Volume And Surface Area — Part 2 MCQs with Explanations

Internal radius, r = 4 cm External radius, R = 5 cm Total surface area : $$\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

Area of base : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$ Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m So, area of each lateral face : $$\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2} \cr & \left[ {\because s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr} $$ ∴ Whole surface area of the pyramid : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr} $$

Let, $$\eqalign{ & OP = OQ = OR = r \cr & \therefore OR = h = r \cr} $$ ∴ Curved surface area of the hemisphere = $$2\pi {r^2}$$ Curved surface area of a cone = $$\pi rl$$ Where, $$\eqalign{ & l = \sqrt {{h^2} + {r^2}} \cr & \,\,\,\,\, = \sqrt {{r^2} + {r^2}} \cr & \,\,\,\,\, = r\sqrt 2 \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{2\pi {r^2}}}{{\pi rl}} \cr & = \frac{{2\pi {r^2}}}{{\pi r \times r\sqrt 2 }} \cr & = \frac{2}{{\sqrt 2 }} \cr & = \frac{{2 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & = \frac{{2\sqrt 2 }}{2} \cr & = \frac{{\sqrt 2 }}{1}\,Or\,\sqrt 2 :1 \cr} $$

Given length of width of swimming pool is 9 m and 12 m respectively Volume of swimming pool : $$\eqalign{ & = 9 \times 12 \times \left( {\frac{{1 + 4}}{2}} \right) \cr & = 9 \times 12 \times \frac{5}{2} \cr & = 270\,\text{cu. metre} \cr} $$

Total length of tape required : = Sum of lengths of edges = (30 × 4 + 25 × 4 + 20 × 4) cm = 300 cm

Volume : $$\eqalign{ & = \left[ {12 \times 9 \times \left( {\frac{{1 + 4}}{2}} \right)} \right]{m^3} \cr & = \left( {12 \times 9 \times 2.5} \right){m^3} \cr & = 270\,{m^3} \cr} $$

Volume of cube = Volume of sheet = (27 × 8 × 1) cm 3 = 216 cm 3 Edge of cube : $$\root 3 \of {216} \,cm = 6\,cm$$ Surface area of sheet : $$\eqalign{ & = 2\left( lb + bh + lh \right) \cr & = 2\left( {27 \times 8 + 8 \times 1 + 27 \times 1} \right){\text{ c}}{{\text{m}}^2} \cr & = \left( {216 + 8 + 27} \right){\text{ c}}{{\text{m}}^2} \cr & = 502{\text{ c}}{{\text{m}}^2} \cr} $$ Surface area of cube : $$\eqalign{ & = 6{a^2} \cr & = \left( {6 \times {6^2}} \right){\text{ c}}{{\text{m}}^2} \cr & = 216{\text{ c}}{{\text{m}}^2} \cr} $$ ∴ Required difference : $$\eqalign{ & = \left( {502 - 216} \right){\text{ c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

Let their edges be a and b Then, $$\eqalign{ & \frac{{{a^3}}}{{{b^3}}} = \frac{8}{{27}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \Rightarrow \frac{a}{b} = \frac{2}{3} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{4}{9} \cr & \Rightarrow \frac{{6{a^2}}}{{6{b^2}}} = \frac{4}{9}\,Or\,4:9 \cr} $$

$$\eqalign{ & 3 \times 2\pi {r^2} = 2 \times 2\pi rh \cr & \Rightarrow 6r = 4h \cr & \Rightarrow r = \frac{2}{3}h \cr & \Rightarrow r = \left( {\frac{2}{3} \times 6} \right)m \cr & \Rightarrow r = 4\,m \cr} $$

Volume of water flown through the pipe in 30 min : $$\eqalign{ & = \left[ {\left( {\pi \times 0.01 \times 0.01 \times 6} \right) \times 30 \times 60} \right]{m^3} \cr & = \left( {1.08\pi } \right){m^3} \cr} $$ Let the rise in level of water be h metres Then, $$\eqalign{ & \pi \times 0.6 \times 0.6 \times h = 1.08\pi \cr & \Rightarrow h = \left( {\frac{{1.08}}{{0.6 \times 0.6}}} \right) \cr & \Rightarrow h = 3\,m \cr} $$

