A cylindrical bucket with base 15 cm with water upto a height of 20 cm a heavy iron spherical ball of radius 9cm is drop into the bucket to submerge completely in the water. Find the incrrase in the level of water.
Radius of cylinder is 15 cm
To find the increase in the level of water when the iron spherical ball is dropped into the bucket, we can use the concept of displacement.
Given:
- The base of the cylindrical bucket has a diameter of 15 cm, so the radius (r) of the base is 7.5 cm.
- The height of the water in the bucket before the ball is dropped is 20 cm.
- The iron spherical ball has a radius (R) of 9 cm.
To calculate the increase in the level of water, we need to find the volume of the iron ball first.
The volume of a sphere can be calculated using the formula:
V = (4/3) * π * R^3
Plugging in the values:
V = (4/3) * 3.14 * 9^3
V ≈ 3053.76 cm^3
Since the iron ball is fully submerged, it will displace an equal volume of water.
The increase in the level of water can be calculated using the formula:
Increase in water level = Volume of ball / Area of base
The area of the base of the cylindrical bucket can be calculated using the formula:
Area = π * r^2
Plugging in the values:
Area = 3.14 * 7.5^2
Area ≈ 176.715 cm^2
Now we can calculate the increase in water level:
Increase in water level = 3053.76 cm^3 / 176.715 cm^2
Calculating the answer:
Increase in water level ≈ 17.28 cm
Therefore, the increase in the level of water when the iron spherical ball is dropped into the bucket is approximately 17.28 cm.
To find the increase in the level of water when the iron spherical ball is dropped into the cylindrical bucket, we need to calculate the volume of the ball and then use the principle of displacement of water.
First, let's calculate the volume of the iron spherical ball using the formula for the volume of a sphere:
Volume of a sphere = (4/3)πr³,
where r is the radius of the sphere.
Given that the radius of the iron ball is 9 cm, the volume of the ball can be calculated as follows:
Volume of the ball = (4/3) × π × 9³ = (4/3) × 3.14 × 729 ≈ 3053.86 cm³.
Since the ball completely submerges in the water, it displaces an equivalent volume of water. This displaced volume of water will cause the increase in the level of water in the cylindrical bucket. Therefore, the increase in the level of water can be calculated as:
Increase in level of water = Volume of the ball / Base area of the bucket.
Given that the base of the bucket has a diameter of 15 cm, the base radius (r) can be calculated as half the diameter:
r = 15 cm / 2 = 7.5 cm.
The base area (A) of the bucket can be calculated using the formula:
Base area of cylinder = πr²,
where r is the base radius.
Base area of the bucket = π × (7.5)² ≈ 176.71 cm².
Using the calculated values, the increase in the level of water can be determined as follows:
Increase in level of water = 3053.86 cm³ / 176.71 cm² ≈ 17.29 cm.
Therefore, the increase in the level of water when the iron spherical ball is dropped into the cylindrical bucket is approximately 17.29 cm.
volume of sphere = (4/3)πr^3
= (4/3)π(9^3)
= 972π cm^3
Increase in water volume = 972π
radius of cylinder = 7.5
so π(7.5)^2 h = 972π
56.25h = 972
h = 17.28 cm
So the water will rise 17.28 cm
check my arithmetic