Question

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub (Take $(\pi=\frac{22}{7})$)

Answer 1

Answer:

The volume of water left in the tube is 616 cm³.

Step-by-step explanation:

SOLUTION :  

Given :  

Radius of the cylindrical tub (r) = 5 cm

Height of the Cylindrical tub ( H) = 9.8 cm

Height of the cone outside the hemisphere (h) = 5 cm

Radius of the hemisphere = 5 cm

Volume of the cylindrical tub , V1  = πr²H

V1 = π( 5)² ×  9.8 = 22/7 × 25 × 9.8 = 22 × 25 × 1.4

V1 = 770 cm³

Volume of the hemisphere , V2 = 2/3πr³

V2 = ⅔ × 22/7 × 3.5³

V2 = 89.79 cm³

Volume of the cone  , V3 = ⅓πr²h

V3 = ⅓ × 22/7 ×3.5² × 5 = (22 × 3.5 × 3.5 ×5)/ (3 × 7) = (22 × 0.5 × 3.5 ×5)/ 3 = 192.5/3 = 64.14 cm³

V3 = 64.14 cm³

Total volume of the solid immersed in the tub = Volume of the cone + Volume of the hemisphere  = V2 + V3

V = 89.79 + 64.14 =  154 cm³

Hence, the total volume of the solid immersed in the tub = 154 cm³

Volume of the water left in the tub = volume of the cylinder - Total volume of the solid immersed in the tub

= 770 - 154 = 616 cm³

Hence, the volume of water left in the tube is 616 cm³.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\huge\bf\pink{\mid{\overline{\underline{Your\: Answer}}}\mid}$

616 cm³

Step-by-step explanation:

Given,

Radius of the cylindrical tub (r) = 5 cm

and,

Height of the Cylindrical tub ( H) = 9.8 cm

also,

Height of the cone outside the hemisphere (h) = 5 cm

and,

Radius of the hemisphere = 5 cm

now,

Volume of the cylindrical tub , V1  = πr²H

=> V1 = π( 5)² ×  9.8

= 22/7 × 25 × 9.8

= 22 × 25 × 1.4

=> V1 = 770 cm³

now,

Volume of the hemisphere , V2 = 2/3πr³

=> V2 = ⅔ × 22/7 × 3.5³

=> V2 = 89.79 cm³

again,

Volume of the cone  , V3 = ⅓πr²h

V3 = ⅓ × 22/7 ×3.5² × 5

= (22 × 3.5 × 3.5 ×5)/ (3 × 7)

= (22 × 0.5 × 3.5 ×5)/ 3 = 192.5/3 = 64.14 cm³

=> V3 = 64.14 cm³

Total volume of the solid immersed in the tub = Volume of the cone + Volume of the hemisphere  = V2 + V3

=> V = 89.79 + 64.14

=  154 cm³

therefore,

the total volume of the solid immersed in the tub = 154 cm³

Volume of the water left in the tub = volume of the cylinder - Total volume of the solid immersed in the tub

= 770 - 154 = 616 cm³

Hence, the volume of water left in the tube is 616 cm³.