Mensuration - Area and volume of Cylinder, Cone and Sphere

A metallic sphere of radius $10.5\\text{ cm}$ is melted and then recast into small cones, each of radius $3.5\\text{ cm}$ and height $3\\text{ cm}$. Find the number of cones formed.

1. Volume of the sphere = $\\frac{4}{3} \\pi R^3 = \\frac{4}{3} \\times \\pi \\times (10.5)^3$\
2. Volume of one small cone = $\\frac{1}{3} \\pi r^2 h = \\frac{1}{3} \\times \\pi \\times (3.5)^2 \\times 3 = \\pi \\times (3.5)^2$\
3. Let the number of cones be $n$. By the principle of conservation of volume: $n \\times \\text{Volume of one cone} = \\text{Volume of sphere}$\
4. $n \\times (\\pi \\times 3.5 \\times 3.5) = \\frac{4}{3} \\times \\pi \\times 10.5 \\times 10.5 \\times 10.5$\
5. $n = \\frac{4 \\times 10.5 \\times 10.5 \\times 10.5}{3 \\times 3.5 \\times 3.5}$\
6. $n = 4 \\times 3 \\times 3 \\times 3.5 = 126$

When a solid is melted and recast, the total volume remains the same. We calculate the volume of the large sphere and divide it by the volume of a single small cone to find the total number of cones.

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is $19\\text{ cm}$ and the diameter of the cylinder is $7\\text{ cm}$. Find the total surface area and the volume of the solid.

1. Radius of cylinder and hemispheres $r = \\frac{7}{2} = 3.5\\text{ cm}$\
2. Height of the cylinder $H = \\text{Total height} - 2 \\times r = 19 - (3.5 + 3.5) = 12\\text{ cm}$\
3. Volume = Volume of cylinder + $2 \\times$ Volume of hemisphere
   $V = \\pi r^2 H + 2 \\times \\frac{2}{3}\\pi r^3 = \\pi r^2 (H + \\frac{4}{3}r)$
   $V = \\frac{22}{7} \\times (3.5)^2 \\times (12 + \\frac{4}{3} \\times 3.5) = 38.5 \\times (12 + 4.67) \\approx 641.67\\text{ cm}^3$\
4. Total Surface Area (TSA) = CSA of cylinder + $2 \\times$ CSA of hemisphere
   $TSA = 2\\pi rH + 4\\pi r^2 = 2\\pi r(H + 2r)$
   $TSA = 2 \\times \\frac{22}{7} \\times 3.5 \\times (12 + 7) = 22 \\times 19 = 418\\text{ cm}^2$

For composite solids, the height of the central cylinder is found by subtracting the radii of the two hemispherical ends from the total height. The surface area includes only the curved parts because the flat circular faces are joined together internally.

1. Volume of sand in the cylindrical bucket: $V_1 = \pi r_1^2 h_1 = \pi \times (18)^2 \times 32$
2. Volume of the conical heap: $V_2 = \frac{1}{3} \pi r_2^2 h_2 = \frac{1}{3} \pi \times r_2^2 \times 24 = 8 \pi r_2^2$
3. Equating the volumes ($V_1 = V_2$): $\pi \times 18 \times 18 \times 32 = 8 \pi r_2^2$ $r_2^2 = \frac{18 \times 18 \times 32}{8} = 18 \times 18 \times 4$ $r_2 = \sqrt{18^2 \times 2^2} = 18 \times 2 = 36\text{ cm}$
4. Slant height $l$ of the heap: $l = \sqrt{r_2^2 + h_2^2} = \sqrt{36^2 + 24^2}$ $l = \sqrt{1296 + 576} = \sqrt{1872} = 12\sqrt{13}\text{ cm} \approx 43.27\text{ cm}$

The volume of sand remains the same when transferred from the cylinder to the cone. We solve for the unknown radius $r$ using the volume equality and then use the Pythagorean theorem for slant height.

1. Dimensions: Radius of hemisphere and cone $r = \frac{4}{2} = 2\text{ cm}$ Height of cone $h = 2\text{ cm}$
2. Volume of the toy = Volume of cone + Volume of hemisphere $V = \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3$ $V = \frac{1}{3} \pi (2)^2 (2) + \frac{2}{3} \pi (2)^3$ $V = \frac{8\pi}{3} + \frac{16\pi}{3} = \frac{24\pi}{3} = 8\pi$
3. Taking $\pi = 3.14$: $V = 8 \times 3.14 = 25.12\text{ cm}^3$

The toy is a composite solid. Its total volume is the sum of the volumes of its individual geometric components (cone and hemisphere) sharing the same base radius.