Cylinder, Cone and Sphere (Surface Area and Volume) — Important Questions

Answer: (b) $\dfrac{4}{3}\pi r^3$.

The volume of a sphere is $\dfrac{4}{3}\pi r^3$; the option $\dfrac{2}{3}\pi r^3$ is the volume of a hemisphere.

Answer: (b) $3\pi r^2$.

Total surface $=$ curved surface $+$ flat circular base $=2\pi r^2+\pi r^2=3\pi r^2$.

Answer: (b) $3:1$.

$V_{\text{cyl}}=\pi r^2h$ and $V_{\text{cone}}=\tfrac13\pi r^2h$, so the ratio is $\pi r^2h:\tfrac13\pi r^2h=3:1$.

Answer: (b) $2.5\,\text{cm}$.

Volume is conserved, so $r^3=1.5^3+2^3+x^3$: $3^3=27$, $1.5^3=3.375$, $2^3=8$. Thus $x^3=27-3.375-8=15.625$, giving $x=\sqrt[3]{15.625}=2.5\,\text{cm}$.

Answer: (a) Recasting keeps the total volume unchanged (R), which is exactly why the combined volume of the cones equals the volume of the original sphere (A); so R correctly explains A.

$V=\pi r^2h=\frac{22}{7}\times7^2\times10=\frac{22}{7}\times49\times10.$
$V=22\times7\times10=1540\,\text{cm}^3.$

Slant height:
$l=\sqrt{r^2+h^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13\,\text{cm}.$

Curved surface area:
$\pi rl=3.14\times5\times13=204.1\,\text{cm}^2.$

Volume is conserved: number of cones $=\dfrac{\text{volume of sphere}}{\text{volume of one cone}}$.

Volume of sphere:
$\frac{4}{3}\pi r^3=\frac{4}{3}\pi(10.5)^3=\frac{4}{3}\pi\times1157.625=1543.5\,\pi\ \text{cm}^3.$

Volume of one cone:
$\frac{1}{3}\pi R^2 h=\frac{1}{3}\pi(3.5)^2(3)=\frac{1}{3}\pi\times12.25\times3=12.25\,\pi\ \text{cm}^3.$

Number of cones:
$\frac{1543.5\,\pi}{12.25\,\pi}=126.$

Hence $126$ cones are formed.

Radius $r=3.5\,\text{cm}$; cone height $h=12\,\text{cm}$.

Slant height of cone:
$l=\sqrt{r^2+h^2}=\sqrt{3.5^2+12^2}=\sqrt{12.25+144}=\sqrt{156.25}=12.5\,\text{cm}.$

The total surface $=$ curved surface of cone $+$ curved surface of hemisphere (the flat faces join and are not exposed):
$=\pi rl+2\pi r^2=\pi r(l+2r)=\frac{22}{7}\times3.5\times(12.5+7).$
$=\frac{22}{7}\times3.5\times19.5=11\times19.5=214.5\,\text{cm}^2.$

Radius $r=7\,\text{cm}$. Length of the cylindrical part:
$h=104-2r=104-14=90\,\text{cm}.$

Total surface $=$ curved surface of cylinder $+$ curved surfaces of two hemispheres (which together make one sphere):
$=2\pi rh+4\pi r^2=2\pi r(h+2r).$
$=2\times\frac{22}{7}\times7\times(90+14)=2\times22\times104=4576\,\text{cm}^2.$

Volume of the hollow shell $=$ volume of the recast cone (volume conserved).

Volume of shell:
$\frac{4}{3}\pi(R^3-r^3)=\frac{4}{3}\pi(5^3-3^3)=\frac{4}{3}\pi(125-27)=\frac{4}{3}\pi\times98=\frac{392}{3}\pi\ \text{cm}^3.$

Volume of cone with base radius $7\,\text{cm}$ and height $h$:
$\frac{1}{3}\pi(7)^2 h=\frac{49}{3}\pi h.$

Equating the volumes:
$\frac{49}{3}\pi h=\frac{392}{3}\pi\ \Rightarrow\ 49h=392\ \Rightarrow\ h=\frac{392}{49}=8\,\text{cm}.$

Hence the height of the cone is $8\,\text{cm}$.

Let the depth of rainfall be $d$ metres. The volume of rain on the roof equals the volume of water in the vessel.

Volume of water in the cylindrical vessel (radius $=1\,\text{m}$, height $3.5\,\text{m}$):
$\pi r^2h=\frac{22}{7}\times1^2\times3.5=\frac{22}{7}\times3.5=11\,\text{m}^3.$

Volume of rain on the roof:
$22\times20\times d=440\,d\ \text{m}^3.$

Equate:
$440\,d=11\ \Rightarrow\ d=\frac{11}{440}=0.025\,\text{m}.$

Convert to centimetres:
$d=0.025\times100=2.5\,\text{cm}.$

Hence the rainfall depth is $2.5\,\text{cm}$.

(i) Slant height of cone:
$l=\sqrt{r^2+h_c^2}=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25\,\text{m}.$

(ii) Canvas $=$ curved surface of cylinder $+$ curved surface of cone:
$2\pi rh+\pi rl=\frac{22}{7}\times7\,(2\times4+25)=22\times(8+25)=22\times33=726\,\text{m}^2.$

(iii) Cost $=726\times50=\text{Rs }36{,}300.$

(iv) Volume of air $=$ volume of cylinder $+$ volume of cone:
$\pi r^2h_{\text{cyl}}+\frac{1}{3}\pi r^2h_c=\frac{22}{7}\times49\left(4+\frac{24}{3}\right)=22\times7\times(4+8)=154\times12=1848\,\text{m}^3.$