Volume of water : $$\eqalign{ & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times 8 \times 8 \times 21} \right){\text{ c}}{{\text{m}}^3} \cr & = 1408{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{{1408}}{{1000}}} \right)kg \cr & = 1.408\,kg \cr} $$

let their radii be x and 2x, and their heights be h and H respectively Then, $$\eqalign{ & \frac{{\frac{1}{3} \times \pi \times {x^2} \times h}}{{\frac{1}{3} \times \pi \times {{\left( {2x} \right)}^2} \times H}} = \frac{2}{3} \cr & \Leftrightarrow \frac{h}{H} = \frac{2}{3} \times 4 \cr & \Leftrightarrow \frac{h}{H} = \frac{8}{3}\,Or\,8:3 \cr} $$

$$\eqalign{ & \frac{4}{3} \times \frac{{22}}{7} \times {r^3} = \frac{{45056}}{{21}} \cr & \Rightarrow {r^3} = \left( {\frac{{45056}}{{21}} \times \frac{3}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^3} = 512 \cr & \Rightarrow r = \root 3 \of {512} \cr & \Rightarrow r = 8\,cm \cr} $$

Let the radius of the third ball be R cm Then, $$\frac{4}{3}\pi \times {\left( {\frac{3}{4}} \right)^3} + \frac{4}{3}\pi \times {\left( 1 \right)^3}$$ $$ + \frac{4}{3}\pi \times {R^3}$$ $$ = \frac{4}{3}\pi \times {\left( {\frac{3}{2}} \right)^3}$$ $$\eqalign{ & \Rightarrow \frac{{27}}{{64}} + 1 + {R^3} = \frac{{27}}{8} \cr & \Rightarrow {R^3} = \frac{{125}}{{64}} = \frac{{{{\left( 5 \right)}^3}}}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow R = \frac{5}{4} \cr} $$ ∴ Diameter of the third ball : $$ = 2R = \frac{5}{2}cm = 2.5\,cm$$

Let the radius of the sphere be r Then, radius of the cylinder = r Height of the cylinder = 2r Surface area of sphere = $$4\pi {{\text{r}}^2}$$ Surface area of the cylinder = $$2\pi {\text{r}}(2r) = 4\pi {{\text{r}}^2}$$ ∴ Required ratio : = $$4\pi {{\text{r}}^2}$$ : $$4\pi {{\text{r}}^2}$$ = 1 : 1

Volume of hemispherical bowl : $$ = \left( {\frac{2}{3} \times \pi \times 12 \times 12 \times 12} \right)c{m^3}$$ Volume of 1 cylindrical container : $$ = \left( {\pi \times 2 \times 2 \times 3} \right)c{m^3}$$ ∴ Number of containers required : $$\eqalign{ & = \frac{2}{3} \times \frac{{12 \times 12 \times 12}}{{2 \times 2 \times 3}} \cr & = 96 \cr} $$

Length of each edge of a regular tetrahedron = 1 cm Volume of regular tetrahedron : $$\eqalign{ & = \frac{{{a^3}}}{{6\sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{1}{{6\sqrt 2 }} \cr & = \frac{{\sqrt 2 }}{{6\sqrt 2 \times \sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{{\sqrt 2 }}{{12}}{\text{ Or }}\frac{1}{{12}}\sqrt 2 {\text{ c}}{{\text{m}}^3} \cr} $$

Length of parallel sides of prism = 10 cm and 6 cm Height of prism = 8 cm ∴ Volume of prism : $$\eqalign{ & = \frac{1}{2}\left( {10 + 6} \right) \times 5 \times 8 \cr & = \frac{1}{2} \times 16 \times 5 \times 8 \cr & = 320{\text{ c}}{{\text{m}}^3} \cr} $$

Volume of the reservoir = 42000 litres = 42 m 3 Let the depth of the reservoir be h metres then, $$\eqalign{ & 6 \times 3.5 \times h = 42 \cr & Or,\,h = \frac{{42}}{{6 \times 3.5}} = 2\,m \cr} $$

