Circular Motion

The maximum speed of cyclist on turn of unbanked road without slipping is given as $v_{\text{max}} = \sqrt{\mu g R} = \sqrt{0.2 \times 10 \times 2} = 2 \text{ms}^{-1} \; [ \because \mu=0.2$ (given) $]$ Given, speed $=7 \text{km}/\text{h}$ $=\;\frac{7000}{3600} \text{ms}^{-1} =\;\frac{70}{36}=1.94 \text{ms}^{-1}$ As given speed is lesser than $v_{\text{max}}$, so the cyclist will not slip. Therefore, Statement I is true. As per Statement II, angle of banking, $\theta=45^{\circ}$ We know that, for banked road, $V_{\text{max}} = \sqrt{\;\frac{g R(\mu+\tan\theta)}{1-\mu\tan\theta}}$ and $\; v_{\text{min}} = \sqrt{\;\frac{g R(\tan\theta-\mu)}{1+\mu\tan\theta}}$ $\Rightarrow \; v_{\text{max}} = \sqrt{\;\frac{10 \times 2(0.2+\tan 45^{\circ})}{1-0.2 \tan 45^{\circ}}}$ and $\; v_{\text{min}} = \sqrt{\;\frac{10 \times 2(1-0.2)}{1+0.2}}$ and $\; v_{\text{min}} = \sqrt{\;\frac{10 \times 2(1-0.2)}{1+0.2}}$ $\Rightarrow \; v_{\text{max}} = 5.47 \text{ms}^{-1}$ and $v_{\text{min}} = 3.65 \text{ms}^{-1}$ $\because \; v=18.5 \text{km}/\text{h} =\;\frac{18.5 \times 1000}{5600}=5.13 \text{ms}^{-1}$ $[ \because$ given speed $=18.5 \text{kmh}^{-1}]$ As, $v_{\text{min}} < v < v_{\text{max}}$, so the cyclist will not slip. $\therefore$ Statement II is also true. Hence, option (d) is the correct.