The centre O of sphere will be the centroid of the  ABCD 
∴   OA= \(\frac { 1 }{ 3 } \) AB
∴  AB=3(OA)=3a
or sin 30^o = [in rt ∠dΔOKB]
⇒  \(\frac { 1 }{ 2 } =\frac { a }{ OB } \) \(\frac { 1 }{ 2 } =\frac { a }{ OB } \)OB=2a
⇒ AB=OA+OB=a+2a=3a
Now, In rt. ∠dΔABC  ,∠ABC=30^o ,∠BAC=90^o
 ⇒ \(\frac { AC }{ AB } =tan\quad { 30 }^{ o }\) 
⇒ Volume of cone BCD= \(\frac { 1 }{ 3 } \pi (AC)^{ 2 }\times AB\)
= \(\frac { 1 }{ 3 } \pi (a\sqrt { 3 } )^{ 2 }\times 3a\)
=3 \(\pi \) a²
and volume of sphere =\(\frac { 4 }{ 3 } \)  \(\pi \) a³
water remaining in the cone
=\(\frac { 4 }{ 3 } \)  \(\pi \) a³ =\(\frac { 5\pi }{ 3 } \) a³

Let the level of water raise in the tank in x hours = 15000x metres
Length of the water flow in x hours = 15000x metres
 Diameter of the pipe = 14 cm
radius = \(\frac { 14 }{ 2 } =7cm=\frac { 7 }{ 100 } cm\)
Volume of water = \(\pi r^{ 2 }h\)
Volume of water flow in x hours in the pond  =  \(\frac { 22 }{ 7 } \times \frac { 7 }{ 100 } \times \frac { 7 }{ 100 } \times 15000x\)
According to the question 
\(50\times 44\times \frac { 21 }{ 100 } =\frac { 22 }{ 7 } \times \frac { 7 }{ 100 } \times \frac { 7 }{ 100 } \times 15000x\)
\(\Rightarrow 22\times 21=\frac { 154\times 15 }{ 10 } x\Rightarrow x=\frac { 22\times 21\times 10 }{ 154\times 15 } =2\)
Hence, the level of water in the pond will rise by 21 cm in 2 hours. 

Given,spherical glass vessel is a combination of a sphere as its base and a cylinder as its neck.

For cylindrical portion,
Height of the cylinder, h₁ = 8 cm
Radius of the cylinder,
\(r_1=\frac{2}{2}=1 \mathrm{~cm}\)
For spherical portion,
Radius of the sphere,
\(r_2=\frac{8.5}{2} \mathrm{~cm}\)
\(\therefore\) Volume of water filled in a spherical glass vessel = Volume of the cylinder + Volume of the sphere
\(\begin{aligned} & =\pi r_1^2 h_1+\frac{4}{3} \pi r_2^{3} \\ \end{aligned}\)
\(\begin{aligned} & =3.14 \times 1 \times 1 \times 8+\frac{4}{3} \times 3.14 \times \frac{8.5}{2} \times \frac{8.5}{2} \times \frac{8.5}{2} \\ \end{aligned}\)
\(\begin{aligned} & =25.12+\frac{1928.3525}{6} \end{aligned}\)
= 25.12 + 321.39 = 346.51
So, the correct answer is 346.51 cm³.

Total cost of covering the walls with wallpaper =  Rs 362.50
Cost of paper per m = Rs 1.25
\(\Rightarrow \) Length of paper = \(\frac { Rs\quad 362.50 }{ Rs\quad 1.25 } =290m\)
Area of paper required = area of 4 walls =area Of 2 doors =area of I almirah 
\(\Rightarrow 290\times \frac { 40 }{ 100 } =\left[ 2h(8+6)-2\times 2\times 1-3\times 2 \right] m^{ 2 }\)
\(\Rightarrow 29\times 4=(28h-4-6)m^{ 2 }\)
\(\Rightarrow \) 116=28h-10\(\Rightarrow \) 126\(\Rightarrow \) h= \(\frac { 126 }{ 28 } =4.5m\)

Here, toy is a combination of a hemisphere and a cone.

