A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a conical block of iron whose radius of the base is 3.5 cm and height is 6 cm is wholly immersed in water, by how much will the water level rise ?
Answers
Given :- A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a conical block of iron whose radius of the base is 3.5 cm and height is 6 cm is wholly immersed in water, by how much will the water level rise ?
Solution :-
As we know that, rise in the level of water is equal to the volume of the conical block of iron .
Let us assume that, water rises to the height of H cm .
So,
→ Volume of cylinder of height H = Volume of conical block of iron .
we know that,
- Volume of cylinder = π * (radius)² * Height .
- Volume of cone = (1/3) * π * (radius)² * Height .
Given that,
- Diameter of cylinder = 14cm .
- radius of cylinder = (diameter/2) = 14/2 = 7cm.
- radius of base of conical block = 3.5 cm. = (7/2) cm.
- height of conical block = 6 cm.
- height of cylinder = H cm.
Putting all values we get,
→ Volume of cylinder of height H = Volume of conical block of iron .
→ π * (7)² * H = (1/3) * π * (7/2)² * 6
π will be cancel from both sides,
→ 3 * (7)² * H = (7)² / 4 * 6
(7)² will be cancel from both sides,
→ 3 * 4 * H = 6
→ 12H = 6
→ H = (6/12)
→ H = (1/2)
→ H = 0.5 cm.
Hence, the water will rise to the height of 0.5 cm.
Learn more :-
the diameter of the spherical football is four times of the diameter of the spherical tennis ball. find the ratio of the...
https://brainly.in/question/25245922
Answer:
this is the ducking #$%^ !answer
Step-by-step explanation:
Volume of the Rectangular Block = 22 * 18 * 7
Radius of the Cistern = 7 cm
Area of the Cylinder = Π * r� * h = 22/7 * 7 * 7 * h
22/7 * 7 * 7 * h = 22 * 18 * 7; h = 18 cm