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(B) Given that $v = ky^\beta$,where $k$ is a constant. Since $v = \frac{dy}{dt}$,we have $\frac{dy}{dt} = ky^\beta$,which implies $y^{-\beta} dy = k d

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(B) Given that $v = ky^\beta$,where $k$ is a constant. Since $v = \frac{dy}{dt}$,we have $\frac{dy}{dt} = ky^\beta$,which implies $y^{-\beta} dy = k dt$. Integrating both sides from $0$ to $L$ for distance and $0$ to $T$ for time: $\int_0^L y^{-\beta} dy = \int_0^T k dt$. This gives $\frac{L^{1-\beta}}{1-\beta} = kT$,so $T = \frac{L^{1-\beta}}{k(1-\beta)}$. The average velocity is $\langle v \rangle = \frac{L}{T} = \frac{L}{L^{1-\beta} / (k(1-\beta))} = k(1-\beta) L^\beta$. We are given $\langle v \rangle \propto L^{1/3}$,so comparing the exponents of $L$,we get $\beta = 1/3$.

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