An aeroplane flying with uniform speed horizontally one $\mathrm{km}$ above the ground is observed at an elevation of $60^{\circ}$. After $10 \mathrm{~s}$ if the elevation is observed to be $30^{\circ}$, then the speed of the plane (in $\mathrm{km} / \mathrm{h}$ ) is

• A

  $\frac{240}{\sqrt{3}}$
• B

  $200 \sqrt{3}$
• Correct answer
  C

  $240 \sqrt{3}$
• D

  $\frac{120}{\sqrt{3}}$

Solution

$\begin{array}{llll}<br />\tan 60^{\circ} & =\frac{D P}{A P} \\<br />\Rightarrow \sqrt{3} & =\frac{1}{A P} \\<br />\Rightarrow A P & =\frac{1}{\sqrt{3}} & {[\because E Q=D P=1]}<br />\end{array}$
In $\Delta E A Q$,
$\begin{aligned}<br />\tan 30^{\circ} & =\frac{E Q}{A P+P Q} \\<br />\Rightarrow \frac{1}{\sqrt{3}} & =\frac{1}{\frac{1}{\sqrt{3}}+P Q} \Rightarrow \frac{1}{\sqrt{3}}+P Q=\sqrt{3} \\<br />\Rightarrow P Q & =\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}} \mathrm{~km} \\<br />\therefore \text { Speed of plane } & =\frac{\text { Distance }}{\text { Time }}=\frac{2 / \sqrt{3}}{\frac{10}{60 \times 60}}=\frac{2}{\sqrt{3}} \times \frac{3600}{10} \\<br />& =240 \sqrt{3} \mathrm{~km} / \mathrm{h}<br />\end{aligned}$

Hence, the correct answer is option: 'C'.