Practice MCQ Questions and Answer on Volume and Surface Area

Solution:

 Clearly, the cylinder formed by rolling the paper along its length has height 18 cm and circumference of base 30 cm i.e., $$\eqalign{ & h = 18{\text{ cm and }} \cr & {\text{2}}\pi r = 30 \cr & Or,r = \frac{{30}}{2} \times \frac{7}{{22}} \cr & Or,r = \frac{{105}}{{22}} \cr} $$ ∴ Volume : $$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times \frac{{105}}{{22}} \times \frac{{105}}{{22}} \times 18} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{14175}}{{11}}{\text{ c}}{{\text{m}}^3} \cr} $$ The cylinder formed by rolling the paper along its breadth has height 30 cm and circumference of base 18 cm i.e., $$\eqalign{ & h = 30{\text{ cm and }} \cr & {\text{2}}\pi r = 18 \cr & Or,r = \frac{{18}}{2} \times \frac{7}{{22}} \cr & Or,r = \frac{{63}}{{22}} \cr} $$ ∴ Volume : $$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times \frac{{63}}{{22}} \times \frac{{63}}{{22}} \times 30} \right){\text{ c}}{{\text{m}}^3} \cr & = \frac{{8505}}{{11}}{\text{ c}}{{\text{m}}^3} \cr} $$ Required ratio : $$\eqalign{ & = \frac{{14175}}{{11}}:\frac{{8505}}{{11}} \cr & = 5:3 \cr} $$

Solution:

 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 : $$ = 2R = \frac{5}{2}cm = 2.5\,cm$$

Solution:

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

Solution:

 Let their heights be 2h and 3h and radii be r and R respectively Then, $$\eqalign{ & \pi {r^2}\left( {2h} \right) = \pi {R^2}\left( {3h} \right) \cr & \Rightarrow \frac{{{r^2}}}{{{R^2}}} = \frac{3}{2} \cr & \Rightarrow \frac{r}{R} = \frac{{\sqrt 3 }}{{\sqrt 2 }}\,i.e.,\sqrt 3 :\sqrt 2 \cr} $$

Solution:

 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 = 6x3 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 = 108x3 cu.units Increase in volume : = (108x3 - 6x3) cu.units = (102x3) cu.units ∴ Required ratio : $$\eqalign{ & = \frac{{102{{\text{x}}^3}}}{{6{{\text{x}}^3}}} \cr & = 17{\text{ }}\left( {{\text{Times}}} \right) \cr} $$

Solution:

 Volume : $$\eqalign{ & = \pi {r^2}h \cr & = \left( {\frac{{22}}{7} \times 14 \times 14 \times 3.5} \right){m^3} \cr & = 2156\,{m^3} \cr} $$ ∴ Cost of the cylinder : $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {2156 \times 50} \right) \cr & = {\text{Rs}}{\text{. 107800}} \cr} $$

Solution:

 Volume of cube = Volume of sheet = (27 × 8 × 1) cm3 = 216 cm3 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 : $$\eqalign{ & = \left( {502 - 216} \right){\text{ c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

Solution:

 Volume of sphere : $$\eqalign{ & = \left( {\frac{4}{3}\pi \times {6^3}} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {288\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$ Volume of each cone : $$\eqalign{ & = \left( {\frac{1}{3}\pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^{\text{3}}} \cr & = \left( {12\pi } \right){\text{c}}{{\text{m}}^{\text{3}}} \cr} $$ ∴ Number of cone : $$\eqalign{ & = \frac{{288\pi }}{{12\pi }} \cr & = 24 \cr} $$

Solution:

 Radius, r = 12 m Height of conical part, h = (16 - 11) m = 5 m Slant height of conical part, $$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\, = \sqrt {{{\left( {12} \right)}^2} + {{\left( 5 \right)}^2}} \cr & \,\,\,\, = \sqrt {169} \cr & \,\,\,\, = 13\,m \cr} $$ Height of cylindrical part, H = 11 m Area of canvas required : = Curved surface area of cylinder + Curved surface area of cone$$\eqalign{ & = 2\pi rH + \pi rl \cr & = \left[ {\frac{{22}}{7}\left( {2 \times 12 \times 11 + 12 \times 13} \right)} \right]{{\text{m}}^2} \cr & = \left[ {\frac{{22}}{7}\left( {264 + 156} \right)} \right]{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 420} \right){{\text{m}}^2} \cr & = 1320\,{{\text{m}}^2} \cr} $$

Solution:

 Let the length of each edge of small cube be a1 and that of larger cube be a2 Then, $$\eqalign{ & 6a_1^2 = 96\, \cr & \Rightarrow a_1^2 = 16 \cr & \Rightarrow {a_1} = 4\,and \cr & 6a_2^2 = 384 \cr & \Rightarrow a_2^2 = 64 \cr & \Rightarrow {a_2} = 8 \cr} $$ ∴ Number of cubes formed : $$\eqalign{ & = \frac{{{\text{Volume of larger cube}}}}{{{\text{Volume of smaller cube}}}} \cr & = \left( {\frac{{8 \times 8 \times 8}}{{4 \times 4 \times 4}}} \right) \cr & = 8 \cr} $$