let their radii be x and 2x, and their heights be h and H respectively
Then,
$$\eqalign{ & \frac{{\frac{1}{3} \times \pi \times {x^2} \times h}}{{\frac{1}{3} \times \pi \times {{\left( {2x} \right)}^2} \times H}} = \frac{2}{3} \cr & \Leftrightarrow \frac{h}{H} = \frac{2}{3} \times 4 \cr & \Leftrightarrow \frac{h}{H} = \frac{8}{3}\,Or\,8:3 \cr} $$

$$\eqalign{ & \frac{4}{3} \times \frac{{22}}{7} \times {r^3} = \frac{{45056}}{{21}} \cr & \Rightarrow {r^3} = \left( {\frac{{45056}}{{21}} \times \frac{3}{4} \times \frac{7}{{22}}} \right) \cr & \Rightarrow {r^3} = 512 \cr & \Rightarrow r = \root 3 \of {512} \cr & \Rightarrow r = 8\,cm \cr} $$

Let the length of the wire be h
Then,
$$\eqalign{ & \pi \times \frac{3}{{20}} \times \frac{3}{{20}} \times h = \frac{4}{3}\pi \times 4 \times 4 \times 4 \cr & \Rightarrow h = \left( {\frac{{4 \times 4 \times 4 \times 4 \times 20 \times 20}}{{3 \times 3 \times 3}}} \right)cm \cr & \Rightarrow h = \left( {\frac{{102400}}{{27}}} \right)cm \cr & \Rightarrow h = 3792.5\,cm \cr & \Rightarrow h = 37.9\,m \cr} $$

Internal radius, r = 4 cm
External radius, R = 5 cm
Total surface area :
$$\eqalign{ & = 2\pi {R^2} + 2\pi {r^2} + \pi \left( {{R^2} - {r^2}} \right) \cr & = 3\pi {R^2} + \pi {r^2} \cr & = \left[ {\pi \left( {3 \times 25 + 16} \right)} \right]{\text{ c}}{{\text{m}}^2} \cr & = \left( {\frac{{22}}{7} \times 91} \right){\text{c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

Total length of tape required :
= Sum of lengths of edges
= (30 × 4 + 25 × 4 + 20 × 4) cm
= 300 cm

Volume :
$$\eqalign{ & = \left[ {12 \times 9 \times \left( {\frac{{1 + 4}}{2}} \right)} \right]{m^3} \cr & = \left( {12 \times 9 \times 2.5} \right){m^3} \cr & = 270\,{m^3} \cr} $$

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

Area of base :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
$$\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2} \cr & \left[ {\because s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr} $$
∴ Whole surface area of the pyramid :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr} $$

Given length of width of swimming pool is 9 m and 12 m respectively
Volume of swimming pool :
$$\eqalign{ & = 9 \times 12 \times \left( {\frac{{1 + 4}}{2}} \right) \cr & = 9 \times 12 \times \frac{5}{2} \cr & = 270\,\text{cu. metre} \cr} $$

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 :
$$\eqalign{ & = \left( {502 - 216} \right){\text{ c}}{{\text{m}}^2} \cr & = 286{\text{ c}}{{\text{m}}^2} \cr} $$

Let their edges be a and b
Then,
$$\eqalign{ & \frac{{{a^3}}}{{{b^3}}} = \frac{8}{{27}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^3} = {\left( {\frac{2}{3}} \right)^3} \cr & \Rightarrow \frac{a}{b} = \frac{2}{3} \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{4}{9} \cr & \Rightarrow \frac{{6{a^2}}}{{6{b^2}}} = \frac{4}{9}\,Or\,4:9 \cr} $$

Let the radius of the cylinder and the hemisphere be r cm
Diameter of the cylinder = (2r) cm
Height of the cylinder = (4 × 2r) cm = 8r cm
Total length of the solid :
= (8r + r + r) cm = 10r cm
⇒ 10r = 35
⇒ r = 3.5

∴ Surface area of the solid :
= Curved surface area of the cylinder + 2 × (Curved surface area of the hemisphere)
$$ = \left( {2 \times \frac{{22}}{7} \times 3.5 \times 28 + 2 \times 2 \times \frac{{22}}{7} \times 3.5 \times 3.5} \right){\text{ c}}{{\text{m}}^2}$$
$$\eqalign{ & = \left( {616 + 154} \right){\text{ c}}{{\text{m}}^2} \cr & = 770{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & 3 \times 2\pi {r^2} = 2 \times 2\pi rh \cr & \Rightarrow 6r = 4h \cr & \Rightarrow r = \frac{2}{3}h \cr & \Rightarrow r = \left( {\frac{2}{3} \times 6} \right)m \cr & \Rightarrow r = 4\,m \cr} $$

