Volume And Surface Area — Part 1 MCQs with Explanations

$$\eqalign{ & {\text{Clearly}}, \cr & {\text{We have }}r = 3\,{\text{cm}}\,{\text{and}}\,h = 4\,{\text{cm}} \cr & \therefore \,\,\,{\text{Volume}} \cr & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^3} \cr & = 12\pi \,{\text{c}}{{\text{m}}^3} \cr} $$

$$\eqalign{ & {\text{1}}\,{\text{hectare}} = 10000\,{m^2} \cr & {\text{So,}}\,{\text{Area}} = \left( {1.5 \times 10000} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15000\,{m^2} \cr & {\text{Depth}} = \frac{5}{{100}}m = \frac{1}{{20}}m \cr & \therefore {\text{Volume}} = \left( {{\text{Area}}\, \times \,{\text{Depth}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {15000 \times \frac{1}{{20}}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 750\,{m^3} \cr} $$

$$\eqalign{ & 2\left( {15 + 12} \right) \times h = 2\left( {15 \times 12} \right) \cr & \Rightarrow h = \frac{{180}}{{27}}m = \frac{{20}}{3}m \cr & \therefore {\text{Volume}} = \left( {15 \times 12 \times \frac{{20}}{3}} \right){m^3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200\,{m^3} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{wire}}\,{\text{be h}} \cr & {\text{Radius}} = \frac{1}{2}mm = \frac{1}{{20}}cm.\,{\text{Then}}, \cr & \Rightarrow \frac{{22}}{7} \times \frac{1}{{20}} \times \frac{1}{{20}} \times h = 66 \cr & \Rightarrow h = {\frac{{66 \times 20 \times 20 \times 7}}{{22}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8400\,cm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 84\,meters \cr} $$

$$\eqalign{ & {\text{External}}\,{\text{radius}} = 4\,cm \cr & {\text{Internal}}\,{\text{radius}} = 3\,cm \cr & {\text{Volume}}\,{\text{of}}\,{\text{iron}} \cr & = \left( {\frac{{22}}{7} \times \left[ {{{\left( 4 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times 21} \right)c{m^3} \cr & = \left( {\frac{{22}}{7} \times 7 \times 1 \times 21} \right)c{m^3} \cr & = 462\,c{m^3} \cr & \therefore {\text{Weight}}\,{\text{of}}\,{\text{iron}} = \left( {462 \times 8} \right)gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3696\,gm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3.696\,kg \cr} $$

Volume of water displaced = (3 x 2 x 0.01) m 3 = 0.06 m 3 . ∴ Mass of man = Volume of water displaced x Density of water = (0.06 x 1000) kg = 60 kg.

$$\eqalign{ & {\text{Total}}\,{\text{Volume}}\,{\text{of}}\,{\text{water}}\,{\text{displaced}} \cr & = \left( {4 \times 50} \right){m^3} = 200\,{m^3} \cr & \therefore {\text{Rise}}\,{\text{in}}\,{\text{water}}\,{\text{level}} \cr & = \left( {\frac{{200}}{{40 \times 20}}} \right)m \cr & = 0.25\,m \cr & = 25\,cm \cr} $$

$$\eqalign{ & l = 10\,m \cr & h = 8\,m \cr & So,\,r = \sqrt {{l^2} - {h^2}} = \sqrt {{{\left( {10} \right)}^2} - {8^2}} = 6\,m \cr & \therefore {\text{Curved}}\,{\text{surface}}\,{\text{area}} \cr & \pi \,rl = \left( {\pi \times 6 \times 10} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60\pi \,{m^2} \cr} $$

$$\eqalign{ & {\text{Area}}\,{\text{of}}\,{\text{the}}\,{\text{wet}}\,{\text{surface}} \cr & = \left[ {2\left( {lb + bh + lh} \right) - lb} \right] \cr & = 2\left( {bh + lh} \right) + lb \cr & = \left[ {2\left( {4 \times 1.25 + 6 \times 1.25} \right) + 6 \times 4} \right]{m^2} \cr & = 49\,{m^2} \cr} $$

