Physics - Heat & Thermo Question with Solution | TestHub

(P) Spherical: H = constant,$\frac{3{\mathrm{T}}_0-{\mathrm{T}}_{\mathrm{P}}}{{\mathrm{R}}_1}=\frac{{\mathrm{T}}_{\mathrm{P}}-{\mathrm{T}}_0}{{\mathrm{R}}_2}$with$\mathrm{R}=\frac{{\mathrm{r}}_2-{\mathrm{r}}_1}{4\pi \mathrm{K}\left({\mathrm{r}}_1\times {\mathrm{r}}_2\right)}$⇒$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=1\Rightarrow {\mathrm{T}}_{\mathrm{P}}=2{\mathrm{T}}_0$→ P-2

(Q) Cylindrical:$\mathrm{R}=\frac{\ln \left(\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\right)}{2\pi \mathrm{Kl}}$⇒$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{\ln \left(\frac{2\mathrm{R}}{\mathrm{R}}\right)}{\ln \left(\frac{3\mathrm{R}}{2\mathrm{R}}\right)}=\frac{0.7}{0.4}$;$\left(3{\mathrm{T}}_0-{\mathrm{T}}_{\mathrm{P}}\right)4=7\left({\mathrm{T}}_{\mathrm{P}}-{\mathrm{T}}_0\right)\Rightarrow 19{\mathrm{T}}_0=11{\mathrm{T}}_{\mathrm{P}}\Rightarrow {\mathrm{T}}_{\mathrm{P}}=\frac{19{\mathrm{T}}_0}{11}$→ Q-5

(R) Rod:$\frac{3{\mathrm{T}}_0-{\mathrm{T}}_{\mathrm{P}}}{{\mathrm{R}}_1}=\frac{{\mathrm{T}}_{\mathrm{P}}-{\mathrm{T}}_0}{{\mathrm{R}}_2}$with$\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{1}{2}\Rightarrow 6{\mathrm{T}}_0-2{\mathrm{T}}_{\mathrm{P}}={\mathrm{T}}_{\mathrm{P}}-{\mathrm{T}}_0\Rightarrow {\mathrm{T}}_{\mathrm{P}}=\frac{7{\mathrm{T}}_0}{3}$→ R-4

(S) T-shape, junction law at P:$\frac{{\mathrm{T}}_{\mathrm{P}}-3{\mathrm{T}}_0}{\mathrm{R}}+\frac{{\mathrm{T}}_{\mathrm{P}}-{\mathrm{T}}_0}{\mathrm{R}}+\frac{{\mathrm{T}}_{\mathrm{P}}-2{\mathrm{T}}_0}{2\mathrm{R}}=0\Rightarrow {\mathrm{T}}_{\mathrm{P}}=2{\mathrm{T}}_0$→ S-2

Hence (A) P-2, Q-5, R-4, S-2.