Answer: Option C. -> 27 : 8
Let the radii of the two spheres be 3r and 2r respectively
Then, required ratio :
$$\eqalign{
& = \frac{{\frac{4}{3}\pi {{\left( {3r} \right)}^3}}}{{\frac{4}{3}\pi {{\left( {2r} \right)}^3}}} \cr
& = \frac{{27}}{8} \cr
& = 27:8 \cr} $$

Answer: Option C. -> $$\frac{8}{{3\sqrt 3 }}{r^3}$$
Clearly, the diagonal of the largest possible cube will be equal to the diameter of the sphere
Let the edge of the cube be a
$$\eqalign{
& \sqrt 3 a = 2r \cr
& \Rightarrow a = \frac{2}{{\sqrt 3 }}r \cr} $$
Volume :
$$\eqalign{
& = {a^3} \cr
& = {\left( {\frac{2}{{\sqrt 3 }}r} \right)^3} \cr
& = \frac{8}{{3\sqrt 3 }}{r^3} \cr} $$

Answer: Option A. -> Rs. 77
Radius of a hemispherical bowl = 3.5 cm
Inner and outer surface area of the bowl :
$$\eqalign{
& = 4\pi {r^2} \cr
& = 4 \times \frac{{22}}{7} \times 3.5 \times 3.5 \cr
& = 154{\text{ sq}}{\text{.cm}} \cr} $$
Total cost of painting at the rate of Rs. 5 per 10 sq.cm :
$$\eqalign{
& = 154 \times \frac{5}{{10}} \cr
& = {\text{Rs}}{\text{. 77}} \cr} $$

Answer: Option A. -> 108
$$\eqalign{
& {\text{Number of spheres :}} \cr
& = \frac{{{\text{Volume of cone}}}}{{{\text{Volume of 1 sphere}}}} \cr
& = \frac{{\frac{1}{3}\pi \times 12 \times 12 \times 24}}{{\frac{4}{3}\pi \times 2 \times 2 \times 2}} \cr
& = 108 \cr} $$

Answer: Option D. -> 770 cm2
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)
$$ = \left( {2 \times \frac{{22}}{7} \times 3.5 \times 28 + 2 \times 2 \times \frac{{22}}{7} \times 3.5 \times 3.5} \right){\text{ c}}{{\text{m}}^2}$$
$$\eqalign{
& = \left( {616 + 154} \right){\text{ c}}{{\text{m}}^2} \cr
& = 770{\text{ c}}{{\text{m}}^2} \cr} $$

Answer: Option B. -> 15 m
Required length :
$$\eqalign{
& = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 5 \right)}^2}} \,m \cr
& = \sqrt {225} \,m \cr
& = 15\,m \cr} $$

Answer: Option A. -> 3080 m3
Let the radius of the cylinder be r and height be h
Then, r + h = 19
Again, total surface area of cylinder = $$\left( {2\pi rh + 2\pi {r^2}} \right)$$
Now,
$$\eqalign{
& 2\pi r\left( {h + r} \right) = 1672 \cr
& \Rightarrow 2\pi r \times 19 = 1672 \cr
& \Rightarrow 38\pi r = 1672 \cr
& \therefore \pi r = \frac{{1672}}{{38}} = 44\,m \cr
& \therefore r = \frac{{44 \times 7}}{{22}} = 14\,m \cr} $$
∴ Height = 19 - 14 = 5 m
Volume of cylinder :
$$\eqalign{
& = \pi {r^2}h \cr
& = \frac{{22}}{7} \times 14 \times 14 \times 5 \cr
& = 22 \times 2 \times 14 \times 5 \cr
& = 3080\,{m^3} \cr} $$

Answer: Option B. -> 32 cm
Let length = x cm
Then,
$$\eqalign{
& x \times 28 \times 5 \times \frac{{25}}{{1000}} = 112 \cr
& \Rightarrow x = \left( {112 \times \frac{{1000}}{{25}} \times \frac{1}{{28}} \times \frac{1}{5}} \right) \cr
& \Rightarrow x = 32\,cm \cr} $$

Answer: Option B. -> 284 cm2
Clearly,
$$l$$ = (20 - 4) cm = 16 cm
b = (15 - 4) cm = 11 cm and
h = 2 cm
∴ Outer surface area of the box :
$$\eqalign{
& = \left[ {2\left( {l + b} \right) \times h} \right] + lb \cr
& = \left[ {\left\{ {2\left( {16 + 11} \right) \times 2} \right\} + 16 \times 11} \right] \cr
& = \left( {108 + 176} \right) \cr
& = 284{\text{ c}}{{\text{m}}^2} \cr} $$

Answer: Option D. -> 1, 2 and 3
Clearly, we have :
V1 = x3,
V2 = (2x)3 = 8x3
V3 = (3x)3 = 27x3
V4 = (4x)3 = 64x3
(i) V1 + V2 + 2V3
= x3 + 8x3 + 2 × 27x3
= 63x3 < V4
(ii) V1 + 4V2 + V3
= x3 + 4 × 8x3 + 27 x3
= 60x3 < V4
(iii) 2(V1 + V3) + V2
= 2 (x3 + 27x3) + 8x3
= 64x3 = V4