1. A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water. If the radius of the cylindrical tub is 5 cm and its height is 10.5 cm, find the volume of water left in the cylindrical tub. Use pi = 22/7
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We have a composite solid made of a hemisphere with a cone mounted on top. Both the hemisphere and cone have a radius of 3.5 centimeters, and the cone has a height of 4 centimeters. This solid is placed in a cylindrical tub with radius 5 centimeters and height 10.5 centimeters. We need to find the volume of water remaining in the tub after the solid is completely submerged.
Let's start by calculating the volume of the hemisphere. The formula for the volume of a hemisphere is two-thirds pi r cubed. With radius 3.5 centimeters, which equals 7 over 2, we calculate r cubed as 343 over 8. Substituting into the formula gives us 343 pi over 12 cubic centimeters for the hemisphere volume.
Now let's calculate the cone volume using the formula one-third pi r squared h. With radius 7 over 2 and height 4, we get 49 pi over 3 cubic centimeters. The total volume of the solid is the sum of hemisphere and cone volumes: 343 pi over 12 plus 49 pi over 3. Converting to a common denominator of 12, this equals 539 pi over 12 cubic centimeters.
Next, we calculate the cylinder volume using pi R squared H. With radius 5 and height 10.5, we get 525 pi over 2 cubic centimeters. The volume of water left equals the cylinder volume minus the solid volume. This gives us 525 pi over 2 minus 539 pi over 12. Converting to a common denominator of 12, we get 2611 pi over 12 cubic centimeters.