To solve this problem, we need to understand how the impact of the liquid emerging from the hole affects the vessel and the necessary friction required to prevent it from sliding.
1. Concept of impact force: The liquid emerging from the hole at the bottom of the vessel carries momentum. As the liquid exits, it exerts a force on the vessel, creating an impact force. This force can cause the vessel to slide if there is insufficient friction between the vessel and the horizontal surface.
2. Rate of liquid flow: The flow rate QQ of the liquid exiting the hole is given by Torricelli's Law:
Q=avQ = a v
where:
o aa is the cross-sectional area of the hole,
o vv is the velocity of the liquid emerging from the hole.
3. Velocity of the liquid: The velocity of the liquid can be determined using the principle of fluid dynamics, where the velocity vv of the liquid leaving the hole is related to the height hh of the liquid column:
v=2ghv = \sqrt{2gh}
where gg is the acceleration due to gravity, and hh is the height of the liquid above the hole.
4. Impact force: The rate of change of momentum of the liquid, which is the impact force FimpactF_{\text{impact}} acting on the vessel, is given by:
Fimpact=dPdt=ddt(m⋅v)=ddt(ρavh)=ρa⋅dvdthF_{\text{impact}} = \frac{dP}{dt} = \frac{d}{dt} (m \cdot v) = \frac{d}{dt} (\rho a v h) = \rho a \cdot \frac{dv}{dt} h
Here, ρ\rho is the density of the liquid, and we assume the height of the liquid changes slowly.
5. Friction to resist sliding: The force that resists the sliding of the vessel is the frictional force FfrictionF_{\text{friction}}, which is given by:
Ffriction=μNF_{\text{friction}} = \mu N
where:
o μ\mu is the coefficient of friction,
o NN is the normal force on the vessel, which is equal to the weight of the vessel and the liquid inside it.
For the vessel to just start sliding, the frictional force must be equal to the impact force. Therefore, the minimum coefficient of friction μ\mu necessary to prevent sliding is determined by the ratio of the impact force to the normal force:
μ=FimpactN\mu = \frac{F_{\text{impact}}}{N}
Given the setup and simplifications, the correct answer is:
B) a4\frac{a}{4}.
