Question

A solid consisting of a right circular cone of height 12 cm and radius 6 cm standing on a hemisphere of radius 6 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of the water displaced out of the cylinder, if the radius of the cylinder is 6 cm and height is 18 cm.

fast guys its urgent answer me​

Answer 1

$\sf{Given:}$

For cone:

• height = 12 cm
• radius = 6 cm

For hemisphere:

• radius = 6 cm

For cylinder:

• radius = 6 cm
• height = 18 cm

$\sf{Find:}$

Volume of the water displaced by the cylinder.

$\sf{Solution:}$

Volume of water displaced by cylinder = Volume of cylinder - (Volume of circular cone + Volume of hemisphere)

=> πr²h - (1/3πr²h + 2/3πr³)

=> 22/7 × (6)² × 18 - [1/3 × 22/7 × (6)² × 12 + 2/3 × 22/7 × (6)³]

=> 22/7 × 36 [(18) - (1/3 × 12) + (2/3 × 6)]

=> 22/7 × 36 [18 - (4 + 4)]

=> 22/7 × 36 (18 - 8)

=> 22/7 × 36 (10)

=> 22/7 × 360

=> 1131.428 cm³

•°• 1131.428 cm³ of the water displaced by the cylinder.

Answer 2

Answer:

★1131.428 cm³★

Step-by-step explanation:

»Radius is same for Cone, Cylinder and Hemisphere = 6 cm.

»H(Height) of Cone = 12 cm.

»H(Height of Cylinder= 18 cm.

According to Question :-

⇒Vol. of Water displaced by cylinder= Vol. of Cylinder -(Vol. of Hemisphere+ Vol. of Cone)

Now, I'm Going to tell Some Basic Formulas Which will be applied :-

⇒Vol. of Cylinder = πr²h

⇒ Vol. of Circular Cone = 1/3 πr²h

⇒ Vol. of Hemisphere= 2/3πr³.

Now Applying this Formula to find Volume of Displaced water from Cylinder.

$⇒\pi {r}^{2} h - (\frac{1}{3}\pi {r}^{2} h + \frac{2}{3} \pi {r}^{3} )$

⇒ 22/7 ×(6)²×18-[1/3×22/7×(6)²×12+2/3×227×(6)³]

⇒22/7 × 36[(18)-(1/3×12)+(2/3×6)]

⇒ 22/7 × 36 [18- (4+4)]

⇒22/7 × 36 [18 - 8]

⇒ 22/7 × 360 ⇒ 1131.428 cm³.

____________________________