Answer: Option C. -> 2 hours
Volume of water flown into the tank :$$\eqalign{
& = \left( {50 \times 44 \times 0.07} \right){{\text{m}}^3} \cr
& = 154\,{{\text{m}}^3} \cr} $$
Volume of water flowing through the pipe in 1 hour :
$$\eqalign{
& = \left( {\frac{{22}}{7} \times 0.07 \times 0.07 \times 5000} \right){{\text{m}}^3} \cr
& = 77\,{\text{m}} \cr} $$
∴ Required time $$ = \left( {\frac{{154}}{{77}}} \right) = 2{\text{ hours}}$$

Answer: Option B. -> 51 minutes 12 seconds
Volume flown in conical vessel :
$$\eqalign{
& = \frac{1}{2}\pi \times {\left( {20} \right)^2} \times 24 \cr
& = 3200\pi \cr} $$
Volume flown in 1 minute :
$$\eqalign{
& = \left( {\pi \times \frac{{2.5}}{{10}} \times \frac{{2.5}}{{10}} \times 1000} \right) \cr
& = 62.5\pi \cr} $$
∴ Time taken :
$$\eqalign{
& = \left( {\frac{{3200\pi }}{{62.5\pi }}} \right) \cr
& = 51\operatorname{minutes} 12\operatorname{seconds} \cr} $$

Answer: Option C. -> 36%
Let original radius = r and original height = h
Original volume = $$\pi {r^2}h$$
New radius = 125% of r = $$\frac{5r}{4}$$
Let new height = H
Then,
$$\eqalign{
& \pi {r^2}h = \pi {\left( {\frac{{5r}}{4}} \right)^2} \times H \cr
& Or,\,H = \frac{{16}}{{25}}h \cr} $$
Decrease in height :
$$\eqalign{
& = \left( {h - \frac{{16h}}{{25}}} \right) \cr
& = \frac{{9h}}{{25}} \cr} $$
∴ Decrease % :
$$\eqalign{
& = \left( {\frac{{9h}}{{25}} \times \frac{1}{h} \times 100} \right)\% \cr
& = 36\% \cr} $$

Answer: Option D. -> 8 times
Let the original radius be r
Then, original volume = $$\frac{4}{3}\pi {r^3}$$
New area = 2r
∴ New volume :
$$\eqalign{
& = \frac{4}{3}\pi {\left( {2r} \right)^3} \cr
& = 8 \times \frac{4}{3}\pi {r^3} \cr
& = 8 \times {\text{Original volume}} \cr} $$

Answer: Option D. -> 16 : 9
Let their radii be R and r
Then,
$$\eqalign{
& \frac{{\frac{4}{3}\pi {R^3}}}{{\frac{4}{3}\pi {r^3}}} = \frac{{64}}{{27}} \cr
& \Rightarrow {\left( {\frac{R}{r}} \right)^3} = \frac{{64}}{{27}} \cr
& \Rightarrow {\left( {\frac{R}{r}} \right)^3} = {\left( {\frac{4}{3}} \right)^3} \cr
& \Rightarrow \frac{R}{r} = \frac{4}{3} \cr} $$
Ratio of surface areas :
$$\eqalign{
& = \frac{{4\pi {R^2}}}{{4\pi {r^2}}} = {\left( {\frac{R}{r}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{4}{3}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{9} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16:9 \cr} $$

Answer: Option D. -> 17 : 8
Let the radius of the cylinder be r
Then, radius of the sphere = 2r
$$\eqalign{
& \frac{{{\text{Volume of cylinder}}}}{{{\text{Volume of sphere}}}} = \frac{3}{2} \cr
& \Rightarrow \frac{{\pi {r^2}h}}{{\frac{4}{3}\pi {{\left( {2r} \right)}^3}}} = \frac{3}{2} \cr
& \Rightarrow \frac{h}{r} = 16 \cr
& \Rightarrow h = 16r \cr} $$
∴ Required ratio :
$$\eqalign{
& \frac{{{\text{Total surface area of cylinder}}}}{{{\text{Surface area of sphere}}}} \cr
& = \frac{{2\pi r.\left( {16r} \right) + 2\pi {r^2}}}{{4\pi {{\left( {2r} \right)}^2}}} \cr
& = \frac{{34\pi {r^2}}}{{16\pi {r^2}}} \cr
& = \frac{{17}}{8}\,Or\,17:8 \cr} $$

Answer: Option A. -> 33.1%
If R is the radius of sphere, volume of the sphere = $$\frac{4}{3}\pi {R^3}$$
When radius of sphere is increased by 10%
New volume :
$$\eqalign{
& = \frac{4}{3}\pi {\left( {1.1R} \right)^3} \cr
& = \frac{4}{3}\pi {R^3}\left( {1.331} \right) \cr} $$
Difference :
$$\eqalign{
& = \frac{4}{3}\pi {R^3}\left( {1.331} \right) - \frac{4}{3}\pi {R^3} \cr
& = \frac{4}{3}\pi {R^3}\left( {1.331 - 1} \right) \cr
& = \frac{4}{3}\pi {R^3}\left( {0.331} \right) \cr} $$
Increase % :
$$\eqalign{
& = \frac{{\frac{4}{3}\pi {R^3}\left( {0.331} \right)}}{{\frac{4}{3}\pi {R^3}}} \times 100 \cr
& = 33.1\% \cr} $$

Answer: Option C. -> 216 cm2
Volume of the cuboid :
$$\eqalign{
& = \left( {9 \times 8 \times 6} \right){\text{ c}}{{\text{m}}^3} \cr
& = 432{\text{ c}}{{\text{m}}^3} \cr} $$
Volume of the cube :
$$\eqalign{
& = \left( {\frac{1}{2} \times 432} \right){\text{ c}}{{\text{m}}^3} \cr
& = 216{\text{ c}}{{\text{m}}^3} \cr} $$
$$\eqalign{
& {a^3} = 216 \cr
& a = \root 3 \of {216} \cr
& a = 6 \cr
& 6{a^2} = 6 \times {6^2} \cr
& 6{a^2} = 216 \text{ cm}^2 \cr} $$

Answer: Option C. -> 5400 m/hr
Volume flown in 5 hours :
$$\eqalign{
& = \left( {225 \times 162 \times \frac{{20}}{{100}}} \right){{\text{m}}^3} \cr
& = 7290\,{{\text{m}}^3} \cr} $$
Volume flown in 1 hour :
$$\eqalign{
& = \left( {\frac{{7290}}{5}} \right){{\text{m}}^3} \cr
& = 1458\,{{\text{m}}^3} \cr} $$
∴ Required speed :
$$\eqalign{
& = \left( {\frac{{1458}}{{0.60 \times 0.45}}} \right){\text{m/hr}} \cr
& = 5400\,{\text{m/hr}} \cr} $$

Answer: Option D. -> None of these
Volume of the cube = 83 cu.m = 512 cu.m