Statement I : A cyclist is moving on an unban | vedclass

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(D) Statement $I :$ For an unbanked road,the maximum safe speed is $v_{\max} = \sqrt{\mu Rg}$.Given $\mu = 0.2$,$R = 2 \, m$,$g = 9.8 \, m/s^2$.$v_{\m

Vedclass · Accepted Answer

Both statement $I$ and statement $II$ are true.

Vedclass · Answer

(D) Statement $I :$ For an unbanked road,the maximum safe speed is $v_{\max} = \sqrt{\mu Rg}$.Given $\mu = 0.2$,$R = 2 \, m$,$g = 9.8 \, m/s^2$.$v_{\max} = \sqrt{0.2 	imes 2 	imes 9.8} = \sqrt{3.92} \approx 1.98 \, m/s$.The cyclist's speed is $7 \, km/h = 7 	imes \frac{5}{18} \approx 1.944 \, m/s$.Since $1.944 \, m/s Statement $II :$ For a banked road,the maximum safe speed is $v_{\max} = \sqrt{Rg \left[ \frac{	an 	heta + \mu}{1 - \mu 	an 	heta} ight]}$.Given $	heta = 45^{\circ}$,$	an 45^{\circ} = 1$,$\mu = 0.2$,$R = 2 \, m$,$g = 9.8 \, m/s^2$.$v_{\max} = \sqrt{2 	imes 9.8 	imes \left[ \frac{1 + 0.2}{1 - 0.2 	imes 1} ight]} = \sqrt{19.6 	imes \frac{1.2}{0.8}} = \sqrt{19.6 	imes 1.5} = \sqrt{29.4} \approx 5.42 \, m/s$.The cyclist's speed is $18.5 \, km/h = 18.5 	imes \frac{5}{18} \approx 5.14 \, m/s$.Since $5.14 \, m/s < 5.42 \, m/s$,the cyclist will not slip. Thus,Statement $II$ is true.

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