Clearly, l = (48 - 16)m = 32 m, b = (36 -16)m = 20 m, h = 8 m. ∴ Volume of the box = (32 x 20 x 8) m 3 = 5120 m 3

$$\eqalign{ & \frac{{\pi \,{r^2}h}}{{2\pi \,rh}} = \frac{{924}}{{264}}\, \cr & \Rightarrow r = {\frac{{924}}{{264}} \times 2} = 7m \cr & {\text{And}},\,2\pi rh = 264 \cr & \Rightarrow h = {264 \times \frac{7}{{22}} \times \frac{1}{2} \times \frac{1}{7}} = 6m \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = \frac{{2r}}{h} = \frac{{14}}{6} = 7:3 \cr} $$

Let the thickness of the bottom be x cm Then, $${\mkern 1mu} {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} = $$ $$8000 \times $$ $$1000$$ $$ \Rightarrow 320 \times 250 \times \left( {110 - x} \right) = 8000 \times 1000$$ $$\eqalign{ & \Rightarrow \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}} = 100 \cr & \Rightarrow x = 10\,{\text{cm}} = 1\,{\text{dm}} \cr} $$

$$\eqalign{ & h = 14\,cm,\,r = 7\,cm \cr & {\text{So}},\,l = \sqrt {{{\left( 7 \right)}^2} + {{\left( {14} \right)}^2}} = \sqrt {245} = 7\sqrt 5 \,cm \cr & \therefore {\text{Total}}\,{\text{surface}}\,{\text{area}} \cr & = \pi \,rl + \pi \,{r^2} \cr & = \left( {\frac{{22}}{7} \times 7 \times 7\sqrt 5 + \frac{{22}}{7} \times 7 \times 7} \right)c{m^2} \cr & = \left[ {154\left( {\sqrt 5 + 1} \right)} \right]c{m^2} \cr & = \left( {154 \times 3.236} \right)c{m^2} \cr & = 498.35\,c{m^2} \cr} $$

$$\eqalign{ & {\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}} \cr & = \left( {{3^3} + {4^3} + {5^3}} \right) = 216\,c{m^3} \cr & {\text{Let}}\,{\text{the}}\,{\text{edge}}\,{\text{of}}\,{\text{the}}\,{\text{large}}\,{\text{cube}}\,{\text{be}}\,a \cr & So,\,{a^3} = 216\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 6\,cm \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = {\frac{{6 \times \left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}}} \cr & = \frac{{50}}{{36}} \cr & = 25:18 \cr} $$

$$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{bricks}} \cr & = \frac{{{\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{wall}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{1}}\,{\text{brick}}}} \cr & = {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \cr & = 6400 \cr} $$

Surface area : $$\eqalign{ & = {2\left( {7 \times 11 + 11 \times 13 + 7 \times 13} \right)} \cr & = {2 \times 311} \cr & = 622{\text{ c}}{{\text{m}}^2} \cr} $$

let l, b and h denote the length, breadth and depth of Meeta's lunch box Then, length of Rita's lunch box : $$ = 110\% {\text{ of }}l = \frac{{11l}}{{10}}$$ Breadth of Rita's lunch box : $$ = 110\% {\text{ of }}b = \frac{{11b}}{{10}}$$ Depth of Rita's lunch box : $$ = 80\% {\text{ of }}h = \frac{{4h}}{5}$$ ∴ Ratio of the capacities of Rita's and Meeta's lunch boxes : $$\eqalign{ & = \frac{{11l}}{{10}} \times \frac{{11b}}{{10}} \times \frac{{4h}}{5}:lbh \cr & = \frac{{121}}{{125}}:1 \cr & = 121:125 \cr} $$

