R = 28 cm 

r = 21 cm 

h = 15 cm 

Capacity of the bucket = $\frac{1}{3}\pi h(R^2+r^2+Rr)$ 

= $\frac{1}{3}\times \frac{22}{7}\times 15\times \left[\right(28{)}^2+(21{)}^2+(28\left)\right(21\left)\right]c{m}^3$ 

= $\frac{22}{7}\times 5\times [784+441+588]c{m}^3$ 

= $\frac{22}{7}\times 5\times 1813c{m}^3$ 

= 22 × 5 × 259 $c{m}^3$

= 28490 $c{m}^3$

= $\frac{28490}{1000}$ liters 

= 28.49 liters 

This leaves only the following surfaces for the combined solid-

1. One flat surface of coin

2. Curved surface of coin

3. Curved surface area of cone

These 3 areas add up together to give total surface area of combined solid.

The total surface area of the tent:

$T=2\pi r{h}_1+\pi rl\pi =\frac{22}{7};r=2.1\text{m}{h}_1=\text{height of the cylinder}=4\text{m}{h}_2=\text{height of the cone}=2.8\text{m}l=\text{slant height of the cone}l=\sqrt{{2.8}^2+{2.1}^2}=3.5\text{m}$

So,

$T=(2\times \frac{22}{7}\times 2.1\times 4+\frac{22}{7}\times 2.1\times 3.5)T=75.9{\text{m}}^2$

Therefore, the total cost of canvas at ₹100 per sq meter = $75.9\times 100=₹7590$.

Total surface area of the bucket
$=TSA=\pi \times l\times (r_1+r_2)+\pi \times ({r_1}^2+{r_2}^2)$  

Where $\pi =\frac{22}{7},$
$l=$ Slant height of the bucket  
$=\sqrt{(h{)}^2+(r_1-r_2{)}^2}$
$r_1=28cm$ and $r_2$  = 7 cm 
$l=$ Slant height of the bucket
 $=\sqrt{(45{)}^2+(21{)}^2}cm=49.6cm$
Therefore, 
$TSA=[\frac{22}{7}\times 49.6\times (28+7)+\frac{22}{7}\times ({28}^2+7^2)]=8074c{m}^2$

If any solid is reshaped into a new solid then volume always remains constant.

In one minute 192.5 litres of water flows.
So, The Volume of water that flows in one hour = (192.50 $\times$ 60) liters.  [$\because 1\text{hour}=60\text{mins}$ ]
Volume in ${\text{cm}}^3$ = (192.5 $\times$ 60 $\times$ 1000) ${\text{cm}}^3$ [$\because 1\text{litre}=1000{\text{cm}}^3$ ]
Inner radius of the pipe = 3.5 cm.

Let the length of column of water that flows in 1 hour be h cm.

Then,$\frac{22}{7}$ $\times$ 3.5 $\times$ 3.5 $\times$ h = 192.5 $\times$ 60 $\times$ 1000

h= 300000 cm = 3 km

Hence, the rate of flow = 3 km per hour.

Given
r = 7m
h = 1m
Surface area of hemisphere = 2$\pi$$r^2$ 
$=2\times \frac{22}{7}\times (7{)}^2$ = 308 $m^2$.

For calculating the surface area of a cone we need to calculate its slant height,
l = $\sqrt{r^2+h^2}$

l = $\sqrt{49+1}=\sqrt{50}$ m.

Surface area of cone = $\pi$rl= $\frac{22}{7}\times 7\times \sqrt{50}=155.39m^2$.

So, area of cloth required = (308 + 155.39)$m^2$ = 463.39 $m^2$

Given, Radius of Cylinder, $r=7\text{m}$ 
            length of roller, $h=2\text{m}$ 
            Width of road,  $B=2\text{m}$
            Length of road be ${}^{\prime }{L}^{\prime }$ 
            Number of revolution taken = 25
Roller rolls on its curved surface area.
$\therefore$  Surface area of   =   CSA of cylinder
   one revolution  
                                 $=2\pi rh=2\times \frac{22}{7}\times 7\times 2=88{\text{m}}^2$

Total area covered in 25  revolution  $=25\times 88=2200{\text{m}}^2$

This area will be equal to the area of the road levelled
$\text{Area of road =}L\times B=2200\Rightarrow L\times 2=2200\Rightarrow L=1100\text{m}$
$\therefore$ Length of the level road is 1100 m.

The shape obtained when a semi circular disc is rotated by 360 degrees is a sphere. If it was rotated by 180 degrees instead of 360 degrees, then we obtain a hemisphere.  The line segment AB will be diameter of the sphere formed.

Let number of lead shots = n 

Height of cone $=h=8cm$

Radius of cone $=r_1=5cm$ 

Volume of water present in the cone = Volume of cone $=\frac{1}{3}.\pi .(r_1{)}^2.h$ 

$=(\frac{1}{3}\times \frac{22}{7}\times 5^2\times 8c{m}^2)$
Volume of water flown out $=(\frac{1}{4}\times \frac{1}{3}\times \frac{22}{7}\times 5^2\times 8)c{m}^3$

 Radius of lead shot $=r=0.5cm$

Volume of each lead shot ... $=\frac{4}{3}.\pi .r^3=(\frac{4}{3}\times \frac{22}{7}\times (0.5{)}^3)c{m}^3$

 According to given situation, n lead shots are thrown into cone such that 1/4th of water present in cone flows out.

It means volume of n lead shots = volume of water flown out

$\Rightarrow$ $(n\times \frac{4}{3}\times \frac{22}{7}\times (0.5{)}^3)=(\frac{1}{4}\times \frac{1}{3}\times \frac{22}{7}\times 5^2\times 8)$

$\Rightarrow$ $n=\frac{1}{4}\times \frac{1}{3}\times 5^2\times 8\times \frac{1}{\left({0.5}^3\right)}\times \frac{3}{4}$

$\Rightarrow$ $n=100$