MCQ Questions on Mensuration 3D for Competitive Exams

Volume of vessel = Volume of roof
π × r² × h = 22 × 20 × x   (where x is rainfall in cm)
$$\eqalign{ & \Rightarrow \frac{{22}}{7} \times \frac{{100 \times 100 \times 350}}{{22 \times 20 \times 100 \times 100}} = {\text{x}} \cr & \Rightarrow {\text{x}} = 2.5{\text{ cm}} \cr} $$

Lateral surface area of prism = 120
Base perimeter × height = 120
3 × (side) × height = 120
(Perimeter of equilateral Δ = 3 × side)
Side × height = $$\frac{{120}}{3}$$  = 40 . . . . . (i)
Volume of prism = 40√3
Area of base × height = 40√3
$$\frac{{\sqrt 3 }}{4}$$ (side)² × height = 40√3
(side)² × height = $$ = \frac{{40\sqrt 3 \times 4}}{{\sqrt 3 }}$$   = 160 . . . . . (ii)
Dividing equation (ii) by equation (i)
$$\frac{{{{\left( {{\text{side}}} \right)}^2} \times {\text{height}}}}{{{\text{side}} \times {\text{height}}}} = \frac{{160}}{{40}}$$
side = 4 cm

Let the side of cube is 3 m

Surface are of cube = 6 × 3² = 54
Surface are of one cuboid = 2($$l$$b + bh + h$$l$$)
= 2(1 × 3 + 3 × 3 + 3 × 1)
= 2(3 + 9 + 3)
= 30
Cube : 2 Cuboid = 54 : 2 × 30 = 9 : 10

$$\eqalign{ & {\text{Volume of lead}} = \frac{4}{3}\pi {r^3} \cr & {\text{Volume of gold}} = \frac{4}{3}\pi {R^3} - \frac{4}{3}\pi {r^3} \cr & {\text{According to the question,}} \cr & \frac{4}{3}\pi {R^3} - \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {r^3} \cr & \frac{4}{3}\pi {R^3} = \frac{8}{3}\pi {r^3} \cr & {R^3} = 2{r^3} \cr & {R^3} = 2{\left( 2 \right)^3} \cr & R = \root 3 \of 2 \times 2 \cr & R = 1.259 \times 2 \cr & R = 2.518 \cr & \therefore {\text{Thickness}} = R - r \cr & = 2.518 - 2 \cr & = 0.518{\text{ cm}} \cr} $$

Volume ⇒ $$\frac{1}{3}$$πr²h
⇒ $$\frac{1}{3}$$π × 12 × 12 × 9
⇒ 144 × 3π
⇒ 432π

$$\eqalign{ & {\text{Volume of a cone}} = \frac{1}{3}\pi {r^2}h \cr & \frac{1}{3}\pi {r^2}\left( {24} \right) = 1232{\text{ c}}{{\text{m}}^2} \cr & {r^2} = \frac{{1232 \times 3 \times 7}}{{24 \times 22}} \cr & {r^2} = 7 \times 7 \cr & r = \sqrt {7 \times 7} \cr & r = 7{\text{ cm}} \cr & l = \sqrt {{r^2} + {h^2}} \cr & = \sqrt {{7^2} + {{24}^2}} \cr & = \sqrt {625} \cr & = 25 \cr & {\text{Curved surface area}} = \pi rl \cr & = \frac{{22}}{7} \times 7 \times 25 \cr & = 550{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & \frac{{2\pi rh}}{{\pi {r^2}}} = \frac{5}{1} \cr & \frac{{2h}}{r} = \frac{5}{1} \cr & r:h = 2:5 \cr} $$

External radius R = 4 cm
Internal radius r = 3 cm
Volume of iron used = πR²h - πr²h
= πh(R² - r²)
= πh(R + r)(R - r)
= $$\frac{{22}}{7}$$ × 20 × (4 + 3) × (4 - 3)
= $$\frac{{22}}{7}$$ × 20 × 7 × 1
= 440 cm³

$$\eqalign{ & {\text{Volume of sphere}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3}\pi {\left( 9 \right)^3} \cr & = 972\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{Volume of cone}} = \frac{1}{3}\pi {r^2}h \cr & = \frac{1}{3}\pi {\left( 9 \right)^2}9 \cr & = 243\pi {\text{ c}}{{\text{m}}^3} \cr & \Rightarrow \% {\text{ of wasted wood}} \cr & = \frac{{\left( {972 - 243} \right)\pi }}{{972\pi }} \times 100 \cr & = 75\% \cr} $$

We know that
A parallelopiped has vertices (v) = 8
edge (e) = 12
face (f) = 6
Put into equation (v - e + f)
⇒ 8 - 12 + 6
⇒ 2