Two conducting slabs of heat conductivity \(K_{1} ~\text{and}~K_{2}\) are joined as shown in figure. If the temperature at the ends of the slabs are \(\theta_{1}~\text{and}~\theta_{2} \ (\theta_{1}   >   \theta_{2} ),  \) then the final temperature \( \left(\theta\right)_{m} \) of the junction will be:

                

1.	\(\frac{K_{1} \theta_{1}   +   K_{2} \theta_{2}}{K_{1}   +   K_{2}}\)	2.	\(\frac{K_{1} \theta_{2}   +   K_{2} \theta_{1}}{K_{1}   +   K_{2}}\)
3.	\(\frac{K_{1} \theta_{2}   +   K_{2} \theta_{1}}{K_{1}   -   K_{2}}\)	4.	None

A cup of coffee cools from \(90^{\circ}\text{C}\) $\mathrm{to}$ \(80^{\circ}\text{C}\) in \(t\) minutes, when the room temperature is \(20^{\circ}\text{C}.\)$\mathrm{}$ The time taken by a similar cup of coffee to cool from \(80^{\circ}\text{C}\) $\mathrm{to}$ \(60^{\circ}\text{C}\) at room temperature same at \(20^{\circ}\text{C}\) is:
1. \(\frac{10}{13}t\) 
2. \(\frac{5}{13}t\)
3. \(\frac{13}{10}t\)
4. \(\frac{13}{5}t\)

A slab of stone with an area \(0.36~\text{m}^{2}\)${\mathrm{}}^{}$ and thickness of \(0.1~\text{m}\) is exposed on the lower surface to steam at \(100​​^\circ\text{C}.\)$$ A block of ice at \(0^{\circ}\text{C}\) rests on the upper surface of the slab. In one hour \(4.8~\text{kg}\) of ice is melted. The thermal conductivity of the slab will be:
(Given latent heat of fusion of ice \(= 3.36\times10^{5}~\text{JKg}^{-1}\)${\mathrm{}}^{}$)
1. \(1.29~\text{J/m/s/}^{\circ}\text{C}\)
2. \(2.05~\text{J/m/s/}^{\circ}\text{C}\)
3. \(1.02~\text{J/m/s/}^{\circ}\text{C}\)
4. \(1.24~\text{J/m/s/}^{\circ}\text{C}\)

The temperature at which the Celsius and Fahrenheit thermometers agree (to give the same numerical value) is:

Which of the curves in the figure represents the relation between Celsius and Fahrenheit temperature?

1.		2.
3.		4.

A piece of iron is heated in a flame. If it becomes dull red first, then becomes reddish yellow, and finally turns to white hot, the correct explanation for the above observation is possible by using: