Question

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

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

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 $\angle POQ = \angle AOB = 31'$
$=\left(\frac{31}{60}\right)^{\circ}=\frac{31}{60}\times\frac{\pi}{180}$ 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 $O$.
$\therefore \frac{31\pi}{60 \times 180}=\frac{l}{r}=\frac{1}{r}\,\left(\because \theta=\frac{l}{r}\right)$
$\Rightarrow r=\frac{60 \times 180}{31\pi}$
$\Rightarrow r=\frac{60 \times 180 \times 7}{31 \times 22}$
$\approx 110.9\,cm$