MCQ Questions on Mensuration 3D for Competitive Exams

Radius of semi-circular sheet = r ⇒ $$\frac{{28}}{2}$$
r = 14 cm
Circumference of sheet = πr = 14π cm
Sheet is folded to form a cone
Let radius of cone = r₁

∴ The circumference of base of cone ⇒ Circumference of sheet
∴ 2πr₁ = 14π
r₁ = 7 cm
∴ Radius of cone = 7 cm
Slant height = Radius of semi-circular sheet
r = 14 cm
$$\eqalign{ & \therefore {\text{Height}} = \sqrt {{{\left( {14} \right)}^2} - {{\left( 7 \right)}^2}} \cr & = \sqrt {147} \cr & = 12{\text{ cm}}\,\,\,\left( {{\text{approx}}} \right) \cr} $$

$$\eqalign{ & 7.7{\text{ Ltr/sec}} \cr & r = 10.5{\text{ cm}} \cr & {\text{V}} = \frac{2}{3}\pi {r^3} \times \frac{2}{3} \cr & = \frac{{\frac{2}{3} \times \frac{2}{3} \times \frac{{22}}{7} \times 10.5 \times 10.5 \times 10.5 \times 1000}}{{7.7 \times 60 \times 6000}} \cr & = \frac{{20 \times 105}}{{36}} \cr & = \frac{{175}}{3}{\text{ hours}} \cr} $$

$$\eqalign{ & {\text{Slant height of the cone }}\left( l \right) \cr & = \sqrt {{6^2} + {8^2}} \cr & = 10{\text{ cm}} \cr & {\text{Required ratio}} = \frac{{2\pi rh}}{{\pi rl}} \cr & = \frac{{2h}}{l} \cr & = \frac{{2 \times 8}}{{10}} \cr & = 8:5 \cr} $$

$$\eqalign{ & 2\pi rh = 264{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \pi {r^2}h = 924{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{On dividing}}:\frac{{2\pi rh}}{{\pi {r^2}h}} = \frac{{264}}{{924}} \cr & \frac{2}{r} = \frac{{264}}{{924}} \cr & r = \frac{{924 \times 2}}{{264}} \cr & r = 7{\text{ cm}} \cr & {\text{Diameter}} = 2r = 2 \times 7 = 14{\text{ cm}} \cr & {\text{Putting, }}r = 7{\text{ in equation }}\left( {\text{i}} \right) \cr & 2\pi rh = 264 \cr & h = \frac{{264 \times 7}}{{2 \times 22 \times 7}} \cr & h = 6{\text{ cm}} \cr & {\text{Required ratio}} = \frac{{2r}}{h} = \frac{{14}}{6} = \frac{7}{3} \cr} $$

In a solid by Euler's formula
Faces + Vertices - 2 = Edges
⇒ Faces + 12 - 2 = 30
⇒ Faces = 30 - 10 = 20

According to the question,
Diameter = 84 cm
Radius = 42 cm = 0.42 m
Height = 120 cm = 1.2 m
CSA of cylinder = 2πrh
$$ = 2 \times \frac{{22}}{7} \times 0.42 \times 1.2$$
Total area covered to planed in 500 rounds
$$\eqalign{ & = 2 \times \frac{{22}}{7} \times 0.42 \times 1.2 \times 500 \cr & {\text{Total cost}} = \left( {2 \times \frac{{22}}{7} \times 0.42 \times 1.2 \times 500} \right) \times 1.50 \cr & = {\text{Rs}}{\text{. }}2376 \cr} $$

$$\eqalign{ & 2.1{\text{ m}} \times 1.5{\text{ m}} \times h = 630{\text{ lt}} \cr & \frac{{21}}{{10}}{\text{m}} \times \frac{{15}}{{10}}{\text{m}} \times h = \frac{{630}}{{1000}}{{\text{m}}^3} \cr} $$
\[\left[ \begin{gathered} \because 1\,{{\text{m}}^3} = 1000{\text{ lt}} \hfill \\ {\text{1000 c}}{{\text{m}}^3} = 1{\text{ lt}} \hfill \\ \end{gathered} \right]\]
$$h = \frac{1}{5}{\text{m}} = 0.20{\text{ metre}}$$

$$\eqalign{ & {\text{V}} = \frac{1}{3}\pi \left( {{R^2} + {r^2} + Rr} \right)h \cr & {\text{V}} = \frac{1}{3} \times \frac{{22}}{7}\left( {25 + 9 + 15} \right) \times 10.5 \cr & {\text{V}} = 11 \times 49 \cr & {\text{V}} = 539{\text{ c}}{{\text{m}}^2} \cr} $$

\[\begin{array}{*{20}{c}} {}&{{\text{Before}}}&{}&{{\text{After}}}&{} \\ {{\text{Radius}} \to }&5&:&{11}&{120\% \to \frac{{ + 6}}{5}} \\ {{\text{Height}} \to }&5&:&3&{40\% \to \frac{{ - 2}}{5}} \\ {{\text{Volume}} \to }&{{5^2} \times 5}&:&{{{11}^2} \times 3}&{} \end{array}\]
         
$$\% \,{\text{Increase}} \Rightarrow \frac{{238}}{{125}} \times 100 = 190.4\% $$

$$\eqalign{ & {r_1}:{r_2} = 3:2 \cr & {v_1}:{v_2} = 27:16 \cr & {\text{Height}} \to {h_1}:{h_2} \cr & \frac{{9 \times {h_1}}}{{4 \times {h_2}}} = \frac{{27}}{{16}} \cr & \frac{{{h_1}}}{{{h_2}}} = \frac{3}{4} \cr & {h_1}:{h_2} = 3:4 \cr} $$