Mensuration 3D — Part 1 MCQs with Explanations

r : h = 3 : 4 $$l$$ : r = 5 : 3 $$\eqalign{ & {\text{Curved surface area}} = \pi rl \cr & 240\pi = \pi \times 3x \times 5x \cr & {x^2} = 16 \cr & x = 4 \cr & \therefore r = 3 \times 4 = 12 \cr & {\text{Volume}} = \frac{1}{3}\pi {r^2}h \cr & = \frac{1}{3}\pi \times 144 \times 16 \cr & = 768\pi \cr} $$

$$\eqalign{ & {\text{Required volume removed}} \cr & = \pi {r^2}h - \frac{1}{3}\pi {r^2}h \cr & = \frac{2}{3}\pi {r^2}h \cr & = \frac{2}{3} \times \frac{{22}}{7} \times .6 \times .6 \times 1.4 \cr & = 1.056{\text{ c}}{{\text{m}}^3} \cr} $$

Let the radius of hemisphere and sphere be 'r' and 'R' $$\eqalign{ & \Rightarrow \frac{4}{3}\pi {R^3} = \frac{2}{3}\pi {r^3} \cr & \frac{{{R^3}}}{{{r^3}}} = \frac{1}{2} \cr & \frac{R}{r} = \frac{1}{{\root 3 \of 2 }} \cr & \Rightarrow {\text{Ratio of curved surface area}} \cr & = \frac{{4\pi {R^2}}}{{2\pi {r^2}}} \cr & = \frac{{2{R^2}}}{{{r^2}}} \cr & = \frac{{2 \times 1}}{{{{\left( {\root 3 \of 2 } \right)}^2}}} \cr & = \frac{2}{{{{\left( 2 \right)}^{\frac{2}{3}}}}} \cr & \Rightarrow \frac{R}{r} = \frac{{{2^{\frac{1}{3}}}}}{1} \cr} $$

Volume of earth taken out = 40 × 30 × 12 = 14400 m 3 Area of rectangular field = 1000 × 30 = 30000 m 2 Area of region of tank = 40 × 30 = 1200 m 2 Remaining area = 30000 - 1200 = 28800 m 2 Increase in height $$ = \frac{{14400}}{{28800}} = 0.5{\text{ m}}$$

Volume of cuboid = $$l$$bh 12 × b × h - 18 = 12 × b × (h - 0.3) 12bh - 18 = 12bh - 3.6b b = 5 m

$$\eqalign{ & \frac{4}{3}\pi {R^3} = \pi {r^2}h \cr & \frac{4}{3}\pi \times 21 \times 21 \times 21 = \pi \times {R^2} \times 7 \cr & {R^2} = 21 \times 21 \times 4 \cr & R = 42{\text{ cm}} \cr & \frac{{{\text{total surface area of rod}}}}{{{\text{total surface area of sphere}}}} \cr & = \frac{{2\pi r\left( {h + r} \right)}}{{4\pi {R^2}}} \cr & = \frac{{42\left( {42 + 7} \right)}}{{2 \times 21 \times 21}} \cr & = \frac{{49}}{{21}} \cr & = \frac{7}{3} \cr & = 7:3 \cr} $$

Total surface area of remaining part 6 = 16πrh + 2πR 2 - 8πr 2 + 2πRH + 8πr 2 = 2π(8rh + R 2 - 4r 2 + RH + 4r 2 ) = 2π(8 × 3.5 × 5 + (14) 2 - 4 × 3.5 × 3.5 + 14 × 15 + 4 × (3.5) 2 ) = 3432

Height of water after ball is immersed = 3.5 × 2 = 7 cm $$\eqalign{ & \Rightarrow {\text{Volume of water}} = \pi {r^2}h - \frac{4}{3}\pi {r^3} \cr & = \pi {r^2}\left( {h - \frac{4}{3}r} \right) \cr & = \frac{{22}}{7} \times 3.5 \times 3.5\left( {7 - \frac{4}{3} \times 3.5} \right) \cr & = 11 \times 3.5\left( {\frac{7}{3}} \right) \cr & = \frac{{269.5}}{3}{\text{ c}}{{\text{m}}^3} \cr} $$ Volume of water before ball was immersed $$\eqalign{ & \Rightarrow \pi {\left( {3.5} \right)^2} \times h = \frac{{269.5}}{3} \cr & h = \frac{{269.5 \times 7}}{{3 \times 3.5 \times 3.5 \times 22}} \cr & h = \frac{7}{3}{\text{ cm}} \cr} $$

