Question

Match the following columns:

Column I	Column II
(a) The radii of the circular ends of  a bucket, in the form of the frustum of a cone of height 30 cm, are 20 cm and 10 cm respectively. The capacity of the bucket is ........cm3. `["Take"  π = 22/7`.]	(p) 2418π
(b) The radii of the circular ends  of a conical bucket of height 15 cm are 20 and 12 cm respectively. The slant height of the bucket is ........ cm.	(q) 22000
(c) The radii of the circular ends of  a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is .........cm2.	(r) 12
(d) Three solid metallic spheres of  radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is ........ cm.	(s) 17

Answer 1

Column I	Column II
(a) The radii of the circular ends of  a bucket, in the form of the frustum of a cone of height 30 cm, are 20 cm and 10 cm respectively. The capacity of the bucket is ........cm3. `["Take"  π = 22/7`.]	(q) 22000
(b) The radii of the circular ends  of a conical bucket of height 15 cm are 20 and 12 cm respectively. The slant height of the bucket is ........ cm.	(s) 17
(c) The radii of the circular ends of  a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. The total surface area of the bucket is .........cm2.	(p) 2418π
(d) Three solid metallic spheres of  radii 3 cm, 4 cm and 5 cm are melted to form a single solid sphere. The diameter of the resulting sphere is ........ cm.	(r) 12

Explanation:

(a) Let R and r be the top and base of the bucket and let h be its height.

Then, R = 20 cm, r = 10 cm and h = 30 cm.

Capacity of the bucket = Volume of the frustum of the cone

`= (pi"h")/3("R"^2 + "r"^2 + "Rr")`

`= 22/7xx1/3xx30xx[(20)^2 + (10^2) + (20xx10)] "cm"^3`

`= 22/7xx[400+100+200]"cm"^3`

`=(220/7xx700)"cm"^3`

= 22000 cm³ 

Hence, (a) ⇒ (q)

(b) Let R and r be the top and base of the bucket and let h be its height.

Then, R = 20 cm, r = 12 cm and h = 15 cm

`"Slant height of the bucket"  l = sqrt(h^2 + ("R"-r)^2) `

`=sqrt((15)^2 + (20-12)^2)`

`=sqrt(225+64)`

`=sqrt(289) `

= 17 cm

Hence, (b) ⇒ (s)

(c) Let R and r be the top and base of the bucket and let be its slant height.

Then, R = 33 cm, r = 27 cm and h = 10 cm

Total surface area of the bucket `= pi ["R"^2 +"r"^2+"l"("R" + r)]`

`= pixx[(33)^2 + (27)^2 + 10xx(33+27)]`

`= pi xx [1089 + 729 + 600]`

`= 2418pi  "cm"^2`

Hence, (c) ⇒ (p)

(d) Let the diameter of the required sphere be d.

Then, volume of the sphere`=4/3 pi"r"^3`

`= 4/3pi("d"/2)^3`

Therefore,

`4/3pi("d"/2)^3 = 4/3pi(3)^3 + 4/3pi(4)^3 + 4/3pi(5)^3`

`=> 4/3pi"d"^3/8 = 4/3pixx[(3)^3+(4)^3 + (5)^3`

`=> "d"^3/8 = 216`

⇒ d³ = 1728

⇒ d³ = 12³ 

⇒ d = 12 cm

Hence, (d) ⇒ (r)