JEE Main Physics Class 11 Laws of Motion Part 2 Questions

The maximum speed of cyclist on turn of unbanked road without slipping is given as
vmax=√µgR=√0.2×10×2=2ms−1‌‌[∵µ=0.2 (given) ]
Given, speed =7km/h
=‌

7000

3600

ms−1=‌

70

36

=1.94ms−1
As given speed is lesser than vmax, so the cyclist will not slip. Therefore, Statement I is true.
As per Statement II, angle of banking, θ=45∘
We know that, for banked road,
Vmax=√‌

gR(µ+tan‌θ)

(1−µ‌tan‌θ)

and ‌‌vmin=√‌

gR(tan‌θ−µ)

1+µ‌tan‌θ

⇒‌‌vmax=√‌

10×2(0.2+tan‌45∘)

1−0.2‌tan‌45∘

and ‌‌vmin=√‌

10×2(1−0.2)

1+0.2

and ‌‌vmin=√‌

10×2(1−0.2)

1+0.2

⇒‌‌vmax=5.47ms−1 and vmin=3.65ms−1
∵‌‌v=18.5km∕h=‌

18.5×1000

5600

=5.13ms−1
[∵ given speed =18.5kmh−1]
As, vmin<v<vmax, so the cyclist will not slip.
∴ Statement II is also true.
Hence, option (d) is the correct.