Side of cubical block = 10 cm ∴ Area of each face of cubical block = 10 2 = 100 cm 2 Radius of Hemisphere = $$\frac{4}{2}$$ = 2 cm Surface area of hemisphere $$\eqalign{ & = 2\pi {r^2} \cr & = 2 \times \frac{{22}}{7} \times {2^2}{\text{ c}}{{\text{m}}^2} \cr & = \frac{{176}}{7}{\text{ c}}{{\text{m}}^2} \cr} $$ Total surface area of hemisphere $$ = 6 \times \frac{{176}}{7} = \frac{{1056}}{7}{\text{ c}}{{\text{m}}^2}$$ Remaining surface area of each face of cubical block $$\eqalign{ & = {10^2} - 2\pi r \cr & = 100 - 2 \times \frac{{22}}{7} \times 2{\text{ c}}{{\text{m}}^2} \cr & = 100 - \frac{{88}}{7}{\text{ c}}{{\text{m}}^2} \cr & = \frac{{612}}{7}{\text{ c}}{{\text{m}}^2} \cr} $$ ∴ Total surface area of 6 remaining cubical block $$ = 6 \times \frac{{612}}{7} = \frac{{3672}}{7}{\text{ c}}{{\text{m}}^2}$$ ∴ Surface area of remaining solid $$\eqalign{ & = \left( {\frac{{1056}}{7} + \frac{{3672}}{7}} \right){\text{c}}{{\text{m}}^2} \cr & = \frac{{4728}}{7}{\text{ c}}{{\text{m}}^2} \cr & = 675\frac{3}{7}{\text{ c}}{{\text{m}}^2} \cr} $$

Base Area = 16π πR 2 = 16π R = 4 Given r = 2 ∵ ΔABC ≅ ΔADE $$\eqalign{ & \therefore \frac{{{\text{BC}}}}{{{\text{DE}}}} = \frac{{{\text{AC}}}}{{{\text{AE}}}} \cr & \frac{4}{8} = \frac{6}{{{\text{AE}}}} \cr & {\text{AE}} = 12 = {\text{L}} \cr} $$ Surface Area of frustum = πRL - πr$$l$$ = π × 4 × 12 - π × 2 × 6 = 48π - 12π = 36π

Radius of semi-circular sheet = r ⇒ $$\frac{{28}}{2}$$ r = 14 cm Circumference of sheet = πr = 14π cm Sheet is folded to form a cone Let radius of cone = r 1 ∴ The circumference of base of cone ⇒ Circumference of sheet ∴ 2πr 1 = 14π r 1 = 7 cm ∴ Radius of cone = 7 cm Slant height = Radius of semi-circular sheet r = 14 cm $$\eqalign{ & \therefore {\text{Height}} = \sqrt {{{\left( {14} \right)}^2} - {{\left( 7 \right)}^2}} \cr & = \sqrt {147} \cr & = 12{\text{ cm}}\,\,\,\left( {{\text{approx}}} \right) \cr} $$

$$\eqalign{ & 7.7{\text{ Ltr/sec}} \cr & r = 10.5{\text{ cm}} \cr & {\text{V}} = \frac{2}{3}\pi {r^3} \times \frac{2}{3} \cr & = \frac{{\frac{2}{3} \times \frac{2}{3} \times \frac{{22}}{7} \times 10.5 \times 10.5 \times 10.5 \times 1000}}{{7.7 \times 60 \times 6000}} \cr & = \frac{{20 \times 105}}{{36}} \cr & = \frac{{175}}{3}{\text{ hours}} \cr} $$

$$\eqalign{ & {\text{Slant height of the cone }}\left( l \right) \cr & = \sqrt {{6^2} + {8^2}} \cr & = 10{\text{ cm}} \cr & {\text{Required ratio}} = \frac{{2\pi rh}}{{\pi rl}} \cr & = \frac{{2h}}{l} \cr & = \frac{{2 \times 8}}{{10}} \cr & = 8:5 \cr} $$

