Mensuration 3D — Part 2 MCQs with Explanations

Volume of vessel = Volume of roof π × r 2 × h = 22 × 20 × x (where x is rainfall in cm) $$\eqalign{ & \Rightarrow \frac{{22}}{7} \times \frac{{100 \times 100 \times 350}}{{22 \times 20 \times 100 \times 100}} = {\text{x}} \cr & \Rightarrow {\text{x}} = 2.5{\text{ cm}} \cr} $$

Lateral surface area of prism = 120 Base perimeter × height = 120 3 × (side) × height = 120 (Perimeter of equilateral Δ = 3 × side) Side × height = $$\frac{{120}}{3}$$ = 40 . . . . . (i) Volume of prism = 40√3 Area of base × height = 40√3 $$\frac{{\sqrt 3 }}{4}$$ (side) 2 × height = 40√3 (side) 2 × height = $$ = \frac{{40\sqrt 3 \times 4}}{{\sqrt 3 }}$$ = 160 . . . . . (ii) Dividing equation (ii) by equation (i) $$\frac{{{{\left( {{\text{side}}} \right)}^2} \times {\text{height}}}}{{{\text{side}} \times {\text{height}}}} = \frac{{160}}{{40}}$$ side = 4 cm

Let the side of cube is 3 m Surface are of cube = 6 × 3 2 = 54 Surface are of one cuboid = 2($$l$$b + bh + h$$l$$) = 2(1 × 3 + 3 × 3 + 3 × 1) = 2(3 + 9 + 3) = 30 Cube : 2 Cuboid = 54 : 2 × 30 = 9 : 10

$$\eqalign{ & {\text{Volume of lead}} = \frac{4}{3}\pi {r^3} \cr & {\text{Volume of gold}} = \frac{4}{3}\pi {R^3} - \frac{4}{3}\pi {r^3} \cr & {\text{According to the question,}} \cr & \frac{4}{3}\pi {R^3} - \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {r^3} \cr & \frac{4}{3}\pi {R^3} = \frac{8}{3}\pi {r^3} \cr & {R^3} = 2{r^3} \cr & {R^3} = 2{\left( 2 \right)^3} \cr & R = \root 3 \of 2 \times 2 \cr & R = 1.259 \times 2 \cr & R = 2.518 \cr & \therefore {\text{Thickness}} = R - r \cr & = 2.518 - 2 \cr & = 0.518{\text{ cm}} \cr} $$

Volume ⇒ $$\frac{1}{3}$$πr 2 h ⇒ $$\frac{1}{3}$$π × 12 × 12 × 9 ⇒ 144 × 3π ⇒ 432π

$$\eqalign{ & {\text{Volume of a cone}} = \frac{1}{3}\pi {r^2}h \cr & \frac{1}{3}\pi {r^2}\left( {24} \right) = 1232{\text{ c}}{{\text{m}}^2} \cr & {r^2} = \frac{{1232 \times 3 \times 7}}{{24 \times 22}} \cr & {r^2} = 7 \times 7 \cr & r = \sqrt {7 \times 7} \cr & r = 7{\text{ cm}} \cr & l = \sqrt {{r^2} + {h^2}} \cr & = \sqrt {{7^2} + {{24}^2}} \cr & = \sqrt {625} \cr & = 25 \cr & {\text{Curved surface area}} = \pi rl \cr & = \frac{{22}}{7} \times 7 \times 25 \cr & = 550{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & \frac{{2\pi rh}}{{\pi {r^2}}} = \frac{5}{1} \cr & \frac{{2h}}{r} = \frac{5}{1} \cr & r:h = 2:5 \cr} $$

External radius R = 4 cm Internal radius r = 3 cm Volume of iron used = πR 2 h - πr 2 h = πh(R 2 - r 2 ) = πh(R + r)(R - r) = $$\frac{{22}}{7}$$ × 20 × (4 + 3) × (4 - 3) = $$\frac{{22}}{7}$$ × 20 × 7 × 1 = 440 cm 3

$$\eqalign{ & {\text{Volume of sphere}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3}\pi {\left( 9 \right)^3} \cr & = 972\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{Volume of cone}} = \frac{1}{3}\pi {r^2}h \cr & = \frac{1}{3}\pi {\left( 9 \right)^2}9 \cr & = 243\pi {\text{ c}}{{\text{m}}^3} \cr & \Rightarrow \% {\text{ of wasted wood}} \cr & = \frac{{\left( {972 - 243} \right)\pi }}{{972\pi }} \times 100 \cr & = 75\% \cr} $$