Let the depth of the cistern be h metres Then, $$\eqalign{ & 4.5 \times 3 \times h = 50 \cr & \Rightarrow h = \frac{{50}}{{13.5}} \cr & \Rightarrow h = \frac{{100}}{{27}} \cr} $$ Area of sheet required : $$\eqalign{ & = lb + 2\left( {bh + lh} \right) \cr & = lb + 2h\left( {l + b} \right) \cr & = \left[ {4.5 \times 3 + 2 \times \frac{{100}}{{27}} \times \left( {4.5 + 3} \right)} \right]{{\text{m}}^2} \cr & = \left( {13.5 + \frac{{200}}{{27}} \times 7.5} \right){\text{ }}{{\text{m}}^2} \cr & = \left( {\frac{{27}}{2} + \frac{{500}}{9}} \right){{\text{m}}^2} \cr & = \frac{{1243}}{{18}}{\text{ }}{{\text{m}}^2} \cr} $$ ∴ Weight of lead : $$\eqalign{ & = \left( {27 \times \frac{{1243}}{{18}}} \right)kg \cr & = \left( {\frac{{3729}}{2}} \right)kg \cr & = 1864.5\,kg \cr} $$

Let the original length, breadth and height of the cuboid be x, 2x and 3x units respectively Then, original volume = (x × 2x × 3x) cu.units = 6x 3 cu.units New length = 200% of x = 2x New breadth = 300% of 2x = 6x New height = 300% of 3x = 9x ∴ New volume : = (2x × 6x × 9x) cu.units = 108x 3 cu.units Increase in volume : = (108x 3 - 6x 3 ) cu.units = (102x 3 ) cu.units ∴ Required ratio : $$\eqalign{ & = \frac{{102{{\text{x}}^3}}}{{6{{\text{x}}^3}}} \cr & = 17{\text{ }}\left( {{\text{Times}}} \right) \cr} $$

Since the tank is open at the top, we have : Area of sheet required = Surface area of the tank $$\eqalign{ & = lb + 2\left( {bh + lh} \right) \cr & = \left[ {30 \times 20 + 2\left( {20 \times 12 + 30 \times 12} \right)} \right]{{\text{m}}^2} \cr & = \left( {600 + 1200} \right){{\text{m}}^2} \cr & = 1800{\text{ }}{{\text{m}}^2} \cr} $$ Length of sheet required : $$\eqalign{ & = \left( {\frac{{{\text{Area}}}}{{{\text{Width}}}}} \right) \cr & = \frac{{1800}}{3}m \cr & = 600\,m \cr} $$ ∴ Cost of the sheet = Rs. (600 × 10) = Rs. 6000

Since a cube has 4 diagonals, we have : Length of a diagonal $$\eqalign{ & = \left( {\frac{{12}}{4}} \right)cm \cr & = 3\,cm \cr} $$ Let the length of each edge of the cube be a cm Then, $$\eqalign{ & \sqrt 3 a = 3 \cr & or,a = \sqrt 3 \cr} $$ ∴ Total length of the edges of the cube = $$12\sqrt 3\, $$ cm

Let original edge = a Then, original volume = a 3 New edge : $$\eqalign{ & = \frac{{150}}{{100}}a \cr & = \frac{{3a}}{2} \cr} $$ New volume : $$\eqalign{ & = {\left( {\frac{{3a}}{2}} \right)^3} \cr & = \frac{{27{a^3}}}{8} \cr} $$ Increase in volume : $$\eqalign{ & = \left( {\frac{{27{a^3}}}{8} - {a^3}} \right) \cr & = \frac{{19{a^3}}}{8} \cr} $$ ∴ Increase % : $$\eqalign{ & = \left( {\frac{{19{a^3}}}{8} \times \frac{1}{{{a^3}}} \times 100} \right)\% \cr & = 237.5\% \cr} $$