Volume of the tank : $$\eqalign{ & = \left( {8 \times 100 \times 6 \times 100 \times 2.5 \times 100} \right)c{m^3} \cr & = 120000000\,c{m^3} \cr & = \left( {\frac{{120000000}}{{1000}}} \right){\text{litres}} \cr & = 120000{\text{ litres}} \cr} $$

Surface area : $$\eqalign{ & = \left[ {6 \times {{\left( {\frac{1}{2}} \right)}^2}} \right]{\text{ c}}{{\text{m}}^2} \cr & = \frac{3}{2}{\text{ c}}{{\text{m}}^2} \cr} $$

Volume of the larger cube : $$\eqalign{ & = \left( {{3^3} + {4^3} + {5^3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 216{\text{ c}}{{\text{m}}^3} \cr} $$ Let the edge of the larger cube be a cm $$\eqalign{ & \therefore {a^3} = 216 \cr & \Rightarrow a = 6 \cr} $$ Required ratio : $$\eqalign{ & = \frac{{6\left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}} \cr & = \frac{{6 \times 50}}{{6 \times 36}} \cr & = \frac{{25}}{{18}}\,Or\,25:18 \cr} $$

$$\eqalign{ & {\text{Number of tins}} \cr & = \frac{{{\text{Voulme of the drum}}}}{{{\text{Volume of each tin}}}} \cr & = \frac{{\left( {\frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times 24} \right)}}{{\left( {\frac{{25}}{{10}} \times \frac{{22}}{{10}} \times \frac{{35}}{{10}}} \right)}} \cr & = 1200 \cr} $$

Volume of cistern : $$\eqalign{ & = \left( {\pi \times {{10}^2} \times 15} \right){m^3} \cr & = 1500\pi {m^3} \cr} $$ Volume of water flowing through the pipe in 1 sec : $$\eqalign{ & = \left( {\pi \times 0.25 \times 0.25 \times 5} \right){m^3} \cr & = 0.3125\pi {m^3} \cr} $$ ∴ Time taken to fill the cistern : $$\eqalign{ & = \left( {\frac{{1500\pi }}{{0.3125\pi }}} \right) \cr & = \left( {\frac{{1500 \times 10000}}{{3125}}} \right) \cr & = 4800\,\sec \cr & = \left( {\frac{{4800}}{{60}}} \right)\min \cr & = 80\,\min \cr} $$

External radius = 6 cm Internal radius = (6 - 0.25) = 5.75 cm Volume of material in hollow cylinder : $$\eqalign{ & = \left[ {\frac{{22}}{7} \times \left\{ {{{\left( 6 \right)}^2} - {{\left( {5.75} \right)}^2}} \right\} \times 15} \right]{\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times 11.75 \times 0.25 \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times \frac{{1175}}{{100}} \times \frac{{25}}{{100}} \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{11 \times 705}}{{56}}} \right){\text{c}}{{\text{m}}^3} \cr} $$ Let the length of solid cylinder be h Then, $$\eqalign{ & \frac{{22}}{7} \times 1 \times 1 \times h = \left( {\frac{{11 \times 705}}{{56}}} \right) \cr & \Rightarrow h = \left( {\frac{{11 \times 705}}{{56}} \times \frac{7}{{22}}} \right) \cr & \Rightarrow h = 44.0625\,{\text{cm}} \cr} $$

Let the original radius and height of the cone be r and h respectively Then, Original volume = $$\frac{1}{3}\pi {r^2}h$$ New radius = $$\frac{r}{2}$$ and new hight = 2h New volume : $$\eqalign{ & = \frac{1}{3} \times \pi \times {\left( {\frac{r}{2}} \right)^2} \times 3h \cr & = \frac{3}{4} \times \frac{1}{3}\pi {r^2}h \cr} $$ ∴ Decrease % : $$\eqalign{ & = \left( {\frac{{\frac{1}{4} \times \frac{1}{3}\pi {r^2}h}}{{\frac{1}{3}\pi {r^2}h}} \times 100} \right)\% \cr & = 25\% \cr} $$

