A cyclist leans with the horizontal at angle $30^{\circ}$, while negotiating round a circular road of radius $20 \sqrt{3} \mathrm{~m}$. The speed of the cycle should be

• A

  $7 \sqrt{3} \mathrm{~m} / \mathrm{s}$
• B

  $14 \mathrm{~m} / \mathrm{s}$
• C

  $7 \sqrt{6} \mathrm{~m} / \mathrm{s}$
• Correct answer
  D

  $10 \sqrt{6} \mathrm{~m} / \mathrm{s}$

Solution

According to the question, the situation is as shown below
Balancing the forces, $N \sin 30^{\circ}=m g$ $\ldots(\mathrm{i})$ $\text { and } \quad N \cos 30^{\circ}=\frac{m v^2}{R}$ $\ldots(\mathrm{ii})$ Dividing Eq. (i) by Eq. (ii), we get $\tan 30^{\circ}=\frac{g R}{v^2}=\frac{g R}{\tan 30^{\circ}}$ $\Rightarrow \quad=10 \times 20 \sqrt{3} \times \sqrt{3} \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]$ $=600=\sqrt{600}=10 \sqrt{6} \mathrm{~m} / \mathrm{s}$

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