Given, total height of toy,
AD = 15.5 cm
For hemispherical portion,
Radius, OC = OD = OB = 3.5 cm
For conical portion,
Height, OA = AD - OD
= 15.5 - 3.5 = 12 cm
and radius = 3.5 cm
Now, total surface area of the toy = Curved surface area of cone + curved surface area of hemisphere
\(\begin{aligned} & =\pi r l+2 \pi r^2=\pi r \sqrt{h^2+r^2}+2 \pi r^2 \quad\left[\because l=\sqrt{h^2+r^2}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 3.5 \times \sqrt{(12)^2+(3.5)^2}+2 \times \frac{22}{7} \times(3.5)^2 \\ \end{aligned}\)
\(\begin{aligned} & =11 \sqrt{144+12.25}+22 \times 3.5 \\ \end{aligned}\)
\(\begin{aligned} & =11 \sqrt{156.25}+11 \times 7 \end{aligned}\)
= 11(12.5) + 77
=137.5 + 77
= 214.5 cm²

74.26 cm²

Two cones and a cylinder.

Here, Outer radius of the pipe = \({{10}\over{2}}=5\) cm
Inner radius of the pipe = 5 - 1 = 4 cm
Let h = length of pipe
Volume of the pipe = \({\pi}({5}^{2}-{4}^{2})h\)
\(={{22}\over{7}}\times9\times h\) cm³
\(\because\)  Volume of the pipe = Volume of the cuboid
\(\Rightarrow \)     \(h={{66\times20\times27\times7}\over{22\times9}}\)
= 3 x 20 x 3 x 7 = 1260 cm.
\(\therefore\)   Volume of the pipe = 1260 cm

Required number of cubes
\(={Volume\ of\ spherical\ ball\over Volume\ of\ each\ cube}\)
\(={{4\over3}\times{22\over7}\times{21\over2}\times{21\over2}\times{21\over2}\over1\times1\times1}\)

7 cm

Length of the longest stick that can be placed inside the box = length of its diagonal
Diagonal of the cuboid=\(\sqrt{l^{2}+b^{2}+h^{2}}\)
=\(\sqrt{(12)^{2}+(4)^{2}+(3)^{2}}\)
=\(\sqrt{144+16+9}=\sqrt{169}\)
= 13 cm

Internal radius=\(\frac{24}{2}\)cm
External radius=\(\frac{25}{2}\)cm
Area to be painted=outer C.S.A. + inner C.S.A. + area of the ring.
=2πR²+2πr²+π[R²-r²]
=2x\(\frac{22}{7}\)x\(\frac{25}{2}\)x\(\frac{25}{2}\)+2x\(\frac{22}{7}\)x\(\frac{24}{2}\)x\(\frac{24}{2}\)+\(\frac{22}{7}\)\(\left[\left(\frac{25}{2}\right)^{2}-\left(\frac{24}{2}\right)^{2}\right]\)
=2x\(\frac{22}{7}\)x\(\frac{1}{4}\)[625+576]+\(\frac{22}{7}\)\(\left[\frac{25}{2}-\frac{24}{2}\right]\)\(\left[\frac{25}{2}-\frac{24}{2}\right]\)
=\(\frac{22}{7}\)\(\left[\frac{1201}{2}+\frac{1}{2}\times \frac{49}{2}\right]\)
=\(\frac{22}{7}\)x\(\frac{1}{4}\)[2402+49]
=\(\frac{22}{7}\)x\(\frac{1}{4}\)x2451 cm²
Cost of painting=Rs.0.05x\(\frac{22}{7}\)x\(\frac{1}{4}\)x2451=Rs.96.28

False.