We know that A parallelopiped has vertices (v) = 8 edge (e) = 12 face (f) = 6 Put into equation (v - e + f) ⇒ 8 - 12 + 6 ⇒ 2

Curved area of cylinder = 88 cm 2 2πrh = 88 2 × $$\frac{{22}}{7}$$ × r × h = 88 r × h = 14 Volume of cylinder = πr 2 h = 132 cm 3 $$\frac{{22}}{7}$$ × r 2 h = 132 r 2 h = 42 r × rh = 42 r × 14 = 42 r = 3 r × h = 14 h $$ = \frac{{14}}{3} = 4\frac{2}{3}$$ Height $$ = 4\frac{2}{3}{\text{ cm}}$$

Ratio of volume of sphere × ratio of weight per 1 cc of material of each = Ratio of weight of two sphere $$\eqalign{ & \frac{{\frac{4}{3}\pi r_1^3}}{{\frac{4}{3}\pi r_2^3}} \times \frac{{289}}{{64}} = \frac{8}{{17}} \cr & \frac{{r_1^3}}{{r_2^3}} = \frac{{8 \times 64}}{{17 \times 289}} \cr & \frac{{{r_1}}}{{{r_2}}} = \frac{{8 \times 8 \times 8}}{{17 \times 17 \times 17}} \cr & \frac{{{r_1}}}{{{r_2}}} = \frac{8}{{17}} \cr & {r_1}:{r_2} = 8:17 \cr} $$

Area of trapezium = $$\frac{1}{2}$$ × h(AB + CD) = $$\frac{1}{2}$$ × 8 × (8 + 14) = 4 × 22 = 88 cm 2 Volume of prism = Height of prism × Area of base ⇒ height × 88 = 1056 (given) ⇒ height = $$\frac{{1056}}{{88}}$$ ⇒ height = 12 cm

Total surface area (TSA) of hemisphere = 3πr 2 Radius of hemisphere = $$\sqrt {\frac{{25}}{\pi }} $$ So, TSA = 3 × π × $${\left( {\sqrt {\frac{{25}}{\pi }} } \right)^2}$$ = 75 unit 2

Outer radius = 4 cm Internal radius = 4 - 0.5 = 3.5 cm Length = 2m = 200 cm Volume of cylinder = π × (R 2 - r 2 )h = π × (4 2 - 3.5 2 ) × 200 = π × 7.5 × 0.5 × 200 = 750π

Here h = 4c, Volume of cylinder = πr 2 h $$ = \frac{{4\pi \times \pi {r^2}h}}{{4\pi }}$$ (Multiply 4π both in Numerator & Denominator) $$\eqalign{ & = \frac{{{{\left( {2\pi r} \right)}^2} \times \left( {4c} \right)}}{{4\pi }} \cr & = \frac{{{c^3}}}{\pi } \cr} $$

Let the three sides of the cuboid be $$l$$, b and h ⇒ $$l$$b = bh = h$$l$$ = 12 ⇒$$l$$ 2 b 2 h 2 = 12 × 12 × 12 ⇒$$l$$ 2 b 2 h 2 = 1728 ⇒ $$l$$bh = $$\sqrt {1728} $$ ⇒ $$l$$bh = $$12\sqrt {12} $$ ⇒ $$l$$bh = $$24\sqrt 3 {\text{ c}}{{\text{m}}^3}$$

$$\eqalign{ & {\text{Sphere }}\left( {\text{V}} \right) = \frac{4}{3}\pi {\left( {\frac{R}{2}} \right)^3} = \frac{4}{3}\pi \times \frac{{{R^3}}}{8} \cr & {\text{Hemisphere }}\left( {\text{V}} \right) = \frac{2}{3}\pi {R^3} \cr & {\text{Sphere}}:{\text{Hemisphere}} = \frac{{4{R^3}}}{{3 \times 8}}:\frac{{2{R^3}}}{3} = 1:4 \cr & {\text{Sphere}}:{\text{Remaining Solid}} = 1:3 \cr} $$

