A cyclist riding the bicycle at a speed of 14 3 ms^ -1 takes a tu… - Vidyadip

MCQ

A cyclist riding the bicycle at a speed of $14\sqrt 3 ms^{-1}$takes a turn around a circular road of radius $20\sqrt 3 $ m without skidding. Given $g = 9.8 ms^{-2},$ what is his inclination to the vertical ....... $^o$

A
$30$
B
$90$
C
$45$
✓
$60$

✓

Answer

Correct option: D.

$60$

d
(d)$\theta = {\tan ^{ - 1}}\left( {\frac{{{v^2}}}{{rg}}} \right) = {\tan ^{ - 1}}\left[ {\frac{{{{(14\sqrt 3 )}^2}}}{{20\sqrt 3 \times 9.8}}} \right] = {\tan ^{ - 1}}[\sqrt 3 ]$$ = 60^\circ $

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