Answer: Option A. -> 429$$\,{\text{sq}}{\text{.m}}$$
Volume of cylinder = $$\pi {r^2}h$$
∴ Curved surface area of cylinder = $$2\pi rh$$
$$\eqalign{
& \therefore \frac{{\pi {r^2}h}}{{2\pi rh}} = \frac{{616}}{{352}} \cr
& \Rightarrow r = \frac{{2 \times 616}}{{352}} \cr
& \Rightarrow r = 3.5\,m \cr} $$
∴ Volume of cylinder :
$$\eqalign{
& \Rightarrow \pi {r^2}h = 616 \cr
& \Rightarrow \frac{{22}}{7} \times 3.5 \times 3.5 \times h = 616 \cr
& \Rightarrow 11 \times 3.5 \times h = 616 \cr
& \Rightarrow h = \frac{{616}}{{11 \times 3.5}} \cr
& \Rightarrow h = 16 \cr} $$
∴ Total surface area of the cylinder :
$$\eqalign{
& = 2\pi rh + 2\pi {r^2} \cr
& = 2\pi r\left( {h + r} \right) \cr
& = 2 \times \frac{{22}}{7} \times 3.5\left( {16 + 3.5} \right) \cr
& = 2 \times \frac{{22}}{7} \times 3.5\left( {19.5} \right) \cr
& = 22 \times 19.5 \cr
& = 429\,{\text{sq}}{\text{.m}} \cr} $$

Answer: Option B. -> 75 cm3
Volume of pyramid :$$\eqalign{
& = \frac{1}{3} \times {\text{Area of base}} \times {\text{Height}} \cr
& = \left( {\frac{1}{3} \times 25 \times 9} \right){\text{c}}{{\text{m}}^3} \cr
& = 75\,{\text{c}}{{\text{m}}^3} \cr} $$

Answer: Option B. -> 452.57 cm3
Radius of hemisphere bowl = 6 cm
∴ Volume of hemisphere :
$$\eqalign{
& = \frac{2}{3}\pi {r^3} \cr
& = \frac{2}{3} \times \frac{{22}}{7} \times 6 \times 6 \times 6 \cr
& = \frac{{9504}}{{21}} \cr
& = 452.57{\text{ c}}{{\text{m}}^3} \cr} $$

Answer: Option B. -> Decrease by 25%
Let the radius of a right circular cine be R cm and height be H cm
Volume of right circular cone $$ = \frac{1}{3}\pi {R^2}H{\text{ cu}}{\text{.cm}}$$
When height of right circular cone is increased by 200% and radius of the base is reduce by 50%
New volume :$$\eqalign{
& {\text{ = }}\frac{1}{3}\pi {\left( {\frac{R}{2}} \right)^2}.3H \cr
& = \frac{1}{3}\pi \frac{{{R^2}4}}{4}.3H \cr
& = \frac{{\pi {R^2}H}}{4} \cr} $$
Difference :
$$\eqalign{
& = \pi {R^2}H\left( {\frac{1}{3} - \frac{1}{4}} \right) \cr
& = \frac{1}{{12}}\pi {R^2}H \cr} $$
Decrease percentage :
$$\eqalign{
& = \frac{{\frac{1}{{12}}\pi {R^2}H}}{{\frac{1}{3}\pi {R^2}H}} \times 100 \cr
& = 25\% \cr} $$

Answer: Option B. -> 525 m
$$\eqalign{
& \pi {r^2} = 346.5 \cr
& {r^2} = \left( {346.5 \times \frac{7}{{22}}} \right) \cr
& {r^2} = \frac{{441}}{4} \cr
& {r^2} = \frac{21}{2} \cr
& \therefore l = \sqrt {{r^2} + {h^2}} \cr
& \,\,\,\,\,\,\, = \sqrt {\frac{{441}}{4} + {{\left( {14} \right)}^2}} \cr
& \,\,\,\,\,\,\, = \sqrt {\frac{{1225}}{4}} \cr
& \,\,\,\,\,\,\, = \frac{{35}}{2} \cr} $$
So, area of canvas needed :
$$\eqalign{
& \pi rl = \left( {\frac{{22}}{7} \times \frac{{21}}{2} \times \frac{{35}}{2}} \right){{\text{m}}^2} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{33 \times 35}}{2}} \right){{\text{m}}^2} \cr} $$
∴ Length of canvas :
$$\eqalign{
& = \left( {\frac{{33 \times 35}}{{2 \times 1.1}}} \right){\text{ m}} \cr
& = 525{\text{ m}} \cr} $$

Answer: Option D. -> 300 cm
Total length of tape required :
= Sum of lengths of edges
= (30 × 4 + 25 × 4 + 20 × 4) cm
= 300 cm

Answer: Option B. -> 270 m3
Volume :
$$\eqalign{
& = \left[ {12 \times 9 \times \left( {\frac{{1 + 4}}{2}} \right)} \right]{m^3} \cr
& = \left( {12 \times 9 \times 2.5} \right){m^3} \cr
& = 270\,{m^3} \cr} $$

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

Answer: Option D. -> 4 m
$$\eqalign{
& 3 \times 2\pi {r^2} = 2 \times 2\pi rh \cr
& \Rightarrow 6r = 4h \cr
& \Rightarrow r = \frac{2}{3}h \cr
& \Rightarrow r = \left( {\frac{2}{3} \times 6} \right)m \cr
& \Rightarrow r = 4\,m \cr} $$

Answer: Option C. -> 286 cm2
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 :
$$\eqalign{
& = \left( {502 - 216} \right){\text{ c}}{{\text{m}}^2} \cr
& = 286{\text{ c}}{{\text{m}}^2} \cr} $$