800cm3 of water is poured into an inverted right conical vessel and the water level is 6cm. How much more water has to be added to the vessel such that the water level becomes 9cm?
The volume of two similar shapes is proportional to the cube of their corresponding sides
so ...
V/800 = 9^3 6^3
V = 800(729/216)
= ....
To solve this problem, we need to find the volume of the conical vessel and then determine how much more water needs to be added to change the water level from 6cm to 9cm.
The volume of a cone is given by the formula: V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height of the cone.
Given that the initial water level is 6cm, we need to find the initial height of the cone. Since the cone is inverted, the height of the water is the difference between the height of the vessel and the water level. Therefore, the initial height of the cone is 9cm - 6cm = 3cm.
Now, let's calculate the initial volume of the cone:
V1 = (1/3)πr^2h1
We are given that the initial volume, V1, is 800cm^3, and the initial height, h1, is 3cm. We can rearrange the formula to solve for the radius, r:
r = √(3V1 / (πh1))
Substituting the given values, we have:
r = √(3 * 800 / (π * 3)) ≈ √(1600 / π) ≈ √510 ≈ 22.63cm
Therefore, the radius of the base of the cone is approximately 22.63cm.
Now, we need to calculate the volume of the cone when the water level is 9cm. The height of the cone will be 9cm - 6cm = 3cm (since the difference in the water levels is 3cm).
V2 = (1/3)πr^2h2
We are trying to find the additional volume of water that needs to be added to the cone, which is the difference between the final volume, V2, and the initial volume, V1.
V_additional = V2 - V1
Substituting the given values, we have:
V_additional = ((1/3)πr^2h2) - V1
V_additional = ((1/3)π * (22.63)^2 * 3) - 800
V_additional ≈ 448.12cm^3
Therefore, approximately 448.12cm^3 of additional water needs to be added to the vessel to increase the water level from 6cm to 9cm.