Answer: Option C
In the figure, AB is chimney and CB and DB are its shadow
$$\eqalign{
& \tan {60^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{{BC}} \cr
& \Rightarrow \sqrt 3 = \frac{h}{{BC}} \cr
& \Rightarrow BC = \frac{h}{{\sqrt 3 }}\,.......\,\left( {\text{i}} \right) \cr
& {\text{and}} \cr
& \tan {30^ \circ } = \frac{h}{{DB}} = \frac{h}{{DB + BC}} \cr
& \frac{1}{{\sqrt 3 }} = \frac{h}{{x + BC}} \cr
& x + BC = h\sqrt 3 \cr
& \Rightarrow BC = h\sqrt 3 - x\,.......\,\left( {{\text{ii}}} \right) \cr
& {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr
& \frac{h}{{\sqrt 3 }} = h\sqrt 3 - x \cr
& \Rightarrow \frac{h}{{\sqrt 3 }} - h\sqrt 3 = - x \cr
& x = h\sqrt 3 - \frac{h}{{\sqrt 3 }} \cr
& x = h\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) \cr
& x = h\frac{{3 - 1}}{{\sqrt 3 }} \cr
& x = \frac{{2h}}{{\sqrt 3 }} \cr
& \therefore h = \frac{{\sqrt 3 }}{2}x \cr} $$