaining solid,
2. the volume of remaining solid,
3. the weight of the material drilled out if it weighs 7 gm per cm³.

Answer 1

i. Total surface area of cuboid = 2(lb + bh + lh) 

= 2(42 × 30 + 30 × 20 + 20 × 42) 

= 2(1260 + 600 + 840) 

= 2 × 2700

= 5400 cm² 

Diameter of the cone = 14 cm 

Radius of the cone = `14/2` = 7 cm

Area of circular base = πr²

= `22/7 xx 7 xx 7`

= 154 cm² 

Area of curved surface area of cone = πrl

= `22/7 xx 7 xx sqrt(7^2 + 24^2)`

= `22sqrt(49 + 576)`

= 22 × 25

= 550 cm² 

Surface area of remaining part = 5400 + 550 – 154 = 5796 cm²

ii. Dimensions of rectangular solids = (42 × 30 × 20) cm

Volume = (42 × 30 × 20) = 25200 cm³  

Radius of conical cavity (r) = 7 cm 

Height (h) = 24 cm 

Volume of cone = `1/3pir^2h` 

= `1/3 xx 22/7 xx 7 xx 7 xx 24` 

= 1232 cm³ 

Volume of remaining solid = (25200 – 1232) = 23968 cm³   

iii. Weight of material drilled out

= 1232 × 7 g

= 8624 g

= 8.624 kg