Correct Answer: $10\sqrt{3}$ metres

Solution :
Let AB be the pole. Suppose BD and BC are the lengths of the shadow of the pole when the elevation of the Sun is 30° and 60°, respectively.
Here, AB = 15 metres
In right $\triangle$ABC,
$⇒\tan 60°=\frac{\text{AB}}{\text{BC}}$
$⇒\sqrt{3}=\frac{AB}{\text{BC}}$
$⇒BC= \frac{15}{\sqrt3} =5\sqrt3$ metres .................... (i)
In $\triangle$ABD,
$⇒\tan 30°=\frac{\text{AB}}{\text{BD}}$
$⇒\frac{1}{\sqrt{3}}=\frac{15}{\text{$5\sqrt3$+CD}}$
$⇒5\sqrt3 + CD = 15\sqrt3$
$⇒CD = 10\sqrt3$ metres
Now, if the elevation of the sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high is $10\sqrt{3}$ metres.
Hence, the correct answer is $10\sqrt{3}$ metres.