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)

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, $\tan \theta=\frac{v^{2}}{r g}=\frac{\frac{(34.3)^{2}}{22}}{\frac{343}{2 \pi} \times 9.8}=1$
$=45^{\circ}$