Answer: Option D. -> 1864.5 kg
Let the depth of the cistern be h metres
Then,
$$\eqalign{
& 4.5 \times 3 \times h = 50 \cr
& \Rightarrow h = \frac{{50}}{{13.5}} \cr
& \Rightarrow h = \frac{{100}}{{27}} \cr} $$
Area of sheet required :
$$\eqalign{
& = lb + 2\left( {bh + lh} \right) \cr
& = lb + 2h\left( {l + b} \right) \cr
& = \left[ {4.5 \times 3 + 2 \times \frac{{100}}{{27}} \times \left( {4.5 + 3} \right)} \right]{{\text{m}}^2} \cr
& = \left( {13.5 + \frac{{200}}{{27}} \times 7.5} \right){\text{ }}{{\text{m}}^2} \cr
& = \left( {\frac{{27}}{2} + \frac{{500}}{9}} \right){{\text{m}}^2} \cr
& = \frac{{1243}}{{18}}{\text{ }}{{\text{m}}^2} \cr} $$
∴ Weight of lead :
$$\eqalign{
& = \left( {27 \times \frac{{1243}}{{18}}} \right)kg \cr
& = \left( {\frac{{3729}}{2}} \right)kg \cr
& = 1864.5\,kg \cr} $$

Answer: Option A. -> Rs. 6000
Since the tank is open at the top, we have :
Area of sheet required = Surface area of the tank
$$\eqalign{
& = lb + 2\left( {bh + lh} \right) \cr
& = \left[ {30 \times 20 + 2\left( {20 \times 12 + 30 \times 12} \right)} \right]{{\text{m}}^2} \cr
& = \left( {600 + 1200} \right){{\text{m}}^2} \cr
& = 1800{\text{ }}{{\text{m}}^2} \cr} $$
Length of sheet required :
$$\eqalign{
& = \left( {\frac{{{\text{Area}}}}{{{\text{Width}}}}} \right) \cr
& = \frac{{1800}}{3}m \cr
& = 600\,m \cr} $$
∴ Cost of the sheet
= Rs. (600 × 10)
= Rs. 6000

Answer: Option D. -> 12$$\sqrt 3 $$ cm
Since a cube has 4 diagonals, we have :
Length of a diagonal
$$\eqalign{
& = \left( {\frac{{12}}{4}} \right)cm \cr
& = 3\,cm \cr} $$
Let the length of each edge of the cube be a cm
Then,
$$\eqalign{
& \sqrt 3 a = 3 \cr
& or,a = \sqrt 3 \cr} $$
∴ Total length of the edges of the cube = $$12\sqrt 3\, $$ cm

Answer: Option D. -> 17 Times
Let the original length, breadth and height of the cuboid be x, 2x and 3x units respectively
Then, original volume = (x × 2x × 3x) cu.units = 6x3 cu.units
New length = 200% of x = 2x
New breadth = 300% of 2x = 6x
New height = 300% of 3x = 9x
∴ New volume :
= (2x × 6x × 9x) cu.units
= 108x3 cu.units
Increase in volume :
= (108x3 - 6x3) cu.units
= (102x3) cu.units
∴ Required ratio :
$$\eqalign{
& = \frac{{102{{\text{x}}^3}}}{{6{{\text{x}}^3}}} \cr
& = 17{\text{ }}\left( {{\text{Times}}} \right) \cr} $$

Answer: Option C. -> 237.5%
Let original edge = a
Then, original volume = a3
New edge :
$$\eqalign{
& = \frac{{150}}{{100}}a \cr
& = \frac{{3a}}{2} \cr} $$
New volume :
$$\eqalign{
& = {\left( {\frac{{3a}}{2}} \right)^3} \cr
& = \frac{{27{a^3}}}{8} \cr} $$
Increase in volume :
$$\eqalign{
& = \left( {\frac{{27{a^3}}}{8} - {a^3}} \right) \cr
& = \frac{{19{a^3}}}{8} \cr} $$
∴ Increase % :
$$\eqalign{
& = \left( {\frac{{19{a^3}}}{8} \times \frac{1}{{{a^3}}} \times 100} \right)\% \cr
& = 237.5\% \cr} $$

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

Answer: Option C. -> 498.35 cm2
$$\eqalign{
& h = 14\,cm,\,r = 7\,cm \cr
& {\text{So}},\,l = \sqrt {{{\left( 7 \right)}^2} + {{\left( {14} \right)}^2}} = \sqrt {245} = 7\sqrt 5 \,cm \cr
& \therefore {\text{Total}}\,{\text{surface}}\,{\text{area}} \cr
& = \pi \,rl + \pi \,{r^2} \cr
& = \left( {\frac{{22}}{7} \times 7 \times 7\sqrt 5 + \frac{{22}}{7} \times 7 \times 7} \right)c{m^2} \cr
& = \left[ {154\left( {\sqrt 5 + 1} \right)} \right]c{m^2} \cr
& = \left( {154 \times 3.236} \right)c{m^2} \cr
& = 498.35\,c{m^2} \cr} $$

Answer: Option C. -> 6400
$$\eqalign{
& {\text{Number}}\,{\text{of}}\,{\text{bricks}} \cr
& = \frac{{{\text{Volume}}\,{\text{of}}\,{\text{the}}\,{\text{wall}}}}{{{\text{Volume}}\,{\text{of}}\,{\text{1}}\,{\text{brick}}}} \cr
& = {\frac{{800 \times 600 \times 22.5}}{{25 \times 11.25 \times 6}}} \cr
& = 6400 \cr} $$

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

Answer: Option B. -> 1 dm
Let the thickness of the bottom be x cm
Then,
$${\mkern 1mu} {\left( {330 - 10} \right) \times \left( {260 - 10} \right) \times \left( {110 - x} \right)} = $$         $$8000 \times $$ $$1000$$
$$ \Rightarrow 320 \times 250 \times \left( {110 - x} \right) = 8000 \times 1000$$
$$\eqalign{
& \Rightarrow \left( {110 - x} \right) = \frac{{8000 \times 1000}}{{320 \times 250}} = 100 \cr
& \Rightarrow x = 10\,{\text{cm}} = 1\,{\text{dm}} \cr} $$