Number of cubes with no side painted = (n - 2) 3 Where n is the side of the bigger cube Required number = (6 - 2) 3 = 64

Let the side of new cube is x cm Volume of cuboid = 8 x volume of new cube 18 × 36 × 72 = 8x 3 18 × 18 × 18 = x 3 x = 18 cm $$\eqalign{ & \frac{{{\text{Total surface area of cuboid}}}}{{8 \times {\text{Curved surface area of cube}}}} \cr & = \frac{{2\left( {lb + bh + hl} \right)}}{{8 \times 4{x^3}}} \cr & = \frac{7}{8} \cr} $$

h = 8 cm V = 392π cm 3 πr 2 h = 392π πr 2 × 8 = 392π r 2 = 49 r = 7 Curved surface area = 2πrh = 2 × π × 7 × 8 = 112π cm 2

$$\eqalign{ & \Rightarrow {\text{Circumference of sectors}} = \frac{{\pi r}}{2} \cr & \Rightarrow {\text{Circumference of base of cone of radius}} = 2\pi {r_1} \cr & \frac{{\pi r}}{2} = 2\pi {r_1} \Rightarrow {r_1} = \frac{r}{4} \cr & \therefore {\text{Radius of cone}} = \frac{r}{4} \cr & {\text{Curved surface area of cone}} = \pi {r_1}l \cr & l = {\text{slant height}} \cr & l = r \cr & \therefore {\text{Surface area of cone}} = \pi \times \frac{r}{4} \times r \Rightarrow \frac{{\pi {r^2}}}{4} \cr} $$

$$\eqalign{ & V = \frac{1}{3}\pi {r^2}h \cr & r \to 20\% \to \frac{{ + 1}}{5} \cr & h \to 30\% \to \frac{{ - 3}}{{10}} \cr & {r^2} \Rightarrow {5^2}:{6^2} \cr & h \Rightarrow 10:7 \cr & V \Rightarrow 250:252 \cr & {\text{Increase}} = 2 \cr & {\text{Increase }}\% = \frac{2}{{250}} \times 100 = 0.8\% \cr} $$

$$\eqalign{ & {\text{Diagonal}} = 35\sqrt 3 \cr & \therefore {\text{The length of largest rod}} = {\text{Diagonal}} = {\text{side}}\sqrt 3 \cr & {\text{side}}\sqrt 3 = 35\sqrt 3 \cr & {\text{side}} = \frac{{35\sqrt 3 }}{{\sqrt 3 }} = 35 \cr & {\text{Side of cube}} = 35 \cr} $$ $$\eqalign{ & {\text{Diameter of the sphere }} = {\text{ side of the cube}} \cr & 2 \times {\text{radius}} = {\text{side}} \cr & {\text{radius}} = \frac{{35}}{2}{\text{ cm}} \cr & {\text{Surface area of the sphere}} = 4\pi {r^2} \cr & = 4 \times \frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \cr & = 3850{\text{ }}{{\text{m}}^2} \cr} $$

$$\eqalign{ & a = \frac{{{\text{Volume of cube}}}}{{{\text{Volume of the sphere}}}} \cr & a = \frac{{{1^3}}}{{\frac{4}{3}\pi \times \frac{{{1^3}}}{8}}} \cr & a = \frac{6}{\pi }{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & b = \frac{{{\text{Surface area of sphere}}}}{{{\text{Surface area of cube}}}} \cr & b = \frac{{4\pi \times \frac{{{1^2}}}{4}}}{{6 \times {1^2}}} \cr & b = \frac{\pi }{6}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & a \times b = \frac{6}{\pi } \times \frac{\pi }{6} = 1{\text{ Answer}} \cr} $$

$$\eqalign{ & {\text{Volume of cylinder}} = \pi r_1^2h - \pi r_2^2h \cr & = \pi h\left( {r_1^2 - r_2^2} \right) \cr & = \pi h\left( {{r_1} + {r_2}} \right)\left( {{r_1} - {r_2}} \right) \cr & = \frac{{22}}{7} \times 15 \times \left( {6.75 + 5.25} \right)\left( {6.75 - 5.25} \right) \cr & = \frac{{22}}{7} \times 15 \times 12 \times 1.5 \cr & {\text{Volume}} = \pi r_3^2\frac{h}{2} = \frac{{22}}{7} \times 15 \times 12 \times 1.5 \cr & r_3^2 \times \frac{{15}}{2} = 15 \times 12 \times 1.5 \cr & {r_3} = 6 \cr} $$