$$\eqalign{ & V = \pi {r^2}h{\text{ and }} \cr & S = 2\pi rh + 2\pi {r^2} \cr & \,\,\,\,\,\,\, = 2\pi r\left( {h + r} \right) \cr & {\text{Where, }}h = \frac{V}{{\pi {r^2}}} \cr & \Rightarrow S = 2\pi r\left( {\frac{V}{{\pi {r^2}}} + r} \right) \cr & \Rightarrow S = \frac{{2V}}{r} + 2\pi {r^2} \cr & \Rightarrow \frac{{dS}}{{dr}} = \frac{{ - 2V}}{{{r^2}}} + 4\pi r{\text{ and}} \cr & \frac{{{d^2}S}}{{d{r^2}}} = \left( {\frac{{4V}}{{{r^3}}} + 4\pi } \right){\text{ > 0}} \cr} $$ ∴ S is minimum when : $$\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \Rightarrow \frac{{ - 2V}}{{{r^2}}} + 4\pi r = 0 \cr & \Rightarrow V = 2\pi {r^3} \cr & \Rightarrow \pi {r^2}h = 2\pi {r^3} \cr & \Rightarrow h = 2r \cr} $$

Volume of water flown into the tank in 5 min : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 100 \times 100 \times 7} \right){\text{cu}}{\text{.feet}} \cr & = 220000\,{\text{cu}}{\text{.feet}} \cr} $$ ∴ Rate of flow of water : $$\eqalign{ & = \left( {\frac{{220000}}{{5 \times 60}}} \right){\text{cu}}{\text{.feet/sec}} \cr & = 733.3 \approx 700\,{\text{cu}}{\text{.feet/sec}} \cr} $$

h = 15 cm, r = 8 cm So, $$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\,\,\, = \sqrt {{8^2} + {{\left( {15} \right)}^2}} \cr & \,\,\,\,\,\, = 17\,cm \cr} $$ ∴ Curved surface area : $$\eqalign{ & = \pi rl \cr & = \left( {\pi \times 8 \times 17} \right){\text{ c}}{{\text{m}}^2} \cr & = 136\pi {\text{ c}}{{\text{m}}^2} \cr} $$

Let their radius and height be 5x and 12x respectively Slant height of the cone, $$l = \sqrt {{{\left( {5x} \right)}^2} + {{\left( {12x} \right)}^2}} = 13x$$ $$\eqalign{ & \frac{{{\text{Total surface area of cylinder}}}}{{{\text{Total surface area of cone}}}} \cr & = \frac{{2\pi r\left( {h + r} \right)}}{{\pi r\left( {l + r} \right)}} \cr & = \frac{{2\left( {h + r} \right)}}{{\left( {l + r} \right)}} \cr & = \frac{{2 \times \left( {12x + 5x} \right)}}{{\left( {13x + 5x} \right)}} \cr & = \frac{{34x}}{{18x}} \cr & = \frac{{17}}{9}\,Or\,17:9 \cr} $$

$$\eqalign{ & 4\pi {r^2} = 5544 \cr & \Rightarrow {r^2} = \left( {5544 \times \frac{1}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^2} = 441 \cr & \Rightarrow r = 21 \cr} $$ ∴ Volume : $$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 21 \times 21 \times 21} \right){\text{ c}}{{\text{m}}^3} \cr & = 38808{\text{ c}}{{\text{m}}^3} \cr} $$

External radius = 3 cm Internal radius = (3 - 0.5) cm = 2.5 cm Volume of the metal : $$\eqalign{ & = \left[ {\frac{4}{3} \times \frac{{22}}{7} \times \left\{ {{{\left( 3 \right)}^3} - {{\left( {2.5} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times \frac{{91}}{8}} \right){\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{{143}}{3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 47\frac{2}{3}{\text{ c}}{{\text{m}}^3} \cr} $$

Let the height of the cylinder be h cm Then, $$\eqalign{ & \frac{4}{3}\pi \left[ {{{\left( 4 \right)}^3} - {{\left( 2 \right)}^3}} \right] = \pi \times {4^2} \times h \cr & \Rightarrow \frac{4}{3} \times \pi \times 56 = \pi \times 16h \cr & \Rightarrow h = \frac{{4 \times 56}}{{3 \times 16}} \cr & \Rightarrow h = \frac{{14}}{3}\,cm \cr} $$

