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A drinking straw is dipped in a pan of water to depth d from the surface (see figure below). Now water is sucked into it up to an initial height $h_0$ and then left to oscillate. As a result, its height $y$ from the surface of the water varies periodically. Ignoring damping, the equation for $y$ is ( $g$ is the acceleration due to gravity):
$\ddot{y}+\frac{g}{d} y=0$
$\ddot{y}(y+d)+\frac{g}{d}(y+d)=0$
$\ddot{y}+\frac{\dot{y}^2}{d}+\frac{g}{d}(y+d)=0$
$\ddot{ y }+( y + d )+\dot{ y }^2+ gy =0$
Solution
(D)
Consider the mass of liquid in the straw. The entire liquid is moving with velocity $\dot{ y }$. Applying Newtons law on it.
$F_{\text {thrust }}=u_{\text {rel }} \frac{d m}{d t}=-\rho A \dot{y}^2$
$F_{\text {pressure }}=(\rho g d) A$
$F_{\text {gravity }}=-\rho A(y+d) g$
$F_{\text {net }}=m a$
$\Rightarrow \rho A(y+d) \ddot{y}=-\rho A \dot{y}^2+\rho g d A-\rho A g(y+d)$
$\Rightarrow(y+d) \ddot{y}^2+\dot{y}^2+g y=0$
Ans. is $D$
(Note : we can't apply Bernoulli's theorem because it is not in steady state. Energy is dissipated.)
