Time taken by minute hand to make one circle = 60 minutes.
\(\therefore\) Angle described in 60 minutes = \(360^o\)
Angle described in 5 minutes [i.e., from 7.00 a.m. to 7.05 a.m.] = \({360^o\over 60^o}\times 5=30^o\)
Radius of circle = length of minute hand = \(\sqrt {21}cm\)
Area swept = \({\theta \over 360^o}\times r^2={30^oover 360^o}\times {22\over 7}\times \sqrt{21}\times \sqrt {21}={1\over 12}\times {22\over 7}\times 21 = {11\over 2}\ cm^2=5.5 cm^2\)

length of arc AB = 5\(\pi\) cm
Let AOB = \(\theta\)
Now   i = \(\frac { \theta }{ 360° } \)x 2\(\pi\)r
5\(\pi\) = \(\frac { \theta }{ 180° } \)\(\pi\)r ⇒ \({900\over r}=\theta\)
Now, Area of sector = \(\frac { \theta }{ 360° } \)x 2\(\pi\)r²
20\(\pi\) =  \(\frac { r }{ 360° } \)x 2\(\pi\)r² ⇒ 20 = \(900°\over360°\)r ⇒ r = 8 cm

Let length of the pendulum he I cm.
Length of the arc = 8.8 cm
⇒ \(\frac { \theta \pi i }{ 180 } =8.8cm\Rightarrow \frac { 30°\times \pi \times i }{ 180° } =8.8\)
⇒ \(\pi i=8.8\times 6\Rightarrow \frac { 22 }{ 7 } \times i=52.8cm\)
⇒ \(i=\frac { 52.8\times 7 }{ 22 } cm\quad =\quad 16.8\quad cm\)

Distance covered in 5000 revolutions = 11 km.
Distance covered in 1 revolution
Distance covered in 1 revolution \(={11000\over 5000}m={11\over 5}m\)
Distance covered in 1 revolution = circumference of the wheel
\(\Rightarrow\)\(2\ \pi r ={11\over 5} \Rightarrow 2\times {22\over 7}\times r = {11\over 5}\)
\(\Rightarrow\)  \(r \Rightarrow {11\over 5}\times 7\times {1\over 2\times 22}={7\over 20}m\)
\(\therefore\) Diameter \(=2\times r=2\times {7\over 20}={7\over 10}\times 100 cm=70cm\)

Area of the sector = \({\theta \pi r^2\over 360^O}1={90^o\over 360^o}\times \pi \times (14)^2\) 
= \({1\over 4}\times {22\over 7}\times 14\times 14\)
= \(154\ cm^2\)