Efficiency , \(\eta =1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
where, T₂  = sink temperature
T₁ = source temperature.
\(1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 1 }{ 2 } \quad ...(i)\)
\(\\ 1-\left( \frac { { T }_{ 2 }-100 }{ { T }_{ 1 } } \right) =\frac { 2 }{ 3 } \ ...(ii)\)
From Eq. (i), \(\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 1 }{ 2 } \) and Eq. (ii)
\(\frac { { T }_{ 2 }-100 }{ { T }_{ 1 } } =\frac { 1 }{ 3 } \)
On dividing, we get 
 \(\frac { { T }_{ 2 } }{ { T }_{ 2 }-100 } =\frac { 3 }{ 2 } \Rightarrow { T }_{ 2 }=300K\) 
(ii) Substituting in Eq.(i) T₁ = 600K
(iii) As efficiency, \({ \eta }_{ 2 }\Rightarrow 1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
It equals to 1 only when \(\frac { { T }_{ 2 } }{ { T }_{ 1 } } =0\quad or\quad { T }_{ 2 }=0K\)
But absolute zero is not possible.

(i) Aman loves nature and his helpful boy. He also has presence of mind as he thought of an excellent idea to help cartman.
(ii) \({ L }_{ 1 }=5.231m,{ L }_{ 2 }=5.243m,{ T }_{ 1 }=27 ^{0}C,{ T }_{ 2 }=?\)
As we know 
\(\therefore { T }_{ 2 }-{ T }_{ 1 }=\frac { { L }_{ 2 }-{ L }_{ 1 } }{ { L }_{ 1 }\times \alpha } \)
\(\\ { T }_{ 2 }=\frac { { L }_{ 2 }-{ L }_{ 1 } }{ { L }_{ 1 }\alpha } \)
\({ T }_{ 2 }=\frac { 5.243-5.231 }{ 5.231\times 1.2\times { 10 }^{ -5 } } +27\)
\(\\ =218 \ ^{0} C\)
(iii) Linear expansion of solids i.e increase in length of solid on heating.

Mass of the body,m = 0.04 kg
The position time graph OA from t = 0 to t = 2s is a straight line therefore body is moving with a constant velocity.
Velocity of the body,v=Slope of x-t graph
= \(\frac { 2-0 }{ 2-0 } =1cm/s={ 10 }^{ -2 }m/s\quad [\because 1\quad cm={ 10 }^{ -2 }m]\)
Part AB of position time graph is also a straight line.Therefore velocity of the body
 \({ v }^{ \prime }=\frac { 0-2 }{ 0-2 } =-1\quad cm/s=-{ 10 }^{ -2 }cm/s\) 
Negative sign shows that the direction of velocity is reversed after 2 s and it is being repeated.
A suitable physical context for this motion is a ball moving with a constant velocity of 10^-2 m/s between two walls located at x=0 and at x=2m and rebounded repeadedtly on striking each wall.
magnitude of the impulse imparted to the ball after every two seconds
= chage in momentum of teh ball
= mv - mv^' = m(v - v^')
= \(0.04[{ 10 }^{ -2 }-(-{ 10 }^{ -2 })]=8\times { 10 }^{ -4 }kg-m/s\) 

In pulling a lawn mower, a force F is applied in upward direction, making an angle \(\theta \) with the horizontal [Fig]. Its vertical component in upward direction decreasing the effective weight of the mower.
 
In pushing a lawn mower, a force F is applied in downward direction, making an angle \(\theta \) with the horizontal [Fig]. Its vertical component is in downward direction increasing the effective weight of the mower. Therefore, it is easier to pull a lawn mower than to push it.

The number of strands of hair on our head
If we assume a uniform distribution of strands of hair on head then, the number of strands of hair
= Area of the head/Area of cross-section of hair
The thickness of a strand of hair is measured by an appropriate instrument, if it is obtained
d = 5 x 10^-5 = 5 x 10^-3 cm
Then, area of cross-section of human hair
 \(=\pi \left( \frac { d }{ 2 } \right) ^{ 2 }=\frac { \pi d^{ 2 } }{ 4 } \)
\(\\ =\frac { 3.14\times (5\times 10^{ -3 })^{ 2 } }{ 4 } =\frac { 3.14\times 25 }{ 4 } \times 10^{ -6 }cm^{ 2 }\)
Average radius of human head (r)=8 cm
 \(\because \) Area of human head \(=\pi r^{ 2 }=3.14\times (8)^{ 2 }\)
3.14 x 64 cm²
\(\therefore \)The number of strands of hair \(=\frac { 3.14\times 64 }{ 3.14\times \frac { 25 }{ 4 } \times 10^{ -6 } } \)
 \(\approx 10\times 10^{ 6 }=10^{ 7 }\)

The total mass of rain-bearing clouds over India during the monsoon
If meteorologist record 10 cm of average rainfall during monsoon then
Height of  average rainfall (h) = 10 cm = 0.1 m
Area of India (A) = 3.3 million square km
= 3.3 x 10⁶ square km  [\(\because \) 1 million = 10⁶]
= 3.3 x 10⁶(10³m)²
= 3.3 x 10⁶ x 10⁶m² = 3.3 x 10¹²m²   [\(\because \) 1 km = 10³]
Volume of rain water (V) = Area x height
= Axh = 3.3 x 10¹²m² x 0.1m
=3.3x10¹¹m³
Density of water (\(\rho \) ) = 10³ kg/m³
Mass of rain water (m) = Volume x Density
= m = vx 
= 3.3 x 10¹¹m³ x 10³ kg/m³
[Density = [\(\because \) mass/Volume]
= 3.3 x 10¹⁴ kg
Therefore, total mass of rain bearing clouds over India during the monsoon is 3.3 x 10¹⁴ kg.