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 $c{m}^3$.

Answer 1

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

surface area = 2(lb +bh +lh) = 5400 $c{m}^2$

Volume= lbh = 25200 $c{m}^2$

For cone r = 7 cm
h = 24 cm
l = $\sqrt{(h^2+r^2)}=\sqrt{(576+49)}=25cm$
then curved surface area = $\pi \times r\times l=22\times 25=550c{m}^2$

volume =$\frac{1}{3}\pi r^{2}h=1232c{m}^3$

Area of bottom of cone (circle) =$\pi r^2=\pi \times 49=154c{m}^2$

Then surface area of remaining rectangle = Total surface area of rectangle - Cone's bottom part = 5400 - 154 = 5246 $c{m}^2$

(i) Surface area of remaining solid = surface area of remaining rectangle + Curved surface area of cone = 5246 + 550 =5796 $c{m}^2$

(ii) Volume of remaining solid = volume of solid - volume of cone = 25200 - 1232 = 23968 $c{m}^3$

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