Cups A and B are cone shaped and have heights of #33 cm# and #26 cm# and openings with radii of #14 cm# and #7 cm#, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

Question

Cups A and B are cone shaped and have heights of #33 cm# and #26 cm# and openings with radii of #14 cm# and #7 cm#, respectively.  If cup B is full and its contents are poured into cup A, will cup A overflow?  If not how high will cup A be filled?

Ella Armstrong · Accepted Answer

Solve for the volume of B and use that to help solve for the height of A to get that the water will rise to a height of #6 1/2#cm

Let's start with the equation for the Volume of a cone: #V=1/3pir^2h# We're being asked to determine if the volume of cone B is greater than cone A (will it overflow from the contents of cone B). Just looking at the measurements of the 2 cones, with the height of cone A and it's radius being bigger than cone B, it's pretty clear the volume of cone B is smaller than cone A. So the next part of the question asks how high the water will come up in cone A. So let's first determine the volume of cone B: #V=1/3pi7^2(26)# #V=1/3pi(49)(26)# #V=(49*26)/3pi# - I'm going to leave it in this form for now as we work with cone A. So let's first prove definitively that the volume of cone A is greater than cone B: #V=1/3pi(14)^2(33)# #V=1/3pi(196)(33)# #V=(196*33)/3pi# Again, we can see that cone A has the greater volume: the bigger term of A (196) is greater than the bigger term of B (49), as is the smaller term #(33>26)#, so cone B won't overflow cone A. So how high up will the fluid come up? Let's solve cone A for height with the volume of cone B: #V=1/3pir^2h# #(49*26)/3pi=1/3pi14^2h# #h=(49*26)/3pi*3/(pi14^2)=13/2# So the water will rise to a height of #6 1/2#cm

Savannah Aguilar · Answer

undefined Yes, cup A will overflow when the contents of cup B are poured into it. The volume of cup B's contents can be calculated using the formula for the volume of a cone:

$$ V = \frac{1}{3} \pi r^2 h $$

Substituting the values for cup B:

$$ V_B = \frac{1}{3} \pi (7^2) (26) $$
$$ V_B = \frac{1}{3} \pi (49) (26) $$
$$ V_B = \frac{1}{3} \pi (1274) $$

The volume of cup B's contents is $ \frac{1274}{3} \pi $ cubic cm.

Now, to determine if cup A will overflow, we need to find the volume it can hold. Using the same formula for the volume of a cone:

$$ V_A = \frac{1}{3} \pi (14^2) (33) $$
$$ V_A = \frac{1}{3} \pi (196) (33) $$
$$ V_A = \frac{1}{3} \pi (6468) $$

The volume of cup A is $ \frac{6468}{3} \pi $ cubic cm.

Comparing the volumes, $ \frac{1274}{3} \pi $ cubic cm from cup B is less than $ \frac{6468}{3} \pi $ cubic cm from cup A. Therefore, when cup B's contents are poured into cup A, cup A will not overflow. Instead, it will be filled to its maximum capacity. The remaining volume in cup A can be calculated by subtracting $ \frac{1274}{3} \pi $ from $ \frac{6468}{3} \pi $:

$$ \frac{6468}{3} \pi - \frac{1274}{3} \pi = \frac{5194}{3} \pi $$

So, cup A will be filled to a height corresponding to $ \frac{5194}{3} \pi $ cubic cm above its base.