Answer: Option B. -> 3.5% increase
Let original height = h and original radius = r
New height = 115% of h $$\frac{23h}{20}$$
New radius = 90% of r $$\frac{9r}{10}$$
Original curved surface area = $$2\pi rh$$
New curved surface area :
$$\eqalign{
& = \left( {2\pi \times \frac{{9r}}{{10}} \times \frac{{23h}}{{20}}} \right) \cr
& = \frac{{207}}{{200}} \times 2\pi rh \cr} $$
Increase in curved surface area :
$$\eqalign{
& = \left( {\frac{{207}}{{200}} \times 2\pi rh - 2\pi rh} \right) \cr
& = \frac{7}{{200}} \times 2\pi rh \cr} $$
∴ Increase % :
$$\eqalign{
& = \left( {\frac{7}{{200}} \times 2\pi rh \times \frac{1}{{2\pi rh}} \times 100} \right)\% \cr
& = 3.5\% \cr} $$

Answer: Option D. -> 28 cm
Volume of the tank = 246.4 litres = 246400 cm3
Let the radius of the base be r cm
Then,
$$\eqalign{
& \left( {\frac{{22}}{7} \times {r^2} \times 400} \right) = 246400 \cr
& \Rightarrow {r^2} = \left( {\frac{{246400 \times 7}}{{22 \times 400}}} \right) \cr
& \Rightarrow {r^2} = 196 \cr
& \Rightarrow r = 14 \cr} $$
∴ Diameter of the base = 2r = 28 cm

Answer: Option C. -> 15 m
Let the radius of base = r and height = h
Required floor area :
= (4 × 11) m2 = 44 m2
Required volume :
= (20 × 11) m3 = 220 m3
So,
$$\eqalign{
& \frac{1}{3}\pi {r^2}h = 220 \cr
& \Rightarrow \frac{1}{3} \times 44 \times h = 220 \cr
& \Rightarrow h = \frac{{220 \times 3}}{{44}} \cr
& \Rightarrow h = 15\,m \cr} $$

Answer: Option D. -> 13500
$$\eqalign{
& {\text{Volume of cylinder :}} \cr
& = \left( {\pi \times 3 \times 3 \times 5} \right)c{m^3} \cr
& = 45\pi \,c{m^3} \cr
& {\text{Volume of a cone :}} \cr
& = \left( {\frac{1}{3}\pi \times \frac{1}{{10}} \times \frac{1}{{10}} \times 1} \right)c{m^3} \cr
& = \frac{\pi }{{300}}\,c{m^3} \cr
& \therefore {\text{Number of cones :}} \cr
& = \left( {45\pi \times \frac{{300}}{\pi }} \right) \cr
& = 13500 \cr} $$

Answer: Option B. -> $$\frac{1}{30}$$
Volume of copper rod :
$$\eqalign{
& = \left( {\pi \times \frac{1}{2} \times \frac{1}{2} \times 8} \right){\text{ c}}{{\text{m}}^3} \cr
& = 2\pi \,{\text{ c}}{{\text{m}}^3} \cr} $$
Let the radius of the wire be r cm
Then, volume of wire :
$$\eqalign{
& = \left( \pi {r^2} \times 1800 \right) {\text{cm}}^3 \cr
& = 1800\, \pi {r^2} {\text{ cm}}^3 \cr
& \therefore 1800\pi {r^2} = 2\pi \cr
& \Rightarrow {r^2} = \frac{2}{{1800}} \cr
& \Rightarrow {r^2} = \frac{1}{{900}} \cr
& \Rightarrow r = \sqrt {\frac{1}{{900}}} = \frac{1}{{30}} \cr} $$

Answer: Option A. -> 44 m3
Let the length of each side of the square base be x metres
Then,
$$\eqalign{
& {x^2} + {x^2} = {\left( 3 \right)^2} \cr
& \Rightarrow 2{x^2} = 9 \cr
& \Rightarrow {x^2} = \frac{9}{2} \cr
& \Rightarrow x = \frac{3}{{\sqrt 2 }} \cr} $$
∴ Volume of parallelopiped :
$$\eqalign{
& = \left( {\frac{3}{{\sqrt 2 }} \times \frac{3}{{\sqrt 2 }} \times 10} \right){{\text{m}}^{\text{3}}} \cr
& = \frac{{90}}{2}{{\text{m}}^{\text{3}}} \cr
& = 45\,{{\text{m}}^{\text{3}}} \cr} $$

Answer: Option B. -> $$11\frac{3}{7}{\text{ cm}}$$
Let the drop in the water level be h cm
Then,
$$\eqalign{
& \Rightarrow \frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} \times h = 11000 \cr
& \Rightarrow h = \left( {\frac{{11000 \times 7 \times 4}}{{22 \times 35 \times 35}}} \right){\text{ cm}} \cr
& \Rightarrow h = \frac{{80}}{7}{\text{ cm}} \cr
& \Rightarrow h = 11\frac{3}{7}{\text{ cm}} \cr} $$

Answer: Option C. -> 10 cm
Let the radius and height of the cone be 3x and 4x respectively
Then,
$$\eqalign{
& \frac{1}{3} \times \frac{{22}}{7} \times {\left( {3x} \right)^2} \times 4x = \frac{{2112}}{7} \cr
& \Rightarrow \frac{{264}}{7}{x^3} = \frac{{2112}}{7} \cr
& \Rightarrow {x^3} = \frac{{2112}}{{264}} \cr
& \Rightarrow {x^3} = 8 \cr
& \Rightarrow x = 2 \cr} $$
∴ Radius = 6 cm, Height = 8 cm
Slant height :
$$\eqalign{
& = \sqrt {{6^2} + {8^2}} \,cm \cr
& = \sqrt {100} \,cm \cr
& = 10\,cm \cr} $$

Answer: Option C. -> 1320 m2
Radius, r = 12 m
Height of conical part, h = (16 - 11) m = 5 m
Slant height of conical part,
$$\eqalign{
& l = \sqrt {{r^2} + {h^2}} \cr
& \,\,\,\, = \sqrt {{{\left( {12} \right)}^2} + {{\left( 5 \right)}^2}} \cr
& \,\,\,\, = \sqrt {169} \cr
& \,\,\,\, = 13\,m \cr} $$
Height of cylindrical part, H = 11 m
Area of canvas required :
= Curved surface area of cylinder + Curved surface area of cone$$\eqalign{
& = 2\pi rH + \pi rl \cr
& = \left[ {\frac{{22}}{7}\left( {2 \times 12 \times 11 + 12 \times 13} \right)} \right]{{\text{m}}^2} \cr
& = \left[ {\frac{{22}}{7}\left( {264 + 156} \right)} \right]{{\text{m}}^2} \cr
& = \left( {\frac{{22}}{7} \times 420} \right){{\text{m}}^2} \cr
& = 1320\,{{\text{m}}^2} \cr} $$

Answer: Option C. -> Rs. 1485
Curved surface area :
$$\eqalign{
& = 2\pi rh \cr
& = \left( {2 \times \frac{{22}}{7} \times \frac{7}{2} \times 22.5} \right){{\text{m}}^2} \cr
& = 495{\text{ }}{{\text{m}}^2} \cr} $$
∴ Cost of plastering :
= Rs. (495 × 3)
= Rs. 1485