$$\eqalign{ & 2\pi rh = 264{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \pi {r^2}h = 924{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{On dividing}}:\frac{{2\pi rh}}{{\pi {r^2}h}} = \frac{{264}}{{924}} \cr & \frac{2}{r} = \frac{{264}}{{924}} \cr & r = \frac{{924 \times 2}}{{264}} \cr & r = 7{\text{ cm}} \cr & {\text{Diameter}} = 2r = 2 \times 7 = 14{\text{ cm}} \cr & {\text{Putting, }}r = 7{\text{ in equation }}\left( {\text{i}} \right) \cr & 2\pi rh = 264 \cr & h = \frac{{264 \times 7}}{{2 \times 22 \times 7}} \cr & h = 6{\text{ cm}} \cr & {\text{Required ratio}} = \frac{{2r}}{h} = \frac{{14}}{6} = \frac{7}{3} \cr} $$

In a solid by Euler's formula Faces + Vertices - 2 = Edges ⇒ Faces + 12 - 2 = 30 ⇒ Faces = 30 - 10 = 20

According to the question, Diameter = 84 cm Radius = 42 cm = 0.42 m Height = 120 cm = 1.2 m CSA of cylinder = 2πrh $$ = 2 \times \frac{{22}}{7} \times 0.42 \times 1.2$$ Total area covered to planed in 500 rounds $$\eqalign{ & = 2 \times \frac{{22}}{7} \times 0.42 \times 1.2 \times 500 \cr & {\text{Total cost}} = \left( {2 \times \frac{{22}}{7} \times 0.42 \times 1.2 \times 500} \right) \times 1.50 \cr & = {\text{Rs}}{\text{. }}2376 \cr} $$

$$\eqalign{ & 2.1{\text{ m}} \times 1.5{\text{ m}} \times h = 630{\text{ lt}} \cr & \frac{{21}}{{10}}{\text{m}} \times \frac{{15}}{{10}}{\text{m}} \times h = \frac{{630}}{{1000}}{{\text{m}}^3} \cr} $$ \[\left[ \begin{gathered} \because 1\,{{\text{m}}^3} = 1000{\text{ lt}} \hfill \\ {\text{1000 c}}{{\text{m}}^3} = 1{\text{ lt}} \hfill \\ \end{gathered} \right]\] $$h = \frac{1}{5}{\text{m}} = 0.20{\text{ metre}}$$

$$\eqalign{ & {\text{V}} = \frac{1}{3}\pi \left( {{R^2} + {r^2} + Rr} \right)h \cr & {\text{V}} = \frac{1}{3} \times \frac{{22}}{7}\left( {25 + 9 + 15} \right) \times 10.5 \cr & {\text{V}} = 11 \times 49 \cr & {\text{V}} = 539{\text{ c}}{{\text{m}}^2} \cr} $$

\[\begin{array}{*{20}{c}} {}&{{\text{Before}}}&{}&{{\text{After}}}&{} \\ {{\text{Radius}} \to }&5&:&{11}&{120\% \to \frac{{ + 6}}{5}} \\ {{\text{Height}} \to }&5&:&3&{40\% \to \frac{{ - 2}}{5}} \\ {{\text{Volume}} \to }&{{5^2} \times 5}&:&{{{11}^2} \times 3}&{} \end{array}\] $$\% \,{\text{Increase}} \Rightarrow \frac{{238}}{{125}} \times 100 = 190.4\% $$

$$\eqalign{ & {r_1}:{r_2} = 3:2 \cr & {v_1}:{v_2} = 27:16 \cr & {\text{Height}} \to {h_1}:{h_2} \cr & \frac{{9 \times {h_1}}}{{4 \times {h_2}}} = \frac{{27}}{{16}} \cr & \frac{{{h_1}}}{{{h_2}}} = \frac{3}{4} \cr & {h_1}:{h_2} = 3:4 \cr} $$

