Question

.A cylindrical tub of radius 5cm and height 9.8cm is full of water.A solid in the form of a right circular cone mounted on a hemispherical base ,is completely immersed in the tub.If the radius of the hemisphere is 3.5cm and height of the conical portion is 5cm,find the volume of water which is left in the tub.​

Answer 1

Step-by-step explanation:

Hello...

The tub is half of a cylinder

∴ volume of water = 1/2 x 22/7 x 5² x 9.8 = 385 cm³

Volume of the solid = volume of cone + volume of hemisphere

= 1/3 x 22/7 x 3.5² x 5 + 2/3 x 22/7 x 3.5³

= 64 1/6 + 89 5/6

= 154 cm³

Remaining volume = 385 - 154 = 231 cm³

@hardikrakholiya21

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Answer 2

SOLUTION

Radius of hemisphere=radius of cone=3.5cm

Height of cone= 5cm

Volume of object

=Volume of hemisphere+volume of cone

$= > ( \frac{2}{3} \pi {r}^{3} ) + ( \frac{1}{3} \pi {r}^{2} h) \\ \\ = > ( \frac{2}{3} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \times \frac{35}{10} ) + ( \frac{1}{3} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \times 5) \\ \\ = > ( \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} ( \frac{2}{3} \times \frac{35}{10} + \frac{1}{3} \times 5) \\ \\ = > 11 \times \frac{35}{10} ( \frac{7}{3} + \frac{5}{3}) \\ \\ = > \frac{385}{10} \times \frac{12}{3} \\ \\ = > 38.5 \times 4 \\ = > 154 {cm}^{2}$

Now,

Radius of cylinder= 5cm

Height of cylinder= 9.8cm

Volume of cylinder= πr²h

$= > \frac{22}{7} \times 5 \times 5 \times \frac{98}{10} \\ \\ = > 770 {cm}^{2}$

Now, when the object is placed in the cylindrical tub full of water, the volume of water displaced will be equal to the volume of the object.

Therefore, remaining Volume in the cylindrical tub

=)Volume of tub - volume of object

=) (770- 154)cm²

=) 616cm² [answer]

Hope it helps ☺️❤️