Let the inner radius of the pipe be r metres
Then,
Volume of water flowing through the pipe in 10 minutes :
$$\eqalign{ & = \left[ {\left( {\frac{{22}}{7} \times {r^2} \times 7} \right) \times 10} \right]{m^3} \cr & = \left( {220{r^2}} \right){m^3} \cr & \therefore 220{r^2} = 440 \cr & \Rightarrow {r^2} = 2 \cr & \Rightarrow r = \sqrt 2 \, m \cr } $$

Answer :- A right-angled triangle.

Let the heights of two cones be 7x and 3x and their radii be 6y and 7y respectively
Then,
Ratio of volume :
$$\eqalign{ & = \frac{{\frac{1}{3}\pi \times {{\left( {6y} \right)}^2} \times 7x}}{{\frac{1}{3}\pi \times {{\left( {7y} \right)}^2} \times 3x}} \cr & = \frac{{36 \times 7}}{{49 \times 3}} \cr & = \frac{{12}}{7}\,Or\,12:7 \cr} $$

Volume of parallelepiped :
$$\eqalign{ & = \left( {5 \times 3 \times 4} \right)c{m^3} \cr & = 60\,c{m^3} \cr} $$
Volume of cube :
$$\eqalign{ & = {\left( 4 \right)^3}\,c{m^3} \cr & = 64\,c{m^3} \cr} $$
Volume of cylinder :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 3 \times 3 \times 3} \right)c{m^3} \cr & = 84.86\,c{m^3} \cr} $$
Volume of spare :
$$\eqalign{ & = \left( {\frac{4}{3} \times \frac{{22}}{7} \times 3 \times 3 \times 3} \right) \cr & = 113.14\,c{m^3} \cr} $$
So, Option D is correct decreasing order

Volume of new sphere :
$$\eqalign{ & = \left[ {\frac{4}{3}\pi \times {{\left( 6 \right)}^3} + \frac{4}{3}\pi \times {{\left( 8 \right)}^3} + \frac{4}{3}\pi \times {{\left( {10} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \left\{ {{{\left( 6 \right)}^3} + {{\left( 8 \right)}^3} + {{\left( {10} \right)}^3}} \right\}} \right]{\text{ c}}{{\text{m}}^3} \cr & = \left( {\frac{4}{3}\pi \times 1728} \right){\text{c}}{{\text{m}}^3} \cr & = \left[ {\frac{4}{3}\pi \times {{\left( {12} \right)}^3}} \right]{\text{ c}}{{\text{m}}^3} \cr} $$
Let the radius of the new sphere be R
Then,
$$\eqalign{ & \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi \times {\left( {12} \right)^3} \cr & \Rightarrow R = 12\,cm \cr} $$
∴ Diameter :
$$\eqalign{ & = 2R \cr & = 2 \times 12 \cr & = 24\,cm \cr} $$

Ball is dropped from the height of 36 m when the ball will rise at the third bounce
∴ Required height :
$$\eqalign{ & = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times 36 \cr & = \frac{{32}}{3} \cr & = 10\frac{2}{3}\,m \cr} $$

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

$$\eqalign{ & 6{a^2} = 150 \cr & \Rightarrow {a^2} = 25 \cr & \Rightarrow a = 5 \cr} $$
∴ Volume :
$${a^3} = {5^3} = 125\,{\text{ c}}{{\text{m}}^3}$$

Surface area :
$$\eqalign{ & = {2\left( {7 \times 11 + 11 \times 13 + 7 \times 13} \right)} \cr & = {2 \times 311} \cr & = 622{\text{ c}}{{\text{m}}^2} \cr} $$

let l, b and h denote the length, breadth and depth of Meeta's lunch box
Then, length of Rita's lunch box :
$$ = 110\% {\text{ of }}l = \frac{{11l}}{{10}}$$
Breadth of Rita's lunch box :
$$ = 110\% {\text{ of }}b = \frac{{11b}}{{10}}$$
Depth of Rita's lunch box :
$$ = 80\% {\text{ of }}h = \frac{{4h}}{5}$$
∴ Ratio of the capacities of Rita's and Meeta's lunch boxes :
$$\eqalign{ & = \frac{{11l}}{{10}} \times \frac{{11b}}{{10}} \times \frac{{4h}}{5}:lbh \cr & = \frac{{121}}{{125}}:1 \cr & = 121:125 \cr} $$

