The heat transfer by conduction through a thick cylinder (Q) is given by (where T₁ = Higher temperature, T₂ = Lower temperature, r₁ = Inside radius, r₂ = Outside radius, $$l$$ = Length of cylinder and k = Thermal conductivity)

A. $${\text{Q}} = \frac{{2\pi l{\text{k}}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{2.3\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$

B. $${\text{Q}} = \frac{{2.3\log \left( {\frac{{{r_2}}}{{{r_1}}}} \right)}}{{2\pi l{\text{k}}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}$$

C. $${\text{Q}} = \frac{{2\pi \left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{2.3l{\text{k}}\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$

D. $${\text{Q}} = \frac{{2\pi l{\text{k}}}}{{2.3\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$

Answer: Option A

Solution(By Examveda Team)

The heat transfer by conduction through a thick cylinder (Q) is given by
$${\text{Q}} = \frac{{2\pi l{\text{k}}\left( {{{\text{T}}_1} - {{\text{T}}_2}} \right)}}{{2.3\log \left( {\frac{{{{\text{r}}_2}}}{{{{\text{r}}_1}}}} \right)}}$$
Where T₁ = Higher temperature, T₂ = Lower temperature, r₁ = Inside radius, r₂ = Outside radius, $$l$$ = Length of cylinder and k = Thermal conductivity.