This problem involves finding the volume of a cylindrical can given its diameter. The solid is a right circular cylinder with diameter 10 cm and height 12 cm. Since the volume formula V = πr²h requires radius, we must first convert: radius = diameter/2 = 10/2 = 5 cm. Applying the formula: V = π(5)²(12) = π(25)(12) = 300π cubic centimeters. This represents the can's total capacity. A common mistake is using diameter directly in the formula, giving π(10)²(12) = 1200π, which is four times too large. Always convert diameter to radius before applying cylinder volume formulas.

This problem requires finding the volume of a concrete pillar. The solid is a cylinder with diameter 12 ft and height 5 ft. The volume formula for a cylinder is V = πr²h, but we must first convert diameter to radius: r = 12/2 = 6 ft. Applying the formula: V = π(6)²(5) = π(36)(5) = 180π ft³. The volume represents the amount of concrete needed to form the pillar. A common mistake is using diameter directly in the formula instead of radius, which would give π(12)²(5) = 720π ft³. Always convert diameter to radius by dividing by 2 before applying cylinder volume formulas.

Solving problems with volume formulas involves calculating the space occupied by three-dimensional solids using appropriate mathematical expressions. The solid in this problem is a right circular cone. The correct volume formula for a cone is V = (1/3)πr²h, where r is the radius and h is the height. Applying the formula with r = 4 cm and h = 12 cm gives the expression (1/3)π(4)²(12), matching choice B. This expression correctly computes the volume by accounting for the conical shape's one-third factor of a cylinder's volume. A common distractor misconception is using the cylinder volume formula without the one-third, as in choice A, which overestimates the volume. To transfer this strategy, always identify the solid as a cone before selecting and applying the volume formula.

This problem asks which calculation correctly applies the volume formula for a sphere. The solid is a sphere with radius 6 cm. The volume formula for a sphere is V = (4/3)πr³, where r is the radius. The correct calculation is V = (4/3)π(6³) = (4/3)π(216). This gives the volume in cubic centimeters. Option C represents the formula for a cylinder (πr²h), which is incorrect for a sphere. To solve volume problems accurately, first identify the three-dimensional shape before selecting the appropriate formula.

This problem asks which expression represents the volume of a cylindrical candle. The solid is a cylinder with radius 5 cm and height 12 cm. The volume formula for a cylinder is V = πr²h, where r is the radius and h is the height. The correct expression is π(5²)(12) or π(25)(12). This formula gives the volume in cubic centimeters. Option A represents the lateral surface area formula (2πrh), not volume. To find volume, always use formulas that multiply three dimensions or include squared terms for circular shapes.

The volume of a cylinder is $$V_{cylinder} = \pi r^2 h$$ and the volume of a cone is $$V_{cone} = \frac{1}{3}\pi r^2 h$$. Since they have the same base radius and height, $$V_{cone} = \frac{1}{3} V_{cylinder} = \frac{1}{3} \times 432\pi = 144\pi$$ cubic centimeters. Choice A ($$72\pi$$) represents $$\frac{1}{6}$$ of the cylinder volume. Choice C ($$216\pi$$) represents $$\frac{1}{2}$$ of the cylinder volume. Choice D ($$324\pi$$) represents $$\frac{3}{4}$$ of the cylinder volume.

This problem involves finding the volume of a party cone filled with candy. The solid is a cone with radius 6 cm and height 10 cm. The volume formula for a cone is V = (1/3)πr²h, where r is the radius and h is the height. The correct calculation is V = (1/3)π(6)²(10), which matches option B. This formula gives one-third the volume of a cylinder with the same base and height. Option A incorrectly uses the cylinder formula without the 1/3 factor, while option D uses the surface area formula. To solve volume problems correctly, first identify whether the solid is a cone, cylinder, or sphere before selecting the appropriate formula.

This problem asks for the volume of a spherical balloon. The solid is a sphere with radius 7 inches. The volume formula for a sphere is V = (4/3)πr³, where r is the radius. Applying the formula: V = (4/3)π(7)³ = (4/3)π(343) in³. The correct answer includes the proper cubic units (in³) for volume. Option B incorrectly uses πr², which is the area of a circle, not the volume of a sphere, while option D has the wrong units (in² instead of in³). When calculating sphere volume, remember to cube the radius and multiply by (4/3)π, not just π.

The volume of water displaced equals the rise in water level times the base area of the cylinder. The water level rises from 15 cm to 18 cm, a rise of 3 cm. Volume displaced = $$\pi(5)^2(3) = 75\pi$$ cubic cm. This equals the volume of the sphere: $$\frac{4}{3}\pi r^3 = 75\pi$$. Solving: $$\frac{4}{3}r^3 = 75$$, so $$r^3 = \frac{75 \times 3}{4} = \frac{225}{4}$$, giving $$r = \sqrt[3]{\frac{225}{4}}$$ cm. Choice A omits the $$\pi$$ cancellation step. Choice B results from incorrectly setting $$\frac{4}{3}\pi r^3 = 75$$ instead of $$75\pi$$. Choice D uses an incorrect volume calculation.

This problem asks for the volume expression of a cone-shaped ice cream cone. The solid is a right circular cone with radius 4 cm and height 9 cm. The volume formula for a cone is V = (1/3)πr²h, where r is the radius and h is the height. The correct expression is (1/3)π(4²)(9), which represents one-third of the volume of a cylinder with the same base and height. This formula accounts for the cone's tapering shape from base to apex. A common mistake is using the cylinder formula π(4²)(9), forgetting the factor of 1/3. Remember that a cone's volume is always one-third of a cylinder with the same dimensions.