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# Test: Equation of Continuity

## 10 Questions MCQ Test Physics Class 11 | Test: Equation of Continuity

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This mock test of Test: Equation of Continuity for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Test: Equation of Continuity (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Equation of Continuity quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Equation of Continuity exercise for a better result in the exam. You can find other Test: Equation of Continuity extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### Blood is flowing at the rate of 200 cm3/s in a capillary of cross-section area 0.5 m2. The velocity of flow is mm/s is

Solution:

From the equation of continuity (av = constant), we have
0.5 V = 200 X 10-6
V = 4 X 10-4 m/s
V = 0.4 mm/s.

QUESTION: 2

### Flow of water in hilly area is an example of streamline flow.

Solution:

Streamline and turbulent flow. When the flow of liquid is such that the velocity, v of every particle at any point of the fluid is constant then the flow is said to be steady or streamline flow. The path followed by a particle of the fluid in stream-line flow is called steady or streamline flow.

QUESTION: 3

### A straight or curved path, such that tangent to it at any point gives the direction of flow of liquid at that point is known as

Solution:

Streamline flow is the characteristic that determines how water of a stream will move in the stream channel.

QUESTION: 4

The flow of liquid in which its layer slides over another without mixing, is called

Solution:

The word lamellar literally means line the fluid flows in lines when we see an ideal fluid, each particle follows a fixed line and there are numerous of this type in each layer so they don't lag each other.

QUESTION: 5

Fire fighters have a jet attached to the head of their water pipes. This is done to

Solution:

In case of flowing fluids: product of pressure at any cross-section of pipe (in which fluid is flowing) and area of cross-section is constant. So, (PA) at source = (PA) at exit point and as we know that the cross section of the jet is less than the source (fire hydrant) => pressure at exit point (just outside jet) will be greater resulting more velocity of water at exit point. This way water can cover large distance so fire fighters can maintain a safe distance with fire.
Option B seems correct at first but in a closed pipe amount of water in will always be equal to amount of water out for any interval (considering a solid pipe) as option B is incorrect. Now option A should also be incorrect
As Mass = density × volume.

QUESTION: 6

A non-viscous liquid flows through a hose. Liquid enters with velocity 6.4 m/s and leaves with velocity 2.5 m/s. What is the ratio of radii of the hose where the liquid enters and where it leaves?

Solution:
QUESTION: 7

Deep water runs almost still. What does it explain.

Solution:

The equation of continuity for incompressible fluids states that the product of speed and cross-sectional area is always a constant. If a river becomes much deeper, the cross-sectional area becomes large and so the speed of flow is reduced. This is the origin of the expression “still waters run deep”.

QUESTION: 8

The equation of continuity is a special case of the law of conservation of

Solution:

Because equation of continuity depends on mass and equation of continuity play a major role in mass.

QUESTION: 9

In case of streamlined flow of liquid, the loss of energy is

Solution:

In case of streamlined flow of liquid, the loss of energy is minimum because different layers glide over one another without intermixing. Therefore, there is no collision between the molecules of different layers, and hence minimum energy loss.

QUESTION: 10

A garden sprinkler has 150 small holes, each of 2 mm2 area. If water is supplied at the rate of 0.3 litres/s, then find the average velocity of the spray.

Solution:

We know by the conservation of volume we get that
0.3 L/sec =  150 x 0.02 cm2 x v
Where v is the speed of the spray,
And we know 1L = 1 cm3
Hence we get v = 150 x 0.02 cm2 /  .3 cm3/s
= 100 cm2/s