Volume of cube = 9 3 = 729 Volume of cuboid = 5 × 31 × 13 = 2015 Total volume = 729 + 2015 = 2744 Let volume of new cube = a 3 a 3 = 2744 a = 14 T.S.A = 6 × 14 2 = 6 × 196 The cost of polish = 6 × 196 × 10 = Rs. 11760

Volume of wall = 12 m × 9 m × 90 cm Volume of Bricks in wall = (12 m × 9 m × 90 cm) × $$\frac{{17}}{{18}}$$ Let, number of bricks = n Now, n × 450 × 10 cm × 7.5 cm × 7 cm = (12 m × 9 m × 90 cm) × $$\frac{{17}}{{18}}$$ $$\eqalign{ & {\text{n}} = \frac{{1200 \times 900 \times 90}}{{450 \times 10 \times 7.5 \times 7}} \times \frac{{17}}{{18}} \cr & {\text{n}} = \frac{{160 \times 17}}{7} \cr & {\text{n}} = \frac{{2720}}{7} \cr & {\text{n}} = 388.5 \cr & {\text{n}} = 389{\text{ Answer}} \cr} $$

Ratio of volume of bigger cone and smaller cone = (Ratio of altitude) 2 = (1 : 2 : 3) 3 = (1 : 8 : 27) ∴ Ratio of parts = 1 : 8 - 1 : 27 - 8 = 1 : 7 : 19

$$\eqalign{ & {\text{Length of Arc}} = L \cr & L = 2 \times \frac{{22}}{7} \times \frac{{r \times 120}}{{360}} \cr & L = 2 \times \frac{{22}}{7} \times 10.5 \times \frac{1}{3} \cr & L = 22 \cr} $$ $$\eqalign{ & 2\pi R = 22 \cr & 2 \times \frac{{22}}{7}R = 22 \cr & R = 3.5 \cr & {h^2} + {R^2} = {10.5^2} \cr & {h^2} + {3.5^2} = {10.5^2} \cr & h = 7\sqrt 2 \cr & V = \frac{1}{3}\pi {r^2}h \cr & V = \frac{1}{3}\pi \times 3.5 \times 3.5 \times 7\sqrt 2 \cr & V = \frac{{343\sqrt 2 }}{{12}}\pi \cr} $$

$$\eqalign{ & {\text{Volume of sphere}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3}\pi \times 3 \times 3 \times 3 \cr & = 36\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{If the water level rises by H cm}} \cr & \pi {R^2}H = 36\pi \cr & 6 \times 6 \times H = 36 \cr & H = 1{\text{ cm}} \cr} $$

Total surface area of each part is = 2 × Area of base + Perimeter of base × Height $$\eqalign{ & = 2 \times \frac{1}{2} \times 4\sqrt 2 \times 4\sqrt 2 + \left( {8 + 8\sqrt 2 } \right) \times 8 \cr & = 32 + 64 + 64\sqrt 2 \cr & = 96 + 64\sqrt 2 \cr} $$

Length of each side = 8√2 cm Diagonal = √2a = √2 × 8√2 = 16 OC $$ = \frac{{\text{d}}}{2} = \frac{{16}}{2} = 8$$ Slant edge = EC = 10 cm In ΔOEC EC 2 = OE 2 + OC 2 10 2 = OE 2 + 8 2 OE 2 = 6 2 Height = OE = 6 cm Volume of the pyramid = $$\frac{1}{3}$$ × Area of base × Height = $$\frac{1}{3}$$ × (8√2) 2 × 6 = 256

10 cm ⇒ height Volume of Prism = Base Area × height 160 = a 2 × 10 a = 4 Total surface area = 2Base Area + Base per meter × length = 2 × 16 + 16 × 10 = 32 + 160 = 192

$$\eqalign{ & {R^3} = n \times {r^3} \cr & 11 \times 11 \times 11 = n \times {2^3} \cr & \frac{{1331}}{8} = n \cr & n = 166 \cr & n = 166{\text{ }}\left( {{\text{integer}}} \right) \cr} $$

$$\frac{{{\text{Curved surfaces area of hemisphere}}}}{{{\text{Total surface area of cube}}}} = \frac{{2\pi {r^2}}}{{6{a^2}}}$$ 3r = 21 r = 7 cm Total volume of figure = Volume of hemisphere + Volume of cube = $$\frac{2}{3}$$πr 3 + a 3 = $$\frac{2}{3}$$ × $$\frac{{22}}{7}$$ × 7 × 7 × 7 + 14 3 = 718.66 + 2744 = 3462.67 cm 3

