Cylinder, Cone and Sphere (Surface Area and Volume) — ICSE Class 10 Maths Important Questions

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

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

Answer: (b) $3\pi r^2$3π r^2.

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

Answer: (b) $3:1$3:1.

$V_{\text{cyl}}=\pi r^2h$V_cyl=π r^2h and $V_{\text{cone}}=\tfrac13\pi r^2h$V_cone=13π r^2h, so the ratio is $\pi r^2h:\tfrac13\pi r^2h=3:1$π r^2h:13π r^2h=3:1.

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

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

Answer: (a) Both A and R are true and R is the correct explanation of A.

The two hemispheres together have total surface area $2\times3\pi r^2=6\pi r^2$2×3π r^2=6π r^2, while the sphere had $4\pi r^2$4π r^2. The increase is $6\pi r^2-4\pi r^2=2\pi r^2$6π r^2-4π r^2=2π r^2 (the two new flat faces), which follows from R.

$V=\pi r^2h=\frac{22}{7}\times7^2\times10=\frac{22}{7}\times49\times10.$V=π r^2h=22/7×7^2×10=22/7×49×10.
$V=22\times7\times10=1540\,\text{cm}^3.$V=22×7×10=1540\,cm^3.

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

Curved surface area:
$\pi rl=3.14\times5\times13=204.1\,\text{cm}^2.$π rl=3.14×5×13=204.1\,cm^2.

Volume is conserved: number of cones $=\dfrac{\text{volume of sphere}}{\text{volume of one cone}}$=volume of sphere/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.$4/3π r^3=4/3π(10.5)^3=4/3π×1157.625=1543.5\,π 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.$1/3π R^2 h=1/3π(3.5)^2(3)=1/3π×12.25×3=12.25\,π cm^3.

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

Hence $126$126 cones are formed.

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

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

The total surface $=$= curved surface of cone $+$+ curved surface of hemisphere (the flat faces join and are not exposed):
$=\pi rl+2\pi r^2=\pi r(l+2r)=\frac{22}{7}\times3.5\times(12.5+7).$=π rl+2π r^2=π r(l+2r)=22/7×3.5×(12.5+7).
$=\frac{22}{7}\times3.5\times19.5=11\times19.5=214.5\,\text{cm}^2.$=22/7×3.5×19.5=11×19.5=214.5\,cm^2.

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

Total surface $=$= curved surface of cylinder $+$+ curved surfaces of two hemispheres (which together make one sphere):
$=2\pi rh+4\pi r^2=2\pi r(h+2r).$=2π rh+4π r^2=2π r(h+2r).
$=2\times\frac{22}{7}\times7\times(90+14)=2\times22\times104=4576\,\text{cm}^2.$=2×22/7×7×(90+14)=2×22×104=4576\,cm^2.

External radius $R=4\,\text{cm}$R=4\,cm, internal radius $r=4-1=3\,\text{cm}$r=4-1=3\,cm, length $h=21\,\text{cm}$h=21\,cm.

(a) Volume of metal:
$\pi(R^2-r^2)h=\frac{22}{7}\times(16-9)\times21=\frac{22}{7}\times7\times21=462\,\text{cm}^3.$π(R^2-r^2)h=22/7×(16-9)×21=22/7×7×21=462\,cm^3.

(b) Total surface area $=$= outer curved surface $+$+ inner curved surface $+$+ the two ring-shaped ends:
$2\pi Rh+2\pi rh+2\pi(R^2-r^2)=2\pi(R+r)h+2\pi(R^2-r^2).$2π Rh+2π rh+2π(R^2-r^2)=2π(R+r)h+2π(R^2-r^2).
$=2\times\frac{22}{7}\times7\times21+2\times\frac{22}{7}\times7=924+44=968\,\text{cm}^2.$=2×22/7×7×21+2×22/7×7=924+44=968\,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.$4/3π(R^3-r^3)=4/3π(5^3-3^3)=4/3π(125-27)=4/3π×98=392/3π cm^3.

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

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

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

Let the depth of rainfall be $d$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 $=1\,\text{m}$=1\,m, height $3.5\,\text{m}$3.5\,m):
$\pi r^2h=\frac{22}{7}\times1^2\times3.5=\frac{22}{7}\times3.5=11\,\text{m}^3.$π r^2h=22/7×1^2×3.5=22/7×3.5=11\,m^3.

Volume of rain on the roof:
$22\times20\times d=440\,d\ \text{m}^3.$22×20× d=440\,d m^3.

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

Convert to centimetres:
$d=0.025\times100=2.5\,\text{cm}.$d=0.025×100=2.5\,cm.

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

(i) Slant height of cone:
$l=\sqrt{r^2+h_c^2}=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25\,\text{m}.$l=√r^2+h_c^2=√7^2+24^2=√49+576=√625=25\,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.$2π rh+π rl=22/7×7\,(2×4+25)=22×(8+25)=22×33=726\,m^2.

(iii) Cost $=726\times50=\text{Rs }36{,}300.$=726×50=Rs 36,300.

(iv) Volume of air $=$= volume of cylinder $+$+ volume of cone:
$\pi r^2h_{\text{cyl}}+\frac{1}{3}\pi r^2h_c=\frac{22}{7}\times49\left(4+\frac{24}{3}\right)=22\times7\times(4+8)=154\times12=1848\,\text{m}^3.$π r^2h_cyl+1/3π r^2h_c=22/7×49(4+24/3)=22×7×(4+8)=154×12=1848\,m^3.