Volume of bucket : = 28.490 litres = (28.490 ×1000) cm 3 = 28490 cm 3 Let the height of the bucket be h cm We have : r = 21 cm. and R = 28 cm $$\therefore \frac{\pi }{3}h\left[ {{{\left( {28} \right)}^2} + {{\left( {21} \right)}^2} + 28 \times 21} \right]$$ $$ = 28490$$ $$ \Rightarrow h\left( {784 + 441 + 588} \right) = $$ $$\frac{{28490 \times 21}}{{22}}$$ $$\eqalign{ & \Rightarrow 1813h = 27195 \cr & \Rightarrow h = \frac{{27195}}{{1813}} \cr & \Rightarrow h = 15\,cm \cr} $$

$$\eqalign{ & \frac{4}{3}\pi {r^3} = 4\pi {r^2} \cr & \Rightarrow r = 3 \text{ units} \cr} $$

Let the radius of the sphere be r cm Then, $$\eqalign{ & \frac{4}{3}\pi {r^3} = \pi \times {\left( {0.1} \right)^2} \times 3600 \cr & \Leftrightarrow {r^3} = 36 \times \frac{3}{4} \cr & \Leftrightarrow {r^3} = 27 \cr & \Leftrightarrow r = 3\,cm \cr} $$

Let their radii be R and r Then, $$\eqalign{ & \frac{{\frac{2}{3}\pi {R^3}}}{{\frac{2}{3}\pi {r^3}}} = \frac{{6.4}}{{21.6}} \cr & \Rightarrow {\left( {\frac{R}{r}} \right)^3} = \frac{8}{{27}} \cr & \Rightarrow {\left( {\frac{R}{r}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \Rightarrow \frac{R}{r} = \frac{2}{3} \cr} $$ ∴ Ratio of curved surface area : $$ = \frac{{2\pi {R^2}}}{{2\pi {r^2}}} = {\left( {\frac{R}{r}} \right)^2} = \frac{4}{9}\,or\,4:9$$

Area of the base : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$ ∴ Volume of pyramid : $$\eqalign{ & = \left( {\frac{1}{3} \times \frac{{\sqrt 3 }}{4} \times 4} \right){m^3} \cr & = \left( {\frac{{\sqrt 3 }}{3}} \right){m^3} \cr & = \left( {\frac{{1.732}}{3}} \right){m^3} \cr & = 0.577\,{m^3} \cr} $$

Let the side of cube = 10 cm ∴ Original volume = 10 × 10 × 10 = 1000 cm 3 Now, side of new cube : = 10 - 25% of 10 = 7.5 ∴ New volume = 7.5 × 7.5 × 7.5 = 421.875 cm 3 ∴ Required ratio : $$\eqalign{ & = \frac{{1000}}{{421.875}} \cr & = \frac{{1000000}}{{421875}} \cr & = \frac{{64}}{{27}}\,Or\,64:27 \cr} $$

Ball is dropped from the height of 36 m when the ball will rise at the third bounce ∴ Required height : $$\eqalign{ & = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times 36 \cr & = \frac{{32}}{3} \cr & = 10\frac{2}{3}\,m \cr} $$

Sum of perimeters of the six faces : $$\eqalign{ & = 2\left[ {2\left( {l + b} \right) + 2\left( {b + h} \right) + 2\left( {l + h} \right)} \right] \cr & = 4\left( {2l + 2b + 2h} \right) \cr & = 8\left( {l + b + h} \right) \cr} $$ Total surface area $$ = 2\left( {lb + bh + lh} \right)$$ $$\eqalign{ & \therefore 8\left( {l + b + h} \right) = 72 \cr & \Rightarrow l + b + h = 9 \cr & 2\left( {lb + bh + lh} \right) = 16 \cr & \Rightarrow lb + bh + lh = 8 \cr} $$ Now, $${\left( {l + b + h} \right)^2} = {l^2} + {b^2} + {h^2} + 2$$ $$\left( {lb + bh + lh} \right)$$ $$\eqalign{ & \Rightarrow {\left( 9 \right)^2} = {l^2} + {b^2} + {h^2} + 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 81 - 16 \cr & \Rightarrow {l^2} + {b^2} + {h^2} = 65 \cr} $$ Required length : $$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$

