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Q. Statement I : A cyclist is moving on an unbanked road with a speed of $7\,kmh ^{-1}$ and takes a sharp circular turn along a path of radius of $2\,m$ without reducing the speed. The static friction coefficient is $0.2 .$ The cyclist will not slip and pass the curve $\left( g =9.8 m / s ^{2}\right)$
Statement II : If the road is banked at an angle of $45^{\circ}$, cyclist can cross the curve of $2\,m$ radius with the speed of $18.5\,kmh ^{-1}$ without slipping.
In the light of the above statements,
choose the correct answer from the options given below.

JEE MainJEE Main 2021Laws of Motion

Solution:

Statement I :
$v _{\max }=\sqrt{\mu Rg }=\sqrt{(0.2) \times 2 \times 9.8}$
$v _{\max }=1.97\,\,m / s$
$7\,km / h =1.944\,\,m / s$
Speed is lower than $v _{\max }$, hence it can take safe turn.
Statement II
$v_{\max }=\sqrt{\operatorname{Rg}\left[\frac{\tan \theta+\mu}{1-\mu \tan \theta}\right]}$
$=\sqrt{2 \times 9.8\left[\frac{1+0.2}{1-0.2}\right]}=5.42\,\,m / s$
$18.5\,\,km / h =5.14\,\,m / s$
Speed is lower than $v _{\max }$, hence it can take safe turn.