Solids (Surface Area and Volume of 3-D Solids) — ICSE Class 9 Maths Important Questions

Answer: (c) $125\text{ cm}^3$125 cm^3.

Volume of a cube $=a^3=5^3=125\text{ cm}^3$=a^3=5^3=125 cm^3.

Answer: (b) $158\text{ cm}^2$158 cm^2.

$2(lb+bh+hl)=2(8\times5+5\times3+3\times8)=2(40+15+24)=2(79)=158\text{ cm}^2$2(lb+bh+hl)=2(8×5+5×3+3×8)=2(40+15+24)=2(79)=158 cm^2.

Answer: (b) $440\text{ cm}^2$440 cm^2.

CSA $=2\pi rh=2\times\dfrac{22}{7}\times7\times10=440\text{ cm}^2$=2π rh=2×22/7×7×10=440 cm^2.

Answer: (d) $8$8 times.

New volume $=(2a)^3=8a^3$=(2a)^3=8a^3, i.e. $8$8 times the original volume $a^3$a^3.

Answer: (a) Melting only changes shape, not the amount of material, so volume is conserved and the two volumes are equal. R correctly explains A.

Volume $=lbh=2\times1.5\times1=3\text{ m}^3$=lbh=2×1.5×1=3 m^3.

Since $1\text{ m}^3=1000\text{ L}$1 m^3=1000 L, capacity $=3\times1000=3000\text{ L}$=3×1000=3000 L.

The longest rod fits along the space diagonal of the cuboidal room.

Diagonal $=\sqrt{l^2+b^2+h^2}=\sqrt{12^2+9^2+8^2}=\sqrt{144+81+64}=\sqrt{289}=17\text{ m}$=√l^2+b^2+h^2=√12^2+9^2+8^2=√144+81+64=√289=17 m.

Given $r=7\text{ cm}$r=7 cm, $h=20\text{ cm}$h=20 cm, $\pi=\dfrac{22}{7}$π=22/7.

(i) Total surface area $=2\pi r(r+h)=2\times\dfrac{22}{7}\times7\times(7+20)$=2π r(r+h)=2×22/7×7×(7+20)

$=2\times22\times27=1188\text{ cm}^2$=2×22×27=1188 cm^2.

(ii) Volume $=\pi r^2h=\dfrac{22}{7}\times7^2\times20=\dfrac{22}{7}\times49\times20$=π r^2h=22/7×7^2×20=22/7×49×20

$=22\times7\times20=3080\text{ cm}^3$=22×7×20=3080 cm^3.

Volume $=a^3=343\Rightarrow a=\sqrt[3]{343}=7\text{ cm}$=a^3=343 a=[3]343=7 cm.

Total surface area $=6a^2=6\times7^2=6\times49=294\text{ cm}^2$=6a^2=6×7^2=6×49=294 cm^2.

Diagonal $=a\sqrt3=7\sqrt3\text{ cm}\approx12.12\text{ cm}$=a3=73 cm12.12 cm.

For a hollow cylinder, volume of metal $=\pi h(R^2-r^2)$=π h(R^2-r^2) with $R=5\text{ cm}$R=5 cm, $r=4\text{ cm}$r=4 cm, $h=28\text{ cm}$h=28 cm.

$R^2-r^2=5^2-4^2=25-16=9$R^2-r^2=5^2-4^2=25-16=9.

Volume $=\dfrac{22}{7}\times28\times9=22\times4\times9=792\text{ cm}^3$=22/7×28×9=22×4×9=792 cm^3.

(i) Volume of water $=$= volume of cuboidal tank $=lbh=80\times40\times30=96000\text{ cm}^3$=lbh=80×40×30=96000 cm^3.

(ii) Let the water rise to height $h$h in the cylinder of radius $r=20\text{ cm}$r=20 cm. Since water is only transferred, its volume is unchanged:

$\pi r^2h=96000$π r^2h=96000

$\dfrac{22}{7}\times20^2\times h=96000$22/7×20^2× h=96000

$\dfrac{22}{7}\times400\times h=96000$22/7×400× h=96000

$h=\dfrac{96000\times7}{22\times400}=\dfrac{672000}{8800}=76.36\text{ cm (approx.)}$h=96000×7/22×400=672000/8800=76.36 cm (approx.).

The water rises to about $76.4\text{ cm}$76.4 cm.

The wood is $2\text{ cm}$2 cm thick on both faces along each dimension, so each internal dimension is $2\times2=4\text{ cm}$2×2=4 cm less than the external.

(i) Internal dimensions:

length $=30-4=26\text{ cm}$=30-4=26 cm, breadth $=25-4=21\text{ cm}$=25-4=21 cm, height $=20-4=16\text{ cm}$=20-4=16 cm.

(ii) Internal volume (capacity) $=26\times21\times16=8736\text{ cm}^3$=26×21×16=8736 cm^3.

(iii) External volume $=30\times25\times20=15000\text{ cm}^3$=30×25×20=15000 cm^3.

Volume of wood $=$= external volume $-$- internal volume $=15000-8736=6264\text{ cm}^3$=15000-8736=6264 cm^3.

(i) Radius $r=\dfrac{2.8}{2}=1.4\text{ m}$r=2.8/2=1.4 m.

(ii) Capacity $=\pi r^2h=\dfrac{22}{7}\times1.4^2\times2.5=\dfrac{22}{7}\times1.96\times2.5$=π r^2h=22/7×1.4^2×2.5=22/7×1.96×2.5

$=\dfrac{22}{7}\times4.9=22\times0.7=15.4\text{ m}^3$=22/7×4.9=22×0.7=15.4 m^3.

(iii) Since $1\text{ m}^3=1000\text{ L}$1 m^3=1000 L, capacity $=15.4\times1000=15400\text{ L}$=15.4×1000=15400 L.

(iv) Number of flats $=\dfrac{15400}{700}=22$=15400/700=22 flats.