Question

Find the distance from the eye at which a coin of a diameter $1cm$ be placed so as to hide the full moon, it is being given that the diameter of the moon subtends an angle of $31^{\prime }$ at the eye of the observer,

• A. $110cm$
• B. $108cm$
• C. $110.9cm$
• D. $112cm$

Answer 1

Answer: (C)

The coin will just hide the full moon if the lines joining the observer's eye $O$ to the ends $A$ and $B$ of moon’s diameter touch the coin at the ends $P$ and $Q$ of its diameter.



Here $\angle POQ=\angle AOB=31^{\prime }$
$={\left(\frac{31}{60}\right)}^{\circ }=\frac{31}{60}\times \frac{\pi }{180}$ radian.
Since, this angle is very small, the diameter $PQ$ of the coin can be regarded as an arc of a circle whose centre is $O$ and radius equal to the distance of the coin from $O$.
$\therefore \frac{31\pi }{60\times 180}=\frac{l}{r}=\frac{1}{r}\left(\because \theta =\frac{l}{r}\right)$
$\Rightarrow r=\frac{60\times 180}{31\pi }$
$\Rightarrow r=\frac{60\times 180\times 7}{31\times 22}$
$\approx 110.9cm$