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प्रश्न
The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle θ. If the acceleration is a ms–2, what will be the slope of the free surface?
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उत्तर
The behaviour of a liquid contained in a horizontally accelerated vessel can be understood by understanding the behaviour of a pendulum suspended from the ceiling of a horizontally accelerated trolley.
Every fluid element attains an equilibrium position under the action of gravity and pseudo-force. The free surface of the liquid orients itself perpendicular to the direction of net effective gravity.
tan θ = a/g
Suppose a tanker accelerates along the x-axis with acceleration a, the free surface of the tanker will not be horizontal because pseudo force acts as shown in the diagram.
Consider an elementary particle of the oil of mass m.
The acting forces on the particle with respect to the tanker are shown in the figure alongside.
Now, balancing forces (as the particle is in equilibrium) along the inclined direction of the surface.
ma = pseudo force
mg = weight of small part of the oil.
Along the free surface,
Net force = 0
⇒ ma cos θ = mg sin θ
⇒ a = g tan θ
⇒ θ = tan-1(a/g)
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