Question

The angle through which a cyclist bends when he covers a circular path of $34.3 \,m$ circumference in $\sqrt{22} \sec$ is $\left(g=9.8 \,m / s ^{2}\right)$

• A. $15^{\circ}$
• B. $30^{\circ}$
• C. $60^{\circ}$
• D. $45^{\circ}$

Answer 1

Answer: (D)

$45^{\circ}$

Answer 2

Speed of particle $v=\frac{s}{t}=\frac{343}{\sqrt{22}} m / s$ $r=\frac{s}{2 \pi}=\frac{34.3}{2 \pi} m$ Now, $\tan \theta=\frac{v^{2}}{r g}=\frac{\frac{(34.3)^{2}}{22}}{\frac{343}{2 \pi} \times 9.8}=1$ $=45^{\circ}$