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Question 1: If the radius of a sphere is increased by $10\%$, prove that its volume will increase by $33.1\%$.
Solution: Let original radius be $r$. New radius $R = 1.1r$. Original Volume $V = \frac{4}{3}\pi r^3$. New Volume $V' = \frac{4}{3}\pi (1.1r)^3 = 1.331 V$. Increase $= 0.331 V = 33.1\%$.
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Question 2: A semi-circular sheet of metal of diameter $28 \text{ cm}$ is bent to form an open conical cup. Find the capacity of the cup.
Solution: Radius of sheet $= 14 \text{ cm}$. This becomes slant height $l$ of cone. Circumference of semi-circle $\pi R = 14\pi$ becomes circumference of base of cone $2\pi r$. So $2\pi r = 14\pi \Rightarrow r = 7$. $h = \sqrt{14^2 - 7^2} = \sqrt{147} \approx 12.12$. $V = \frac{1}{3}\pi r^2 h \approx 622 \text{ cm}^3$.
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Question 3: A cone of height $24 \text{ cm}$ and radius of base $6 \text{ cm}$ is made up of modeling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Solution: Volume of Cone $= \frac{1}{3}\pi (6)^2 (24) = 288\pi$. Volume of Sphere $= \frac{4}{3}\pi r^3$. Equating: $\frac{4}{3}\pi r^3 = 288\pi \Rightarrow r^3 = 216 \Rightarrow r = 6 \text{ cm}$.
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Question 4: The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are $12 \text{ cm}$ each. Find the radius of the sphere.
Solution: Cylinder $h=12, d=12 \Rightarrow r=6$. CSA $= 2\pi(6)(12) = 144\pi$. Sphere $4\pi R^2 = 144\pi \Rightarrow R^2 = 36 \Rightarrow R = 6 \text{ cm}$.
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Question 5: A shot-put is a metallic sphere of radius $4.9 \text{ cm}$. If the density of the metal is $7.8 \text{ g/cm}^3$, find the mass of the shot-put.
Solution: Volume $= \frac{4}{3} \times \frac{22}{7} \times (4.9)^3 \approx 493 \text{ cm}^3$. Mass $= \text{Vol} \times \text{Density} = 493 \times 7.8 \approx 3845.44 \text{ g} \approx 3.85 \text{ kg}$.
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Question 6: A wall of length $10 \text{ m}$ was to be built across an open ground. The height of the wall is $4 \text{ m}$ and thickness of the wall is $24 \text{ cm}$. If this wall is to be built up with bricks whose dimensions are $24 \text{ cm} \times 12 \text{ cm} \times 8 \text{ cm}$, how many bricks would be required?
Solution: Wall Vol $= 1000 \times 24 \times 400 \text{ cm}^3$. Brick Vol $= 24 \times 12 \times 8$. Number $= (1000 \times 24 \times 400) / (24 \times 12 \times 8) = 4166.6 \approx 4167$ bricks.
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Question 7: The diameter of a roller is $84 \text{ cm}$ and its length is $120 \text{ cm}$. It takes $500$ complete revolutions to move once over to level a playground. Find the area of the playground in $\text{m}^2$.
Solution: $r=42 \text{ cm}, h=120 \text{ cm}$. CSA $= 2 \times \frac{22}{7} \times 42 \times 120 = 31680 \text{ cm}^2$. Area $= 500 \times 31680 = 15840000 \text{ cm}^2 = 1584 \text{ m}^2$.
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Question 8: A heap of wheat is in the form of a cone whose diameter is $10.5 \text{ m}$ and height is $3 \text{ m}$. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution: $r=5.25, h=3$. Vol $= \frac{1}{3}\pi (5.25)^2 (3) = 86.625 \text{ m}^3$. Slant height $l = \sqrt{3^2 + 5.25^2} \approx 6.05$. CSA $= \pi(5.25)(6.05) \approx 99.825 \text{ m}^2$.
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Question 9: A right triangle ABC with sides $5 \text{ cm}, 12 \text{ cm}$ and $13 \text{ cm}$ is revolved about the side $12 \text{ cm}$. Find the volume of the solid so obtained.
Solution: It forms a cone with $h=12 \text{ cm}$ and $r=5 \text{ cm}$. Volume $= \frac{1}{3}\pi (5)^2 (12) = 100\pi \text{ cm}^3$.
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Question 10: Twenty-seven solid iron spheres, each of radius $r$ and surface area $S$ are melted to form a sphere with surface area $S'$. Find the radius $r'$ of the new sphere and ratio of $S$ and $S'$.
Solution: Vol of 27 spheres $= 27 \times \frac{4}{3}\pi r^3$. New Vol $= \frac{4}{3}\pi (r')^3$. So $r' = 3r$. Ratio $S/S' = (4\pi r^2) / (4\pi (3r)^2) = 1/9$.