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(D) Let the distance moved up the incline be $s$. The acceleration while moving up is $a_{up} = g(\sin 	heta + \mu \cos 	heta)$.Using $v^2 - u^2 = 2

Vedclass · Accepted Answer

$346$

Vedclass · Answer

(D) Let the distance moved up the incline be $s$. The acceleration while moving up is $a_{up} = g(\sin 	heta + \mu \cos 	heta)$.Using $v^2 - u^2 = 2as$,for the upward journey: $0 - v_{0}^2 = -2 a_{up} s \implies s = \frac{v_{0}^2}{2g(\sin 	heta + \mu \cos 	heta)}$.The acceleration while moving down is $a_{down} = g(\sin 	heta - \mu \cos 	heta)$.For the downward journey: $(\frac{v_{0}}{2})^2 - 0 = 2 a_{down} s \implies s = \frac{v_{0}^2}{8g(\sin 	heta - \mu \cos 	heta)}$.Equating the two expressions for $s$: $\frac{v_{0}^2}{2g(\sin 	heta + \mu \cos 	heta)} = \frac{v_{0}^2}{8g(\sin 	heta - \mu \cos 	heta)}$.$4(\sin 	heta - \mu \cos 	heta) = \sin 	heta + \mu \cos 	heta \implies 3 \sin 	heta = 5 \mu \cos 	heta$.$\mu = \frac{3}{5} 	an 30^{\circ} = \frac{3}{5} 	imes \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{5} \approx 0.3464$.Given $\mu = \frac{I}{1000}$,we have $I = 346.4$. The nearest integer is $346$.

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