(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.

Given, A = 1000 cm² , L = 0.4 cm
\(\Delta T=T_{ 1 }-T_{ 2 }=37-(-5)=42^{ 0 }C\)
\(\\ K=2.2\times 10^{ -3 }cal\ s^{ -1 }cm^{ -1 }K^{ -1 }\)
Rate of loss heat,
\(\\ H=\frac { Q }{ T } =\frac { KA(T_{ 1 }-T_{ 2 }) }{ L } =\frac { 2.2\times 10^{ -3 }\times 1000\times 42 }{ 0.4 } \)
\(\\ H=231\ cal\ s^{ -1 }\)

3.6 x 10^-2

20\(\Omega\), 3.8 x 10-3 \(^{0}\)C^-1, - 5.6 x 10 ^-7 \(^{0}\)C^-2

The lower layer of earth's atmosphere reflects infra-red radiations from earth back to the surface of the earth. So the heat radiation received by the earth from the sun during del) time are trapped by the atmosphere. Therefore, if the earth did not have atmosphere, its surface would become too cold to tolerate

A body with good reflectivity is a poor absorber of heat and the poor absorbers of heat are poor emitters of thermal radiations

According to Stefan's Law,
Energy emitted per unit time i.e. power
H = A xT⁴
Here   H₁ = 450 W, T = T₁ and radius r = r₁
and H₂ = ?  T = 2T₁ and r = \(\frac { { r }_{ 1 } }{ 2 } \)
\({ H }_{ 1 }=4{ \pi r }_{ 1 }^{ 2 },\sigma { T }_{ 1 }^{ 4 }\)
and \({ H }_{ 2 }=4\pi \left( \frac { { r }_{ 1 } }{ 2 } \right) ^{ 2 }.\sigma \left( 2{ T }_{ 1 } \right) ^{ 2 }\)
Dividing eq (i) by (ii) we get
\(\frac { { H }_{ 1 } }{ { H }_{ 2 } } =\frac { 4\pi { r }_{ 1 }^{ 2 }.\sigma { T }_{ 1 }^{ 4 } }{ 4\pi \left( \frac { { r }_{ 1 } }{ 2 } \right) ^{ 2 }.\sigma (2{ T }_{ 1 })^{ 4 } } \)
\(\Rightarrow\) \(\frac { 450 }{ { H }_{ 2 } } =\frac { { r }_{ 1 }^{ 2 } }{ \frac { { r }_{ 1 }^{ 2 } }{ 4 } } \times \frac { { T }_{ 1 }^{ 4 } }{ 16{ R }_{ 1 }^{ 4 } } \)
\(\Rightarrow\)  \(\frac { 450 }{ { H }_{ 2 } } =\frac { 1 }{ 4 } \)
\(\therefore\)  \({ H }_{ 2 }=450\times 4=1800w\)
Hence the required power = 1800 W

AS radius of ball, r₀ = 9.0 cm = 0.090 m at 0⁰C, hence its
\(Volume,\ \ V_{ 0 }=\frac { 4 }{ 3 } \pi { r }_{ 0 }^{ 3 }=\frac { 4 }{ 3 } \times 3.14\times (0.090)^{ 3 }\)
\(\\ =3.05\times 10^{ -3 }m^{ 3 }\)
\(\\ Again\ as\ \alpha =1.2\times 10^{ -5 }K^{ -1 },\)
\(\\ \therefore \ \gamma =3\alpha \)
\(\\ =3\times 1.2\times 10^{ -5 }=3.6\times 10^{ -5 }K^{ -1 }\)
\(\\ Moreover\ rise\ in\ temperature\)
\(\\ \triangle T=90^{ 0 }C-0^{ 0 }C=90^{ 0 }C=90K\)
\(\\ Increase\ in\ volume,\ \triangle V=V\gamma \triangle T\)
\(\\ =3.05\times 10^{ -3 }\times 3.6\times 10^{ -5 }\times 90\)
\(\\ =9.88\times 10^{ -6 }m^{ 3 }=9.88cm^{ 3 }\) 

\(11\times { 10 }^{ 5 }N\)

We know that, speed,  \(v\infty \sqrt { T } \)
By formula v = \(\frac { xRT }{ p } \)
Where T is in kelvin 
\(\frac { v_{ t } }{ v_{ 0 } } =\sqrt { \frac { 273+t }{ 273+0 } } =3\)   
\(\\ \Rightarrow \frac { 273+t }{ 273 } =9\Rightarrow \ t=9\times 273-273=2184^{ 0 }C\) 

Let M be the mass of glycerine, its density at 0⁰C, its density at t⁰C.
\(\gamma =\frac { { V }_{ t }-{ V }_{ 0 } }{ { V }_{ 0 } } =\frac { \frac { M }{ { \rho }_{ t } } -\frac { M }{ { { \rho } }_{ 0 } } }{ (M/{ { \rho } }_{ 0 })\triangle T } \)
\(\\ \gamma =\frac { \frac { 1 }{ { \rho }_{ t } } -\frac { 1 }{ { \rho }_{ 0 } } }{ (1/{ \rho }_{ 0 })\triangle T } =\frac { { \rho }_{ 0 }-{ \rho }_{ t } }{ { \rho }_{ 0 }\triangle t } \)

∴ Fractional change in density
\(\\ \frac { { \rho }_{ 0 }-{ \rho }_{ t } }{ { \rho }_{ 0 } } =\gamma \triangle T\)
\(\\ =49\times 10^{ -5 }\times 30=0.0147\)

Heat carried away from a fire sideways mainly by radiation. Above the fire, heat is carried by both radiation and convection of air. But convection carries much more heat than radiation. So, it is much hotter above a fire than by its sides

The purpose of a coolant is to absorb maximum heat with least rise in its own temperature. This is possible only if specific heat is high because Q = mc \(\Delta \)T. For a given value of m and Q, the rise in temperature \(\Delta \)T will be small if c is large. This will prevent different parts of the nuclear reactor from getting too hot.

As both flasks have same height and capacity, the area of the cylindrical wall will be less than that of the square wall. Hence, the thermos flask of circular cross section will transmit less heat as compared to the thermos flask of square cross section and will be better.

The neck expands but not the stopper due to poor conductivity of glass. Thus, the stopper can be taken out easily.

This is because platinum begins evaporate above 1200⁰C.

Electrons contribute largely both towards the flow of electricity and the flow of heat. A good conductor contains a large number of free electrons. So, it is both a good conductor of heat and electricity.

The two bodies may have different masses and different materials i.e., they may have different thermal capacities. In case the two bodies have equal thermal capacities, they would settle at the mean \(\frac { ({ T }_{ 1 }+{ T }_{ 2 }) }{ 2 } \)

When the birds swell their feathers, they are able to enclose air in the feathers. Air, being a poor conductor of heat, so it prevents the loss of heat from the bodies of the birds to the surroundings and as such they do not feel cold in winter.