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