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.

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 = $\dfrac{22}{7}$ x 7 x 7 = 154 cm².

By formula,

⇒ l² = r² + h²

⇒ l² = (7)² + (24)²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$

⇒ l = 25 cm.

Area of curved surface area of cone = πrl

= $\dfrac{22}{7} \times 7 \times 25$ = 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 = $\dfrac{1}{3}πr^2h$

= $\dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 24$

= 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.