Explanation
We know that the heat content is proportional to the mass and specific heat capacity of the substance as well as temperature. As both the spheres are of the same material and are at the same temperature, heat content will be dependent on mass only.
$$ m = \rho \times v$$
where $$ \rho $$ is the density of the sphere and $$v$$ the volume.
And $$ v $$ = $$ \dfrac {4 \pi r^3}{3} $$
The given radius ratio is $$ 1:2 $$.
Hence heat contents ratio is in the ratio $$1:8$$
Let $$A$$ be the area of the both the plates.
Given ,
$$ \dfrac{k_1}{k_2}$$ = $$ \dfrac {2}{3} $$ -------$$(i)$$
where $$k_1$$ and $$k_2$$ are the thermal conductivities of plate 1 and 2 respectively.
And
$$ \dfrac{l_1}{l_2}$$ = $$ \dfrac {2}{3} $$ ------- $$(ii)$$
where $$l_1$$ and $$l_2$$ are the thermal conductivities of plate 1 and 2 respectively.
Let the temperature of the common surface be $$T$$
Equating the rate of heat transfers through both the plates
$$ \dfrac{k_1 A (10-T)}{L_1}$$ = $$ \dfrac{k_2 A (T-0)}{L_2}$$
Using $$ (i) $$ and $$ (ii) $$
We have ,
$$ 10-T = T-0 $$
Hence , $$ T = 5^0C $$
Please disable the adBlock and continue. Thank you.