Among given options air has minimum value of rmal conductivity. Thermal conductivity of air is approximately K_air=0.02.

Q=-KA(dT/dx)
(ML²T^-3) = k (L²) (?)
K = ML²T^-3/L? , K=MLT^-3?^-1.

For one dimensional steady heat flow-
i. Temperature distribution in slab is linear
ii. Temperature distribution in cylinder is logarithmic
iii. Temperature distribution in sphere is hyperbolic.

Thermal diffusivity indicates ease at which energy get diffused in volume of substances. It is defined as ratio of rmal conductivity to heat capacity of substance.
Therefore, ? = k/ρc.

Thermal conductivity is minimum in gases and maximum in solids. Thermal conductivity of copper is higher than air. Hence heat flow will be in copper block rar than air block.

The temperature distribution will vary logarithmically with radius for cylinder.
? = \(\frac{2?Lk(T1-T2)}{(ln?(\frac{r1}{r2})}\).

Let k₁ and k₂ be rmal conductivity of two insulators, k₁ being better insulator.
Q = \(\frac{A?T}{\frac{ln?(\frac{r2}{r1})}{k1}+\frac{ln?(\frac{r3}{r2})}{k2}}\)
If we interchange insulators, value of Q might increase or decrease depending upon thickness of each layer.

Thermal resistance, R_th = \(\frac{L}{KA}=\frac{0.2}{0.5�1}=0.5\frac{K}{W} \)

Heat transfer, Q = \(\frac{(T1-T2)}{R_{th}}=\frac{100}{0.5}\)=200W=0.2kW.

Minimum outer radius should be critical radius. The minimum critical radius is given by,
r_c = \(\frac{k}{h}=\frac{0.12}{4}\)=0.03m=30mm
Hence, outer diameter = r_c = (2�30) mm = 60mm.

In steady state, Heat generated = Heat conducted at surface
q_g � ?R²L = q₀ � 2?RL
q_g = 2q₀/R.