IIT JEE 1964 Chemistry Question Paper with Answer and Solution

(B) Let the total height of the pole be $H$. The lower part is $\frac{H}{3}$ and the upper part is $\frac{2H}{3}$.
Let the point be $P$ at a distance $d = 20 \ m$ from the base.
Let $\alpha$ be the angle subtended by the lower part at $P$,and $\beta$ be the angle subtended by the whole pole at $P$.
The angle subtended by the upper part is $\theta = \beta - \alpha$,where $\tan \theta = \frac{1}{2}$.
From the geometry,$\tan \alpha = \frac{H/3}{d} = \frac{H}{3d}$ and $\tan \beta = \frac{H}{d}$.
Using the formula $\tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha} = \frac{1}{2}$.
Substituting the values: $\frac{\frac{H}{d} - \frac{H}{3d}}{1 + (\frac{H}{d})(\frac{H}{3d})} = \frac{1}{2}$.
$\frac{2H/3d}{1 + H^2/3d^2} = \frac{1}{2} \Rightarrow \frac{2H}{3d} \times \frac{3d^2}{3d^2 + H^2} = \frac{1}{2}$.
$\frac{2Hd}{3d^2 + H^2} = \frac{1}{2} \Rightarrow 4Hd = 3d^2 + H^2$.
$H^2 - 4dH + 3d^2 = 0$.
Given $d = 20$,$H^2 - 4(20)H + 3(20^2) = 0 \Rightarrow H^2 - 80H + 1200 = 0$.
$(H - 20)(H - 60) = 0$.
Thus,$H = 20 \ m$ or $H = 60 \ m$.