Practice MCQ Questions and Answer on Volume and Surface Area

Solution:

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

Solution:

 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 : $$\eqalign{ & = \pi rl \cr & = \left( {\pi \times 8 \times 17} \right){\text{ c}}{{\text{m}}^2} \cr & = 136\pi {\text{ c}}{{\text{m}}^2} \cr} $$

Solution:

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

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:

 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:

 $$\eqalign{ & 4\pi {R^2} = 6{a^2} \cr & \Rightarrow \frac{{{R^2}}}{{{a^2}}} = \frac{3}{{2\pi }} \cr & \Rightarrow \frac{R}{a} = \frac{{\sqrt 3 }}{{\sqrt {2\pi } }} \cr} $$ $$\eqalign{ & \therefore \frac{{{\text{Volume of spere}}}}{{{\text{Volume of cube}}}} \cr & = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}} \cr & = \frac{4}{3}\pi {\left( {\frac{R}{a}} \right)^3} \cr & = \frac{4}{3}\pi \frac{{3\sqrt 3 }}{{2\pi \sqrt {2\pi } }} \cr & = \frac{{2\sqrt 3 }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt {12} }}{{\sqrt {2\pi } }} \cr & = \frac{{\sqrt 6 }}{{\sqrt \pi }} \cr & \text{or, }\sqrt 6 :\sqrt \pi \cr} $$

Solution:

 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 : $$\eqalign{ & = \sqrt {{l^2} + {b^2} + {h^2}} \cr & = \sqrt {65} \cr & = 8.05\,cm \cr} $$

Solution:

 Let the length of base be 3a cm and breadth be 2a cm Total surface area of prism : = [Perimeter of base × height] + [2 × Area of base] = [2 (3a + 2a) × 12 + 2 × 3a × 2a] sq.cm = (120a + 12a2) sq.cm According to the question, 120a + 12a2 = 288 ⇒ a2 + 10a = 24 ⇒ a2 + 10a - 24 = 0 ⇒ a2 + 12a - 2a - 24 = 0 ⇒ a (a + 12) - 2 (a + 12) = 0 ⇒ (a - 2)(a + 12) = 0 ⇒ a = 2 because a $$ \ne $$ -12 ∴ Volume of prism : = Area of base × Height = (3a × 2a × 12)cu.cm = 72a2 cu.cm = (72 × 2 × 2)cu.cm = 288 cu.cm

Solution:

 Let their radii be 3x, 4x and heights be 4y, 3y Ratio of their volumes : $$\eqalign{ & = \frac{{\pi \times {{\left( {3x} \right)}^2} \times 4y}}{{\pi \times {{\left( {4x} \right)}^2} \times 3y}} \cr & = \frac{{36}}{{48}} \cr & = \frac{3}{4}\,Or\,3:4 \cr} $$

Solution:

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