Total thermal resistance is 
$R=\dfrac{t _1}{K _1A _1}+\dfrac{t _2}{K _2A _2}+\dfrac{t _1}{K _1A _1}$

$\large \begin{array}{l} According\, to\, question............. \ Here, \ \, \, \, \, \, m=1200kg,\, \, \, break\, drum\, (m)=10kg \ \, \, \, \, u=100km/h=27.77m/s \ \, \, \, \, { S _{ Steel } }=450J/KgK,\, \, \, \, \, V=0\, m/s \ so,\,  \ Energy\, released\, during\, breaking=\, \, change\, in\, kinetic\, energy \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac { 1 }{ 2 } m{ v^{ 2 } }-\frac { 1 }{ 2 } m{ u^{ 2 } } \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac { 1 }{ 2 } (1200)\, { (0)^{ 2 } }-\frac { 1 }{ 2 } (1200)\, { (27.77)^{ 2 } } \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =0-462937.0374\, J \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =462937.0374\, J \ Now, \ \, \, 60\quad perecent\quad of\, this\, energy\, appears=0.6\times 462937.0374\, J=277762.225 \ For\, change\, in\, temperature:\, \, \, \, ms\Delta \frac { 1 }{ 2 } t=277762.225 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \Delta t=\dfrac { { 277762.225 } }{ { ms } } =\dfrac { { 277762.225 } }{ { 10\times 450 } }  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, \Delta t\, ={ 61.7^{ 0 } }C \ so\, \, the\, correct\, option\, is\, B. \end{array}$

Heat flows from hotter part to colder part and the above arrangement is done to avoid convection currents.

Smoke is lighter than air so it rises above the air, hence it is easier to breath crawling than while standing up.

The molecules of liquids and gases are not rigidly bound to each other and can move freely. This makes the transfer of heat by convection possible.

When birds flap their wings the density of the air decreases nearby since the air particles move apart. This less dense air rises and helps the bird to keep flying. The denser air replaces this space. This movement is known as convection current of air. 

Convection significantly transferring heat upwards. The hot gases from the combustion rises so we get both the radiation heat from the flame to our hand and the convective heat transfer from the hot air to your hand.

Most of heat transfer taking place on earth is due to conduction and convection .

The hot air is less dense (lighter) than the cooler air around it, so it naturally goes upward, making the area above the flame hotter than the area to the sides of it.

Calcination is defined as converting an ore into its oxide by heating strongly , below its melting point either in a limited supply of air or in the absence of air.

The thermal current is nothing but the rate of flow heat,

Conduction and Convection are non-existent in space, so both the statements are wrong. 

Near flames, heat transfer occurs by both convection and radiation (and some conduction too).
But convection occurs only in upward direction. So, above the flame, it is hotter rather than in front of the flame.

A continuous flow of warmer and cooler parts of air is established in which the air molecules are responsible for flow of heat. As the warmer air molecules move up, they are replaced bu cooler air molecules.