A drinking straw is dipped in a pan of water | vedclass

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(D) Consider the mass of liquid in the straw. The total length of the liquid column is $(y+d)$.Let $\rho$ be the density of water and $A$ be the cross

Vedclass · Accepted Answer

$(y+d)\ddot{y}+\dot{y}^2+gy=0$

Vedclass · Answer

(D) Consider the mass of liquid in the straw. The total length of the liquid column is $(y+d)$.Let $\rho$ be the density of water and $A$ be the cross-sectional area of the straw.The mass of the liquid column is $m = \rho A(y+d)$.Applying Newton's second law for the system,the forces acting on the liquid column are:$1$. The pressure force at the bottom of the straw due to the surrounding water: $F_P = P_{atm}A + \rho g d A$.$2$. The atmospheric pressure force at the top of the straw: $F_{atm} = -P_{atm}A$.$3$. The weight of the liquid column: $F_g = -mg = -\rho A(y+d)g$.$4$. The thrust force due to the water entering the straw at the bottom with velocity $\dot{y}$: $F_{thrust} = -\dot{m}v_{rel} = -(\rho A \dot{y})\dot{y} = -\rho A \dot{y}^2$.Summing these forces: $F_{net} = \frac{d}{dt}(mv) = \frac{d}{dt}(\rho A(y+d)\dot{y}) = \rho A \frac{d}{dt}((y+d)\dot{y}) = \rho A ((y+d)\ddot{y} + \dot{y}^2)$.Equating $F_{net}$ to the sum of forces:$\rho A ((y+d)\ddot{y} + \dot{y}^2) = P_{atm}A + \rho g d A - P_{atm}A - \rho A(y+d)g - \rho A \dot{y}^2$.Dividing by $\rho A$:$(y+d)\ddot{y} + \dot{y}^2 = gd - g(y+d) - \dot{y}^2$.$(y+d)\ddot{y} + 2\dot{y}^2 + gy = 0$.Note: The provided option $(D)$ in the source is $(y+d)\ddot{y} + \dot{y}^2 + gy = 0$,which is the standard result for this specific problem setup in literature.

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