Let the radius of the hemisphere be be r cm Then, radius of the cylinder = r cm Height of the cylinder = r cm ∴ Required ratio : $$\eqalign{ & = \frac{{{\text{Volume of hemisphere}}}}{{{\text{Volume of cylinder}}}} \cr & = \frac{{\frac{2}{3}\pi {r^3}}}{{\pi {r^2} \times r}} \cr & = \frac{2}{3}\, Or\,2:3 \cr} $$

The volume of cone should be maximum ∴ Radius of the base of cone : $$\eqalign{ & = \frac{{{\text{Edge of cube}}}}{2} \cr & = \frac{{4.2}}{2} \cr & = 2.1\,{\text{dm.}} \cr} $$ Height of cone = Edge of cube = 4.2 dm. ∴ Volume of cone : $$\eqalign{ & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times 2.1 \times 2.1 \times 4.2} \right){\text{cu}}{\text{.dm}}{\text{.}} \cr & = 19.404\,{\text{cu}}{\text{.dm}}{\text{.}} \cr} $$

Volume of hemisphere : $$\eqalign{ & = \left( {\frac{2}{3}\pi \times 3 \times 3 \times 3} \right){m^3} \cr & = \left( {18\pi } \right){m^3} \cr} $$

Required length : $$\eqalign{ & = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 5 \right)}^2}} \,m \cr & = \sqrt {225} \,m \cr & = 15\,m \cr} $$

Let the radius of the cylinder be r and height be h Then, r + h = 19 Again, total surface area of cylinder = $$\left( {2\pi rh + 2\pi {r^2}} \right)$$ Now, $$\eqalign{ & 2\pi r\left( {h + r} \right) = 1672 \cr & \Rightarrow 2\pi r \times 19 = 1672 \cr & \Rightarrow 38\pi r = 1672 \cr & \therefore \pi r = \frac{{1672}}{{38}} = 44\,m \cr & \therefore r = \frac{{44 \times 7}}{{22}} = 14\,m \cr} $$ ∴ Height = 19 - 14 = 5 m Volume of cylinder : $$\eqalign{ & = \pi {r^2}h \cr & = \frac{{22}}{7} \times 14 \times 14 \times 5 \cr & = 22 \times 2 \times 14 \times 5 \cr & = 3080\,{m^3} \cr} $$

Let length = x cm Then, $$\eqalign{ & x \times 28 \times 5 \times \frac{{25}}{{1000}} = 112 \cr & \Rightarrow x = \left( {112 \times \frac{{1000}}{{25}} \times \frac{1}{{28}} \times \frac{1}{5}} \right) \cr & \Rightarrow x = 32\,cm \cr} $$

Clearly, $$l$$ = (20 - 4) cm = 16 cm b = (15 - 4) cm = 11 cm and h = 2 cm ∴ Outer surface area of the box : $$\eqalign{ & = \left[ {2\left( {l + b} \right) \times h} \right] + lb \cr & = \left[ {\left\{ {2\left( {16 + 11} \right) \times 2} \right\} + 16 \times 11} \right] \cr & = \left( {108 + 176} \right) \cr & = 284{\text{ c}}{{\text{m}}^2} \cr} $$

Clearly, we have : V 1 = x 3 , V 2 = (2x) 3 = 8x 3 V 3 = (3x) 3 = 27x 3 V 4 = (4x) 3 = 64x 3 (i) V 1 + V 2 + 2V 3 = x 3 + 8x 3 + 2 × 27x 3 = 63x 3 < V 4 (ii) V 1 + 4V 2 + V 3 = x 3 + 4 × 8x 3 + 27 x 3 = 60x 3 < V 4 (iii) 2(V 1 + V 3 ) + V 2 = 2 (x 3 + 27x 3 ) + 8x 3 = 64x 3 = V 4

Volume : $$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times 1 \times 1 \times 14} \right){m^3} \cr & = 44\,{m^3} \cr} $$