Total surface area of the figure = \(\frac { 2\pi { r }h+\pi { r }^{ 2 } }{ upper\quad cylinder } +\frac { \pi { r }^{ 2 }+2\pi { r }h }{ lower\quad cylinder } =4\pi { r }h+2\pi { r }^{ 2 }\) 

683.83 cm³

Here, R = 20 cm, r = 8 cm and h = 16 cm
Slant height l = \(\sqrt { { h }^{ 2 }+{ \left( R-r \right) }^{ 2 } } \)
= \(\sqrt { { 16 }^{ 2 }+{ \left( 20-8 \right) }^{ 2 } } \)
= \(\sqrt { 256+144 } =\sqrt { 400 } \)
= 20 cm
Volume of container
= \(\frac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) \)
= \(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 16\left( { 20 }^{ 2 }+{ 8 }^{ 2 }+20\times 8 \right) \)
= \(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 16\left( 400+64+160 \right) \)
= \(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 16\times 624\)
= \(\frac { 73216 }{ 7 } \) = 10459.42 cm³
= 10.45942 litres
= 10.46 litres
\(\therefore\)  Quantity of milk that container can hold = 10.46 litres
cost of milk = Rs 10.46 x 30
= Rs 313.80

Radius of the cylinder = 3 cm
Height = 5 cm
\(\therefore\)  Volume = \(\pi\) X 3 X 3 X 5 cm³
= 45\(\pi\) cm³
Radius to cone = \({{3}\over{2}}\)cm
Height of the cone = \({{8}\over{9}}\) cm
\(\therefore \) Volume of the cone = \({{1}\over{3}}{\pi r}^{2}h\)
\(={{1}\over{3}}\pi\times{{9}\over{4}}\times{{8}\over{9}}={{2{\pi}\over{3}}}\)cm3
Volume of metal left = 45\({\pi}\) cm³ - \({{2\pi}\over{3}}\)cm³
= \({{133 \pi}\over{3}}\) cm³
Ratio = \({{133 \pi}\over{3}}:{{2 \pi}\over{3}}=133:2\)

Volume of kerosene in case
= 20 x 20 x 45 = 25 x 15 x h (Volume in second cane)
\(\Rightarrow\)  h = \(\frac{20\times20\times45}{25\times15}\) = 48 cm

Let radius of the tank be x dm then
height of the tank be 6x dm
Total cost of painting = Rs 237.60
Area to be painted = \(\frac { 237.60\times 100 }{ 60 } \) =396 sq dm
\(\Rightarrow 2\pi r(r+h)=396\Rightarrow 2\times \frac { 22 }{ 7 } \times x(x+6x)=396\Rightarrow x\times 7x=\frac { 396\times 7 }{ 2\times 22 } \)
\(\Rightarrow x^{ 2 }=9\Rightarrow x=3\) dm
\(\Rightarrow \) Radius = 3dm , height = 6X3 dm = 18dm
Volume = \(\frac { 22 }{ 7 } \times 3^{ 2 }\times 18dm^{ 2 }=509.14\quad dm^{ 2 }\)
We have R= external radius = 12.5 cm
r = internal radius =(external radius - thickness ) = (12.5-1) cm = 11.5 cm
h= height of the pipe = 20 cm
\(\therefore \) Total surface area of the pipe = external curved surface + internal curved surface + 2 (area of the base of the ring )
\(2\pi Rh+2\pi rh+2(\pi R^{ 2 }-\pi r^{ 2 })=2\pi (R+r)h+2\pi (R^{ 2 }-r^{ 2 })\\ \)
= \(2\pi (R+r)h+2\pi (R+r)h+2\pi (R^{ 2 }-r^{ 2 })\)
\(=2\times \frac { 22 }{ 7 } \times (12.5+11.5)\times (20+12.5-11.5)cm^{ 2 }\)
\(=2\times \frac { 22 }{ 7 } \times 24\times 21cm^{ 2 }=3168\quad cm^{ 2 }\)