A thick conducting shell of inner radius R and outer radius 2R is surrounding by air at constant temperature $T_0>0{}^{0}C$ . Time taken for temperature of water $\left(at\text{ 0}{}^{0}C\right)$ inside the shell to increase to $\frac{T_0}{2}$ is $\frac{\rho R^{2}s}{x}$ . Find x, if thermal conductivity of shell is  .$k=In2,\rho =density\text{ of shell materials and s = specific heat capacity}$

Let’s first find resistance of shell. Consider a thin shell element of radius  $r$ and thickness $dr$.
  Resistance of element,
   $dR=\frac{dr}{k\cdot 4\pi r^2}$
  Resistance of shell,
   $$\begin{gathered} R^{\text{'}}=\frac{1}{4\pi k}\underset{R}{\overset{2R}{\int }}\frac{1}{r^2}dr=\frac{1}{4\pi k}{[-\frac{1}{r}]}_R^{2R} \\ \Rightarrow R^{\text{'}}=\frac{1}{8\pi kR}\mathrm{.........}\text{(i)} \end{gathered}$$
    Heat will flow from hot air in chamber towards water, leading to rise in temperature of water. If at any instant, temperature of water is $\text{T }{}^0\text{C}$ ,
 Rate of heat flow,
  $$\begin{gathered} \frac{dQ}{dt}=\frac{Temperature\text{ difference }}{Resistance}=\frac{T_0-T}{R^{\text{'}}} \\ ms\frac{dT}{dt}=\frac{T_0-T}{R^{\text{'}}} \\ \Rightarrow \frac{dT}{T_0-T}=\frac{1}{ms{R}^{\text{'}}}dt \end{gathered}$$
   Integrating both sides, we get
  $$\begin{gathered} \underset{0}{\overset{\frac{T_0}{2}}{\int }}\frac{dT}{T_0-T}=\frac{1}{msR\text{'}} \\ \ln T_0-\ln (T_0-\frac{T_0}{2})=\frac{t}{ms{R}^{\text{'}}} \\ \Rightarrow t=ms{R}^{\text{'}}ln2=\frac{ms}{8\pi kR}\ln 2 \end{gathered}$$
     [Using Eq. (ii)]
  $=\frac{\rho \cdot \frac{4}{3}\pi R^{3}s}{8\pi \ln 2R}\cdot \ln 2=\frac{\rho R^{2}s}{6}$
 Comparing with the given value, we get   x = 6