$$\eqalign{ & 6{a^2} = 150 \cr & \Rightarrow {a^2} = 25 \cr & \Rightarrow a = 5 \cr} $$ ∴ Volume : $${a^3} = {5^3} = 125\,{\text{ c}}{{\text{m}}^3}$$

Let the length of each edge of each cube be a Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a Surface area of the cuboid : $$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$ Sum of surface area of 3 cubes : $$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$ ∴ Required ratio : $$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$

Let the inner radius of the pipe be r metres Then, Volume of water flowing through the pipe in 10 minutes : $$\eqalign{ & = \left[ {\left( {\frac{{22}}{7} \times {r^2} \times 7} \right) \times 10} \right]{m^3} \cr & = \left( {220{r^2}} \right){m^3} \cr & \therefore 220{r^2} = 440 \cr & \Rightarrow {r^2} = 2 \cr & \Rightarrow r = \sqrt 2 \, m \cr } $$

Answer :- A right-angled triangle.

Let the heights of two cones be 7x and 3x and their radii be 6y and 7y respectively Then, Ratio of volume : $$\eqalign{ & = \frac{{\frac{1}{3}\pi \times {{\left( {6y} \right)}^2} \times 7x}}{{\frac{1}{3}\pi \times {{\left( {7y} \right)}^2} \times 3x}} \cr & = \frac{{36 \times 7}}{{49 \times 3}} \cr & = \frac{{12}}{7}\,Or\,12:7 \cr} $$

Volume of parallelepiped : $$\eqalign{ & = \left( {5 \times 3 \times 4} \right)c{m^3} \cr & = 60\,c{m^3} \cr} $$ Volume of cube : $$\eqalign{ & = {\left( 4 \right)^3}\,c{m^3} \cr & = 64\,c{m^3} \cr} $$ Volume of cylinder : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 3 \times 3 \times 3} \right)c{m^3} \cr & = 84.86\,c{m^3} \cr} $$ Volume of spare : $$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 3 \times 3 \times 3} \right) \cr & = 113.14\,c{m^3} \cr} $$ So, Option D is correct decreasing order

Volume of new sphere : $$\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( 6 \right)}^3} + \frac{4}{3}\pi \times {{\left( 8 \right)}^3} + \frac{4}{3}\pi \times {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \left\{ {{{\left( 6 \right)}^3} + {{\left( 8 \right)}^3} + {{\left( {10} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3}\pi \times 1728} \right){\text{c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \times {{\left( {12} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr} $$ Let the radius of the new sphere be R Then, $$\eqalign{ & \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \times {\left( {12} \right)^3} \cr & \Rightarrow R = 12\,cm \cr} $$ ∴ Diameter : $$\eqalign{ & = 2R \cr & = 2 \times 12 \cr & = 24\,cm \cr} $$

Let the length of the wire be h Then, $$\eqalign{ & \pi \times \frac{3}{{20}} \times \frac{3}{{20}} \times h = \frac{4}{3}\pi \times 4 \times 4 \times 4 \cr & \Rightarrow h = \left( {\frac{{4 \times 4 \times 4 \times 4 \times 20 \times 20}}{{3 \times 3 \times 3}}} \right)cm \cr & \Rightarrow h = \left( {\frac{{102400}}{{27}}} \right)cm \cr & \Rightarrow h = 3792.5\,cm \cr & \Rightarrow h = 37.9\,m \cr} $$