- Home
- Standard 11
- Physics
4-1.Newton's Laws of Motion
easy
A cyclist riding the bicycle at a speed of $14 \sqrt{3} \,m / s$ takes a turn around a circular road of radius $20 \sqrt{3} \,m$ without skidding. What is his inclination to the vertical?
A$30$
B$45$
C$60$
D$75$
Solution
(c)
$\tan \theta=\frac{v^2}{r g}$
$=\frac{14 \times 14 \times 3}{20 \sqrt{3} \times 10} \simeq \sqrt{3}$
$\theta=60^{\circ}$
$\tan \theta=\frac{v^2}{r g}$
$=\frac{14 \times 14 \times 3}{20 \sqrt{3} \times 10} \simeq \sqrt{3}$
$\theta=60^{\circ}$
Standard 11
Physics
Similar Questions
easy
medium