Let the length of each edge of each cube be a
Then, the cuboid formed by placing 3 cubes adjacently has the dimensions 3a , a and a
Surface area of the cuboid :
$$\eqalign{ & = 2\left[ {3a \times a + a \times a + 3a \times a} \right] \cr & = 2\left[ {3{a^2} + {a^2} + 3{a^2}} \right] \cr & = 14{a^2} \cr} $$
Sum of surface area of 3 cubes :
$$\eqalign{ & = \left( {3 \times 6{a^2}} \right) \cr & = 18{a^2} \cr} $$
∴ Required ratio :
$$\eqalign{ & = 14{a^2}:18{a^2} \cr & = 7:9 \cr} $$

$$\eqalign{ & \frac{{\pi {r^2}h}}{{2\pi rh}} = \frac{{924}}{{264}} \cr & \Rightarrow r = \left( {\frac{{924}}{{264}} \times 2} \right) \cr & \Rightarrow r = 7\,m \cr} $$
And,
$$\eqalign{ & \therefore 2\pi rh = 264 \cr & \Rightarrow h = \left( {264 \times \frac{7}{{22}} \times \frac{1}{2} \times \frac{1}{7}} \right) \cr & \Rightarrow h = 6\,m \cr} $$
∴ Required ratio :
$$ = \frac{{2r}}{h} = \frac{{14}}{6} = 7:3$$

$$\eqalign{ & {\text{Number of tins}} \cr & = \frac{{{\text{Voulme of the drum}}}}{{{\text{Volume of each tin}}}} \cr & = \frac{{\left( {\frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times 24} \right)}}{{\left( {\frac{{25}}{{10}} \times \frac{{22}}{{10}} \times \frac{{35}}{{10}}} \right)}} \cr & = 1200 \cr} $$

Volume of the larger cube :
$$\eqalign{ & = \left( {{3^3} + {4^3} + {5^3}} \right){\text{ c}}{{\text{m}}^3} \cr & = 216{\text{ c}}{{\text{m}}^3} \cr} $$
Let the edge of the larger cube be a cm
$$\eqalign{ & \therefore {a^3} = 216 \cr & \Rightarrow a = 6 \cr} $$
Required ratio :
$$\eqalign{ & = \frac{{6\left( {{3^2} + {4^2} + {5^2}} \right)}}{{6 \times {6^2}}} \cr & = \frac{{6 \times 50}}{{6 \times 36}} \cr & = \frac{{25}}{{18}}\,Or\,25:18 \cr} $$

Total volume of the body :
= Volume of the cylinder + Volume of the cone + Volume of the hemisphere
$$\eqalign{ & = \pi {r^2}.\,r + \frac{1}{3}\pi {r^2}.\,r + \frac{2}{3}\pi {r^3} \cr & = 2\pi {r^3} \cr & = 2\pi {\left( 2 \right)^3} \cr & = 16\pi \cr} $$

Volume of cistern :
$$\eqalign{ & = \left( {\pi \times {{10}^2} \times 15} \right){m^3} \cr & = 1500\pi {m^3} \cr} $$
Volume of water flowing through the pipe in 1 sec :
$$\eqalign{ & = \left( {\pi \times 0.25 \times 0.25 \times 5} \right){m^3} \cr & = 0.3125\pi {m^3} \cr} $$
∴ Time taken to fill the cistern :
$$\eqalign{ & = \left( {\frac{{1500\pi }}{{0.3125\pi }}} \right) \cr & = \left( {\frac{{1500 \times 10000}}{{3125}}} \right) \cr & = 4800\,\sec \cr & = \left( {\frac{{4800}}{{60}}} \right)\min \cr & = 80\,\min \cr} $$

$$\eqalign{ & \frac{{\pi \,{r^2}h}}{{2\pi \,rh}} = \frac{{924}}{{264}}\, \cr & \Rightarrow r = {\frac{{924}}{{264}} \times 2} = 7m \cr & {\text{And}},\,2\pi rh = 264 \cr & \Rightarrow h = {264 \times \frac{7}{{22}} \times \frac{1}{2} \times \frac{1}{7}} = 6m \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = \frac{{2r}}{h} = \frac{{14}}{6} = 7:3 \cr} $$