$$\eqalign{ & r = \frac{{21}}{2} \cr & R = \frac{{39}}{2} \cr & {\text{Curved serface area of Hemisphere}} \cr & = 2\pi {r^2} \cr & = 2 \times \frac{{22}}{7} \times \frac{{21 \times 21}}{{2 \times 2}} \cr & = 693\,{\text{c}}{{\text{m}}^2} \cr & H = \left( {22.5 - 10.5} \right) = 12{\text{ cm}} \cr & {L^2} = {H^2} + {\left( {R - r} \right)^2} \cr & {L^2} = {12^2} + {\left( {\frac{{39}}{2} - \frac{{21}}{2}} \right)^2} \cr & {L^2} = 144 + 81 \cr & L = \sqrt {225} \cr & L = 15{\text{ cm}} \cr & {\text{Curved serface area of Frustum of cone}} \cr & = \pi l\left( {R + r} \right) \cr & = \frac{{22}}{7} \times 15 \times \left[ {\left( {\frac{{39}}{2}} \right) + \left( {\frac{{21}}{2}} \right)} \right] \cr & = \frac{{22}}{7} \times 15 \times 30 \cr & = \frac{{9900}}{7}{\text{ c}}{{\text{m}}^2} \cr & = 1414\frac{2}{7}{\text{ c}}{{\text{m}}^2} \cr & \therefore {\text{Total area of cloth}} \cr & = \left( {693 + 1414\frac{2}{7}} \right){\text{c}}{{\text{m}}^2} \cr & = 2107\frac{2}{7}{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & {\text{Radius of water drop}} = \frac{1}{{20}}{\text{ cm}} \cr & {\text{Volume of a sphere}} = \frac{4}{3}\pi \times \frac{1}{{20}} \times \frac{1}{{20}} \times \frac{1}{{20}} \cr & {\text{Let the radius of cone}} = R \cr & {\text{Height}} = 2R \cr & {\text{According to question}} \cr & \frac{1}{3}\pi \times R \times R \times 2R = \frac{4}{3}\pi \times \frac{1}{{20}} \times \frac{1}{{20}} \times \frac{1}{{20}} \cr & {R^3} = \frac{{2 \times 32000}}{{20 \times 20 \times 20}} \cr & {R^3} = \frac{{64000}}{{20 \times 20 \times 20}} \cr & {R^3} = \frac{{40 \times 40 \times 40}}{{20 \times 20 \times 20}} \cr & R = \frac{{40}}{{20}} \cr & R = 2 \cr & {\text{Height of glass}} = 2R = 2 \times 2 = 4{\text{ cm}} \cr} $$

$$\eqalign{ & {\text{Volume}} = \frac{1}{3}\pi \left( {{R^2} + {r^2} + Rr} \right)h \cr & = \frac{1}{3} \times \frac{{22}}{7} \times \left( {9 + 4 + 6} \right) \times 21 \cr & = 22 \times 19 \cr & = 418{\text{ c}}{{\text{m}}^3} \cr} $$

Height of cone (h) = 9 ft Radius (r) = 7 ft Volume = $$\frac{1}{3}$$πr 2 h = $$\frac{1}{3}$$ × 3.14 × 7 × 7 × 9 = 461.58 cube feet

Areas of three adjacent faces of a cuboidal solid block L × B = 216 B × H = 96 H × L = 144 Multiply all (LBH) 2 = 216 × 96 × 144 LBH = 36 × 4 × 12 LBH = 1728 LBH = 8 × a 3 1728 = 8 × a 3 a 3 = 216 a = 6 Lateral surface area of 3 such cubes = 3 × 4a 2 = 12 × 6 × 6 = 432 cm 2

