Problems Related to Different Solid Objects

When a solid metallic object of one shape is melted down and completely recast into one or more objects of another shape (assuming no loss of metal during the melting process), the total volume remains unchanged:

If a large solid of volume $V_{ ext{large}}$ is melted to form $n$ identical smaller solids, each of volume $V_{ ext{small}}$:

A very common board problem involves melting a sphere or cuboid and drawing it into a thin cylindrical wire of uniform cross-section:

• A wire of circular thickness is geometrically a Right Circular Cylinder.
• Let the radius of the wire be $r_{ ext{wire}}$ and its length (height of cylinder) be $L$.
• Volume of the wire: $\mathbf{V = \pi r_{ ext{wire}}^2 L}$.
• Equating volumes gives the length of the wire: $\mathbf{L = rac{V_{ ext{original}}}{\pi r_{ ext{wire}}^2}}$.
• Unit Alert: Wire diameter is often given in millimeters ($ ext{mm}$) while original solid dimensions are in centimeters ($ ext{cm}$) or meters ($ ext{m}$). Always convert all dimensions to the same unit before calculating!

In composite solids, two or more standard 3D shapes share a common interface. The total volume is simply the sum of the constituent volumes:

When finding the total surface area of a composite solid, do NOT simply add the total surface areas of the individual components!

• Toy (Cone atop Hemisphere): Exposed surface consists of the curved surface of the cone plus the curved surface of the hemisphere. The flat circular base of area $\pi r^2$ is glued shut and disappears! $$\mathbf{ ext{TSA}_{ ext{toy}} = ext{CSA}_{ ext{cone}} + ext{CSA}_{ ext{hemisphere}} = \pi r l + 2\pi r^2 = \pi r(l + 2r)}$$
• Circus Tent (Cylinder surmounted by Cone): Canvas is required only for the curved wall of the cylinder and the sloping roof of the cone (the tent has no canvas on the floor and the roof-ceiling interface is open air!): $$\mathbf{ ext{Canvas Area} = ext{CSA}_{ ext{cyl}} + ext{CSA}_{ ext{cone}} = 2\pi r H + \pi r l = \pi r(2H + l)}$$
• Capsule with Hemispherical Ends: $$\mathbf{ ext{TSA} = ext{CSA}_{ ext{cyl}} + 2 imes ext{CSA}_{ ext{hemisphere}} = 2\pi r H + 2(2\pi r^2) = 2\pi r H + 4\pi r^2 = 2\pi r(H + 2r)}$$ Notice that $4\pi r^2$ is the surface area of a complete sphere!

By Archimedes' Principle, when a solid object is completely immersed in a liquid inside a vessel, the volume of liquid displaced equals the volume of the immersed solid object:

When a solid is dropped into a cylindrical jar of base radius $r_{ ext{jar}}$, the displaced water forms an added cylindrical column of water of height $h_{ ext{rise}}$:

If $N$ identical solid balls (such as lead spheres of radius $r_{ ext{sphere}}$) are immersed: $$\pi r_{ ext{jar}}^2 imes h_{ ext{rise}} = N imes \left(rac{4}{3}\pi r_{ ext{sphere}}^3 ight)$$ $$\mathbf{h_{ ext{rise}} = rac{4 N r_{ ext{sphere}}^3}{3 r_{ ext{jar}}^2}}$$ Notice that $\pi$ cancels out completely from both sides!

In problems involving hollow vessels (such as a hollow cylindrical pipe or a hollow hemispherical bowl), differentiate between:

• Internal Capacity (অভ্যন্তরীণ ধারণক্ষমতা): Depends strictly on internal radius $r_1$: $V_{ ext{capacity}} = \pi r_1^2 h$.
• Volume of Material / Metal (ধাতুর আয়তন): Difference between external volume and internal volume: $$V_{ ext{metal}} = \pi (r_2^2 - r_1^2) h$$
• Mass of the Object: $\mathbf{ ext{Mass} = ext{Volume of Metal} imes ext{Density}}$.

Board exam questions frequently ask for the cost of painting, polishing, or tin-plating. Always identify whether the cost is applied to Surface Area (rate per $ ext{m}^2$ or $ ext{cm}^2$) or to Volume (rate per $ ext{m}^3$ or $ ext{cm}^3$):