Question

A wooden plank of mass ‘m’ and dimensions $l\times b$ floats over a very shallow river of Thickness ‘T’ and viscosity $\eta$. At the speed of the plank is $v_0$. A car is at rest on the bank at this instant the plank is adjacent to the car which starts to move with a constant acceleration If after sometime when the speed of the plank is $\frac{v_0}{2}t$. The car catches up with the plank t. (The car moves with a constant acceleration). Find the speed of the car at this instant when the car is crossing the plank

• A. $\frac{\sqrt{3}{v}_0}{2}$
• B. $\frac{v_0\eta lb}{2T}$
• C. $\frac{v_0}{ln2}$
• D. $\frac{3v_0}{4}$

Answer 1

The correct option is C

$\frac{v_0}{ln2}$

$\frac{-\eta Av}{T}=m\frac{dv}{dt}\frac{-\eta Av}{T}=mv\frac{dv}{dx}\frac{\eta A}{Tm}{\int }_0^{x}dx=-{\int }_{v_0}^{\frac{v_0}{2}}dvx=\frac{v_0}{2}\frac{Tm}{\eta A}\dots \dots \left(I\right)-\frac{\eta Av}{Tm}=\frac{dv}{dt}-\frac{\eta A}{Tm}{\int }_0^{t}dt={\int }_{v_0}^{\frac{v_0}{2}}\frac{dv}{v}t=\frac{\left(ln2\right)Tm}{\eta A}\dots \dots \left(II\right)\frac{1}{2}a{t}^2=\frac{v_{0}Tm}{2\eta A}\left(fromI\right)\left(at\right)t=\frac{v_{0}Tm}{2\eta A}V\frac{\left(ln2\right)Tm}{\eta A}=\frac{v_{0}Tm}{2\eta A}\left(from\right(II)V=\frac{v_0}{ln2}$