Diameter of beaker = 7 cm Radius = $$\frac{7}{2}$$ cm Level of water rises = 5.6 cm Diameter of a marble = 1.4 cm ∴ Radius = $$\frac{{1.4}}{2}$$ = 0.7 cm Let n marbles are dropped so, Volume of n marbles $$ = n \times \frac{4}{3}\pi \times {\left( {0.7} \right)^3}$$ $$\eqalign{ & \Rightarrow n \times \frac{4}{3}\pi \times {\left( {0.7} \right)^3} = n \times {\left( {\frac{7}{2}} \right)^2} \times 5.6 \cr & \Rightarrow n \times \frac{4}{3} \times \frac{7}{{10}} \times \frac{7}{{10}} \times \frac{7}{{10}} = \frac{7}{2} \times \frac{7}{2} \times \frac{{56}}{{10}} \cr & \Rightarrow n = 150 \cr} $$

$$\eqalign{ & \pi {r^2}h = \frac{4}{3}\pi {R^3} \times n \cr & {6^2} \times 5 = \frac{4}{3} \times {1^3} \times n \cr & n = 135 \cr} $$

Let the increase in level = $$x$$ m $$\eqalign{ & \Rightarrow \left( {1000 \times 1000 \times \frac{2}{{100}}} \right) \times \frac{1}{2} = 100 \times 10 \times x \cr & \Rightarrow x = 10{\text{ m}} \cr} $$

Volume of all three cube = 4 3 + 5 3 + 6 3 = 64 + 125 + 216 = 405 cm 3 Now, 62 cm 3 is lost due to improper handling. ∴ Volume of new cube = 405 - 62 = 343 (Side of new cube) 3 = 343 Side of new cube $$ = \root 3 \of {343} = 7$$ Total surface area of new cube = 6 × (side) 2 = 6 × (7) 2 = 6 × 49 = 294 cm 2

$$\eqalign{ & 4\pi \left[ {{{\left( {x + 2.5} \right)}^2} - {x^2}} \right] = 110 \cr & \Rightarrow 4 \times \frac{{22}}{7}\left[ {5x + 6.25} \right] = 110 \cr & \Rightarrow 20x = 10 \cr & \Rightarrow x = \frac{1}{2} \cr & \Rightarrow {\text{Volume}} = \frac{4}{3} \times \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{{11}}{{21}} \cr} $$

Surface area of sphere = 4πr 2 = x cm 2 Volume of sphere = $$\frac{4}{3}$$πr 3 = y cm 3 According to the question, y = 14x ⇒ $$\frac{4}{3}$$πr 3 = 14 × 4πr 2 ⇒ r = 42 cm

$$\eqalign{ & \pi {r^2} = 25 \times 4 \cr & \pi {r^2} = 100{\text{ }}{{\text{m}}^2} \cr & {r^2} = \frac{{100}}{\pi } \cr & {\text{Volume}} = \frac{1}{3}\pi {r^2}h \cr & 80 \times 25 = \frac{1}{3} \times \pi \times \frac{{100}}{\pi } \times h \cr & h = 60{\text{ m}} \cr} $$

Volume of the part (prism) = Area of base × height Area of base (Isosceles Δ) $$\eqalign{ & = \frac{b}{4}\sqrt {4{a^2} - {b^2}} \cr & = \frac{6}{4}\sqrt {4{{\left( 5 \right)}^2} - {{\left( 6 \right)}^2}} \cr & = 12{\text{ c}}{{\text{m}}^2} \cr} $$ Volume of prism = 12 × 8 = 96 cm 3

$$\eqalign{ & {\text{Area of }}\Delta ABC = \frac{{\sqrt 3 }}{4}{a^2} \cr & 16\sqrt 3 = \frac{{\sqrt 3 }}{4}{a^2} \cr & a = 8 \cr & {\text{In }}\Delta ABD, \cr & \frac{1}{2} \times 8 \times MD = 30 \cr & MD = \frac{{15}}{2} \cr & {\text{Now, in }}\Delta MOD, \cr & MO = \frac{a}{{2\sqrt 3 }} = \frac{4}{{\sqrt 3 }} \cr & M{D^2} = M{O^2} + O{D^2} \cr & \frac{{225}}{4} = {\left( {\frac{4}{{\sqrt 3 }}} \right)^2} + O{D^2} \cr & OD = \sqrt {\frac{{225}}{4} + \frac{{16}}{3}} \cr & OD = \sqrt {\frac{{611}}{{12}}} \cr} $$