Each cube has side $\sqrt[3]{64}=4$ cm. Joining two cubes end to end forms a cuboid of dimensions $8\text{ cm}\times4\text{ cm}\times4\text{ cm}$. Surface area $=2(lb+bh+hl)=2(8\times4+4\times4+8\times4)=160$ cm$^2$.
The surface area of the resulting cuboid is $160$ cm$^2$.
Radius $r=7$ cm. The cylinder height is $13-7=6$ cm. Inner surface area $=$ CSA of cylinder $+$ CSA of hemisphere $=2\pi rh+2\pi r^2=2\times\dfrac{22}{7}\times7\times6+2\times\dfrac{22}{7}\times7^2=264+308=572$ cm$^2$.
The inner surface area of the vessel is $572$ cm$^2$.
The cone height is $15.5-3.5=12$ cm and its slant height is $l=\sqrt{12^2+3.5^2}=12.5$ cm. Total surface area $=$ CSA of cone $+$ CSA of hemisphere $=\pi rl+2\pi r^2=\dfrac{22}{7}\times3.5\times12.5+2\times\dfrac{22}{7}\times3.5^2=137.5+77=214.5$ cm$^2$.
The total surface area of the toy is $214.5$ cm$^2$.
The hemisphere must fit on the top face of the cube, so its greatest diameter is 7 cm and $r=3.5$ cm. Surface area $=$ TSA of cube $-$ circular base area $+$ CSA of hemisphere $=6\times7^2-\pi r^2+2\pi r^2=294+\dfrac{22}{7}\times3.5^2=332.5$ cm$^2$.
The greatest diameter is 7 cm, and the surface area of the solid is $332.5$ cm$^2$.
The cube has edge $l$, so its surface area is $6l^2$. A circular area of radius $l/2$ is removed from one face, but the curved surface of the hemispherical depression is added. Therefore area $=6l^2-\pi(l/2)^2+2\pi(l/2)^2=6l^2+\dfrac{\pi l^2}{4}=\left(6+\dfrac{\pi}{4}\right)l^2$.
The surface area of the remaining solid is $\left(6+\dfrac{\pi}{4}\right)l^2$ square units.
Radius $r=2.5$ mm. The two hemispheres together form a sphere, and the cylindrical part has height $14-5=9$ mm. Surface area $=2\pi rh+4\pi r^2=2\times\dfrac{22}{7}\times2.5\times9+4\times\dfrac{22}{7}\times2.5^2=220$ mm$^2$.
The surface area of the capsule is $220$ mm$^2$.
Radius $r=2$ m. Canvas area $=$ CSA of cylinder $+$ CSA of cone $=2\pi rh+\pi rl=2\times\dfrac{22}{7}\times2\times2.1+\dfrac{22}{7}\times2\times2.8=26.4+17.6=44$ m$^2$. Cost $=44\times500=₹22000$.
The canvas area is $44$ m$^2$, and the cost is $₹22000$.
Here $r=0.7$ cm and $h=2.4$ cm. The cone slant height is $l=\sqrt{0.7^2+2.4^2}=2.5$ cm. Total surface area $=$ outer CSA of cylinder $+$ base of cylinder $+$ inner CSA of cone $=2\pi rh+\pi r^2+\pi rl$. Using $\pi=\dfrac{22}{7}$, this is $10.56+1.54+5.5=17.6$ cm$^2$, which rounds to $18$ cm$^2$.
The total surface area of the remaining solid is approximately $18$ cm$^2$.
The exposed area is the curved surface of the cylinder plus the curved surfaces of two hemispherical hollows. Thus TSA $=2\pi rh+2(2\pi r^2)=2\times\dfrac{22}{7}\times3.5\times10+4\times\dfrac{22}{7}\times3.5^2=220+154=374$ cm$^2$.
The total surface area of the article is $374$ cm$^2$.
Volume $=$ volume of hemisphere $+$ volume of cone $=\dfrac{2}{3}\pi r^3+\dfrac{1}{3}\pi r^2h$. With $r=1$ cm and $h=1$ cm, volume $=\dfrac{2}{3}\pi+\dfrac{1}{3}\pi=\pi$ cm$^3$.
The volume of the solid is $\pi$ cm$^3$.
Radius $r=1.5$ cm. Since each cone has height 2 cm, the cylinder height is $12-2-2=8$ cm. Volume $=\pi r^2h+2\left(\dfrac13\pi r^2\times2\right)=\dfrac{22}{7}\times1.5^2\times8+2\times\dfrac13\times\dfrac{22}{7}\times1.5^2\times2=66$ cm$^3$.
The volume of air contained in the model is $66$ cm$^3$.
Radius $r=1.4$ cm. The two hemispherical ends make one sphere of diameter 2.8 cm, so the cylindrical length is $5-2.8=2.2$ cm. Volume of one gulab jamun $=\pi r^2h+\dfrac43\pi r^3=\dfrac{22}{7}\times1.4^2\times2.2+\dfrac43\times\dfrac{22}{7}\times1.4^3=25.0507$ cm$^3$ approximately. Syrup in 45 pieces $=45\times30\%\times25.0507=338.18$ cm$^3$ approximately.
Approximately $338.18$ cm$^3$ of syrup would be found in 45 gulab jamuns.
Volume of cuboid $=15\times10\times3.5=525$ cm$^3$. Volume of one conical depression $=\dfrac13\pi r^2h=\dfrac13\times\dfrac{22}{7}\times0.5^2\times1.4=0.3667$ cm$^3$. Four depressions remove $4\times0.3667=1.4667$ cm$^3$. Volume of wood $=525-1.4667=523.53$ cm$^3$ approximately.
The volume of wood in the pen stand is approximately $523.53$ cm$^3$.
Volume of the cone $=\dfrac13\pi r^2h=\dfrac13\pi\times5^2\times8=\dfrac{200\pi}{3}$ cm$^3$. One-fourth flows out, so displaced volume $=\dfrac{50\pi}{3}$ cm$^3$. Volume of one lead shot $=\dfrac43\pi(0.5)^3=\dfrac{\pi}{6}$ cm$^3$. Number of shots $=\dfrac{50\pi/3}{\pi/6}=100$.
The number of lead shots dropped is $100$.
The lower cylinder has radius 12 cm and height 220 cm; the upper cylinder has radius 8 cm and height 60 cm. Total volume $=3.14\times12^2\times220+3.14\times8^2\times60=111532.8$ cm$^3$. Mass $=111532.8\times8=892262.4$ g $=892.2624$ kg, approximately $892.26$ kg.
The mass of the pole is approximately $892.26$ kg.
Volume of cylinder $=\pi\times60^2\times180=648000\pi$ cm$^3$. Volume of cone $=\dfrac13\pi\times60^2\times120=144000\pi$ cm$^3$. Volume of hemisphere $=\dfrac23\pi\times60^3=144000\pi$ cm$^3$. Water left $=648000\pi-288000\pi=360000\pi=\dfrac{7920000}{7}$ cm$^3$.
The volume of water left is $\dfrac{7920000}{7}$ cm$^3$ or approximately $1131428.57$ cm$^3$.
For the spherical part, $r=4.25$ cm. Volume $=\dfrac43\times3.14\times4.25^3=321.39$ cm$^3$ approximately. The neck is a cylinder of radius 1 cm and height 8 cm, so its volume $=3.14\times1^2\times8=25.12$ cm$^3$. Total capacity $=321.39+25.12=346.51$ cm$^3$, which is close to $345$ cm$^3$.
The calculated capacity is approximately $346.51$ cm$^3$, so the child's measurement of $345$ cm$^3$ is approximately correct.