L.S.A. of cylinder = 2πrh = 2 × $$\frac{{22}}{7}$$ × 5 × 30 = 300π L.S.A. of cones = 2[πr$$l$$] = 2[π × 5 × 13] = 130π T.S.A. of remaining part = 300π + 130π = 430π

$$l$$ = 12 cm b = 8 cm h = 10 cm Cost = Rs. 25/m 2 Area of 4 walls = 2($$l$$ + b) × h = 2(12 + 8) × 10 = 400 m 2 Area of roof = 12 × 8 = 96 m 2 Total Area = 400 + 96 = 496 Cost = 496 × 25 = Rs. 12400

$$\eqalign{ & 4\pi \left[ {{{\left( {R + 4} \right)}^2} - {R^2}} \right] = 464\pi \cr & 4\left( {R + 4 + R} \right)\left( {R + 4 - R} \right) = 464 \cr & 16\left( {2R + 4} \right) = 464 \cr & 2R + 4 = 29 \cr & R = \frac{{25}}{2} \cr & {\text{Volume of sphare}} = \frac{4}{3}\pi {R^3} \cr & = \frac{4}{3}\pi \times {\left( {\frac{{25}}{2}} \right)^3} \cr & = \frac{4}{3} \times \frac{{15625}}{8}\pi \cr & = \frac{{15625}}{6}\pi \cr} $$

(I) 6 × 7 × 8 = 336 (II) (7) 3 = 343 Cube will have more volume.

$$\eqalign{ & {\text{Curved surface area of cone}} = \pi rl \cr & \frac{{22}}{7} \times 21 \times l = 2310 \cr & 66l = 2310 \cr & l = 35 \cr & {\text{Height}} = \sqrt {{{35}^2} - {{21}^2}} \cr & = \sqrt {1225 - 441} \cr & = \sqrt {784} \cr & = 28{\text{ cm}} \cr & {\text{New radius}} = 21 + 21 = 42{\text{ cm}} \cr & {\text{New height}} = 28 - 14 = 14{\text{ cm}} \cr & {\text{New volume}} = \frac{1}{3}\pi {r^2}h \cr & = \frac{1}{3} \times \frac{{22}}{7} \times 42 \times 42 \times 14 \cr & = 22 \times 42 \times 28 \cr & = 25872{\text{ c}}{{\text{m}}^3} \cr & {\text{Capacity}} = \frac{{25872}}{{1000}} = 25.9{\text{ litres}} \cr} $$

Total surface area = perimeter × height + 2 × area of base 216√3 = 6 × 6 × h + 2 × 6 × $$\frac{{\sqrt 3 }}{4}$$ × 6 × 6 216√3 = 36h + 108√3 108√3 = 36h h = 3√3 cm

Let the time taken to fill the tank = $$x$$ hours $$\eqalign{ & \Rightarrow \left( {\pi {r^2}h} \right) \times x = 50 \times 44 \times \frac{7}{{100}} \cr & \Rightarrow x = \frac{{50 \times 44 \times 7 \times 7 \times 100 \times 100}}{{22 \times 7 \times 7 \times 100 \times 5000}} \cr & \Rightarrow x = 2{\text{ hours}} \cr} $$

Note 1 m 3 = 1000 litres 1000 cm 3 = 1 litre V 2 = abc V = $$\sqrt {3 \times 12 \times 16} $$ V = 6 × 4 = 24m 3 = 24000 litres

Volume of cuboid = 32 × 12 × 9 Volume of a cube = 16 × 12 × 9 a 3 = 16 × 12 × 9 a = 12 Area of cuboid : Area of two cube = 2(32 × 12 + 12 × 9 + 9 × 32) : 2 × 6 × 12 2 = 65 : 72

Slant height of cone $$ = \sqrt {{8^2} + {6^2}} = 10{\text{ cm}}$$ Slant height of cone = Radius of sector = 10 cm