Mensuration - Conversion of solids from one shape to another

A metallic sphere of radius $4.2$ cm is melted and recast into the shape of a cylinder of radius $6$ cm. Find the height of the cylinder.

1. Let $r = 4.2$ cm be the radius of the sphere and $R = 6$ cm be the radius of the cylinder. Let $h$ be the height of the cylinder. 2. Since the sphere is melted and recast, $Volume \\text{ of sphere} = Volume \\text{ of cylinder}$. 3. Formula: $\\frac{4}{3} \\pi r^3 = \\pi R^2 h$. 4. Substitute values and cancel $\\pi$: $\\frac{4}{3} \\times (4.2)^3 = 6^2 \\times h$. 5. $\\frac{4 \\times 4.2 \\times 4.2 \\times 4.2}{3} = 36h$. 6. $4 \\times 1.4 \\times 17.64 = 36h \\Rightarrow 5.6 \\times 17.64 = 36h$. 7. $98.784 = 36h \\Rightarrow h = \\frac{98.784}{36} = 2.744$ cm.

The principle used is the conservation of volume. By equating the volume of the original sphere to the volume of the new cylinder, we can solve for the unknown height $h$ of the cylinder.

How many silver coins, $1.75$ cm in diameter and of thickness $2$ mm, must be melted to form a cuboid of dimensions $5.5$ cm $\\times 10$ cm $\\times 3.5$ cm?

1. Cuboid dimensions: $l = 5.5$ cm, $b = 10$ cm, $h = 3.5$ cm. $Volume = 5.5 \\times 10 \\times 3.5 = 192.5 \\text{ cm}^3$. 2. Coin dimensions (Cylinder): Radius $r = \\frac{1.75}{2} = 0.875$ cm, Thickness $h' = 2 \\text{ mm} = 0.2$ cm. 3. Volume of one coin: $V = \\pi r^2 h' = \\frac{22}{7} \\times (0.875)^2 \\times 0.2$. 4. Let $n$ be the number of coins. $n \\times Volume \\text{ of one coin} = Volume \\text{ of cuboid}$. 5. $n \\times \\frac{22}{7} \\times \\frac{7}{8} \\times \\frac{7}{8} \\times \\frac{1}{5} = 192.5$. 6. $n \\times \\frac{11 \\times 7}{32 \\times 5} = 192.5 \\Rightarrow n \\times \\frac{77}{160} = 192.5$. 7. $n = \\frac{192.5 \\times 160}{77} = 2.5 \\times 160 = 400$.

To find the number of coins, we divide the total volume of the resulting cuboid by the volume of a single cylindrical coin. All units were converted to centimeters before solving.

$Volume \text{ of Cone} = \frac{1}{3} \pi r_c^2 h_c$ $Volume \text{ of Cone} = \frac{1}{3} \times \pi \times 6^2 \times 24$ $Volume \text{ of Cone} = 288\pi \text{ cm}^3$ Let the radius of the sphere be $R$. $Volume \text{ of Sphere} = \frac{4}{3} \pi R^3$ Since the cone is recast into a sphere: $\frac{4}{3} \pi R^3 = 288\pi$ $R^3 = \frac{288 \times 3}{4}$ $R^3 = 72 \times 3 = 216$ $R = \sqrt[3]{216} = 6 \text{ cm}$

The volume of the original cone must equal the volume of the new sphere. By equating the two volume formulae, we can cancel out $\pi$ and solve for the unknown radius $R$.

Convert all units to cm: Length of wire $L = 18 \text{ m} = 1800 \text{ cm}$. Radius of rod $r = \frac{1}{2} = 0.5 \text{ cm}$. $Volume \text{ of Rod} = \pi r^2 h = \pi \times (0.5)^2 \times 8 = 2\pi \text{ cm}^3$ Let the radius of the wire be $R$. $Volume \text{ of Wire} = \pi R^2 L = \pi \times R^2 \times 1800$ Since volume remains the same: $1800\pi R^2 = 2\pi$ $R^2 = \frac{2}{1800} = \frac{1}{900}$ $R = \frac{1}{30} \text{ cm}$ $Thickness (Diameter) = 2R = 2 \times \frac{1}{30} = \frac{1}{15} \text{ cm} \approx 0.067 \text{ cm}$

A wire is simply a very long and thin cylinder. The volume of the copper remains constant as it is stretched from a rod into a wire. Ensure all units are consistent (cm) before calculating.