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

We need to evaluate the validity of the two statements regarding a cyclist negotiating a circular curve under different road conditions.

1. Analyze Statement I (Unbanked Road)

For a cyclist to safely cross a flat, unbanked circular curve without slipping, the maximum safe speed ($$v_{\text{max}}$$) is determined by the limiting static friction:

$$v_{\text{max}} = \sqrt{\mu_s g r}$$

Given parameters for Statement I:

• Radius ($$r$$) = $$2\text{ m}$$
• Static friction coefficient ($$\mu_s$$) = $$0.2$$
• Acceleration due to gravity ($$g$$) = $$9.8\text{ m s}^{-2}$$

Calculate $$v_{\text{max}}$$:

$$v_{\text{max}} = \sqrt{0.2 \times 9.8 \times 2} = \sqrt{3.92} \approx 1.98\text{ m s}^{-1}$$

Convert this maximum safe speed to kilometers per hour ($$\text{km h}^{-1}$$):

$$v_{\text{max}} = 1.98 \times \frac{18}{5} = 7.128\text{ km h}^{-1}$$

Since the cyclist’s actual speed ($$7\text{ km h}^{-1}$$) is less than the maximum safe threshold ($$7.128\text{ km h}^{-1}$$), the cyclist will successfully pass the curve without slipping. Thus, Statement I is true.

2. Analyze Statement II (Banked Road)

When a road is banked at an angle $$\theta$$ and the contribution of friction is neglected to find the ideal safe speed ($$v_0$$) without relying on tire grip:

$$v_0 = \sqrt{g r \tan\theta}$$

Given parameters for Statement II:

• Banking angle ($$\theta$$) = $$45^\circ$$
• Radius ($$r$$) = $$2\text{ m}$$

Calculate $$v_0$$:

$$v_0 = \sqrt{9.8 \times 2 \times \tan 45^\circ} = \sqrt{19.6 \times 1} \approx 4.427\text{ m s}^{-1}$$

Convert this optimal speed to kilometers per hour ($$\text{km h}^{-1}$$):

$$v_0 = 4.427 \times \frac{18}{5} \approx 15.94\text{ km h}^{-1}$$

If we factor in the friction ($$\mu_s = 0.2$$) to find the absolute maximum possible speed before sliding up the bank ($$v_{\text{max, banked}}$$):

$$v_{\text{max, banked}} = \sqrt{g r \left( \frac{\tan\theta + \mu_s}{1 - \mu_s \tan\theta} \right)}$$

$$v_{\text{max, banked}} = \sqrt{9.8 \times 2 \times \left( \frac{1 + 0.2}{1 - 0.2 \times 1} \right)} = \sqrt{19.6 \times \frac{1.2}{0.8}} = \sqrt{19.6 \times 1.5} = \sqrt{29.4} \approx 5.422\text{ m s}^{-1}$$

Converting this absolute limit to kilometers per hour ($$\text{km h}^{-1}$$):

$$v_{\text{max, banked}} = 5.422 \times \frac{18}{5} \approx 19.52\text{ km h}^{-1}$$

Since the stated speed of $$18.5\text{ km h}^{-1}$$ is strictly below this limit ($$19.52\text{ km h}^{-1}$$), the cyclist can securely navigate the curve without slipping. Thus, Statement II is true.

Conclusion

Both Statement I and Statement II are true, which matches Option D .