MCQ Questions on Mensuration 3D for Competitive Exams

$$\eqalign{ & \frac{{{\text{Total surface area}}}}{{{\text{Volume}}}} = \frac{1}{7} \cr & \frac{{4\pi {r^2}}}{{\frac{4}{3}\pi {r^3}}} = \frac{1}{7} \cr & r = 21{\text{ cm}} \cr & {\text{Radius of small sphere}} = \frac{1}{6} \times r \cr & = \frac{1}{6} \times 21 \cr & = \frac{7}{2}{\text{ cm}} \cr & {\text{Number of small sphere}} \cr & = \frac{{{\text{Volume of large sphere}}}}{{{\text{Volume of small sphere}}}} \cr & = \frac{{\frac{4}{3}\pi {R^3}}}{{\frac{4}{3}\pi {r^3}}} \cr & = \frac{{21 \times 21 \times 21}}{{\frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}}} \cr & = 27 \times 8 \cr & = 216 \cr & {\text{Curved surface area of small sphere}} \cr & = 216 \times 4\pi {r^2} \cr & = 216 \times 4 \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \cr & = 33264{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & r{\text{ - inradius of incircle of triangle}} \cr & {\text{Perimeter}} = 15{\text{ cm }}\left( {{\text{given}}} \right) \cr & \therefore {\text{Semiperimeter}}\left( S \right) = \frac{{15}}{2}{\text{cm}} \cr & {\text{Inradius of any triangle}} \cr & r \Rightarrow \frac{\Delta }{S} \cr & r = \frac{{{\text{area}}}}{{{\text{semiperimeter}}}} \cr & {\text{Where }}\Delta {\text{ is the area of triangle }} \cr & \therefore r{\text{ }} = {\text{ }}3{\text{ cm }}\left( {{\text{given}}} \right) \cr & \Rightarrow 3 = \frac{{{\text{area of triangle}}}}{{\frac{{15}}{2}}} \cr & \Rightarrow 3 \times \frac{{15}}{2} = {\text{area of triangle}} \cr & \Rightarrow \frac{{45}}{2}{\text{cm}} = {\text{area of triangle}} \cr & \therefore {\text{Volume of prism}} \cr & \Rightarrow 270{\text{ c}}{{\text{m}}^3}\,\left( {{\text{given}}} \right) \cr & \therefore 270 = h \times \frac{{45}}{2} \cr & h = 12{\text{ cm}} \cr} $$

Water displaced by 40 men = 40 × 5 = 200 m³
Rise in level of pool
$$\eqalign{ & = \frac{{200}}{{30 \times 25}} \cr & = \frac{8}{{30}} \times 100 \cr & = 26.66{\text{ cm}} \cr} $$

$$\eqalign{ & {r_1}:{r_2} = 5:6 \cr & {v_1}:{v_2} = 8:9 \cr & \frac{{{v_1}}}{{{v_2}}} = \frac{{\pi r_1^2{h_1}}}{{\pi r_2^2{h_2}}} \cr & \frac{8}{9} = \frac{{25 \times {h_1}}}{{36 \times {h_2}}} \cr & 32:25 = {h_1}:{h_2} \cr} $$

$$\eqalign{ & {\text{Volume of well}} = \pi {r^2}h \cr & = \frac{{22}}{7} \times {2.8^2} \times 15 \cr & {\text{Volume of platform}} = 8 \times 1.5 \times x \cr & {\text{Volume of platform}} = {\text{Volume of well}} \cr & 8 \times 1.5 \times x = \frac{{22}}{7} \times 2.8 \times 2.8 \times 15 \cr & x = 30.8{\text{ m}} \cr} $$

According to the question
Curved surface area = $$\frac{1}{3}$$ × Total surface area
2πrh = $$\frac{1}{3}$$ × 2πr(h + r)
h = 2r . . . . . . (1)
Now total surface area = 2πr(h + r) = 462
By using equation - (1)
2πr(2r + r) = 462
r = 7 cm

Let the radius and height be 5x and 12x
$$\eqalign{ & \Rightarrow \frac{1}{3} \times \pi \times 25{x^2} \times 12x = \frac{{2200}}{7} \cr & \Rightarrow {x^3} = \frac{{2200 \times 7 \times 3}}{{7 \times 22 \times 25 \times 12}} \cr & \Rightarrow x = 1 \cr & \Rightarrow {\text{slant height}} \cr & = \sqrt {{5^2} + {{12}^2}} \cr & = 13{\text{ cm}} \cr} $$

r + h = 20 . . . . . . (i)
Total surface area = 880
2πr(h + r) = 880
2 × $$\frac{{22}}{7}$$ × r × 20 = 880
r = 7, h = 13
Volume = πr²h
= $$\frac{{22}}{7}$$ × 7 × 7 × 13
= 154 × 13
= 2002 cm³

$$\eqalign{ & {\text{Side of square}} = \frac{1}{{\sqrt 2 }} \times 10\sqrt 2 = 10{\text{ cm}} \cr & {\text{Slant height}} = \sqrt {{5^2} + {{12}^2}} = 13{\text{ cm}} \cr & {\text{Lateral surface area}} \cr & = \frac{1}{2} \times {\text{Perimeter of base}} \times {\text{Slant height}} \cr & = \frac{1}{2} \times 40 \times 30 \cr & = 260{\text{ c}}{{\text{m}}^2} \cr} $$

\[\begin{array}{*{20}{c}} {}&{{\text{Sphere}}}&{}&{{\text{Cylinder}}} \\ {{\text{Radius}} \to }&R&:&{3R} \\ {{\text{Height}} \to }& - &{}&{4R} \end{array}\]
$$\eqalign{ & N \times \frac{4}{3}\pi {R^3} = \pi {\left( {3R} \right)^2} \times 4R \cr & N \times \frac{4}{3}{R^3} = 3 \times 4{R^3} \cr & N = 27 \cr} $$