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

The quantities of heat required to raise the temperature of two solid copper spheres of radii \(r_1\) and \(r_2\)${\mathrm{}}_{}$
$$\((r_1 = 1.5~r_2)\)$$through \(1~\text{K}\) are in the ratio:
1. \(\frac{9}{4}\)
2. \(\frac{3}{2}\)
3. \(\frac{5}{3}\)
4. \(\frac{27}{8}\)

Three stars \(A,\) \(B,\) and \(C\) have surface temperatures \(T_A,~T_B\)and \(T_C\) respectively. Star \(A\) appears bluish, star \(B\) appears reddish and star \(C\) yellowish. Hence,

A copper rod of \(88\) cm and an aluminium rod of an unknown length have an equal increase in their lengths independent of an increase in temperature. The length of the aluminium rod is:
\(\left(\alpha_{Cu}= 1.7\times10^{-5}~\text{K}^{-1}~\text{and}~\alpha_{Al}= 2.2\times10^{-5}~\text{K}^{-1}\right)\)
1. \(68~\text{cm}\)
2. \(6.8~\text{cm}\)
3. \(113.9~\text{cm}\)
4. \(88~\text{cm}\)

An object kept in a large room having an air temperature of \(25^\circ \text{C}\) takes \(12\) minutes to cool from \(80^\circ \text{C}\) to \(70^\circ \text{C}.\) The time taken to cool for the same object from \(70^\circ \text{C}\) to \(60^\circ \text{C}\) would be nearly:
1. \(10\) min
2. \(12\) min
3. \(20\) min
4. \(15\) min

A deep rectangular pond of surface area A, containing water (density = \(\rho,\) specific heat capacity = \(s\)), is located in a region where the outside air temperature is at a steady value of \(-26^{\circ}\mathrm{C}\)$$. The thickness of the ice layer in this pond at a certain instant is \(x\). Taking the thermal conductivity of ice as \(k\), and its specific latent heat of fusion as \(L\), the rate of increase of the thickness of the ice layer, at this instant, would be given by:
1. \(\frac{26k}{x\rho L-4s}\)
2. \(\frac{26k}{x^2\rho L}\)
3. \(\frac{26k}{x\rho L}\)
4. \(\frac{26k}{x\rho L+4s}\)