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Mechanical Properties of Fluids Test - 16

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Mechanical Properties of Fluids Test - 16
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  • Question 1
    1 / -0
    Assertion: A raindrop after falling through some height attains a constant velocity

    Reason: At constant velocity, the viscous drag is equal to its weight.
    Solution
    After falling through some height, the raindrops weight will be balanced by the viscous force of the air of the drop. It than attains constant velocity.
  • Question 2
    1 / -0
    Assertion: For the flow to be streamlined, value of critical velocity should be as low as possible
    Reason: Once the actual velocity of flow of liquid becomes greater than the critical velocity, the flow becomes turbulent.
    Solution
    Critical velocity is the velocity of a liquid flow upto which its flow is streamlined and after which its flow becomes turbulent.
    And,reynold's number is directly proportional to critical velocity which determines the streamline flow. If the number is small the flow of liquid is streamline.
    Both the statements are true but reason is not the correct explanation of assretion. 

  • Question 3
    1 / -0
    Assertion: An object falling through a viscous medium eventually attains terminal velocity
    Reason: All the rain drops hit the surface of the earth with the same constant velocity
    Solution
    As an object moves through a viscous medium, the two opposing forces weight and viscous force balance each other and they attain a terminal velocity.
    All rain drops won't hit the earth with same velocity as viscosity of atmosphere is not uniform and all rain drops are not of same mass.
  • Question 4
    1 / -0
    A drop of water of radius r is falling through the air of coefficient of viscosity $$\eta $$ with a constant velocity of v. The resultant force on the drop is:
    Solution
    When a drop is falling at constant velocity, viscous force is balanced by weight, hence resultant force is zero.
  • Question 5
    1 / -0
    Read the following statements and pick the correct choice
    Statement A: With increase in temperature, viscosity of a gas increases and that of a liquid decreases.
    Statement B: If the density of a small sphere is equal to the density of the liquid in which it is dropped, then the terminal velocity of the sphere will be zero.
    Solution

    Viscosity of gas increase with increase in temperature as randomness in molecules increases and number of collisions also increase, while that of a liquid decrease as molecules move away from each other making it less viscous.
    If density of sphere equal that of liquid, the two forces itself balance immedeatly leading to zero terminal velocity.

  • Question 6
    1 / -0
    A small ball is dropped in a viscous liquid. Its fall in the liquid is best described by the figure:

    Solution
    Initially when the ball falls, it starts accelerating due to gravity. But viscous force opposes it. So it attains a terminal velocity and then its acceleration becomes zero then it remains constant which is best described by curve C.

    Hence curve C is the right option.
  • Question 7
    1 / -0
    After terminal velocity is reached the acceleration of a body falling through a viscous fluid is :
    Solution
    One the terminal velocity is reached, both the forces balance each other, i.e.; weight is balanced by the viscous force and buoyancy.
    Hence acceleration is zero.
  • Question 8
    1 / -0
    When a metallic sphere is dropped in a long column of a liquid, the motion of the sphere is opposed by the viscous force of the liquid. If the apparent weight of the sphere equals to the retarding forces on it, the sphere moves down with a velocity called:
    Solution
    Terminal velocity is the speed of the object when net force on the object is zero. If the apparent weight of the sphere equals to the retarding forces on it, then net force on it is zero. So, sphere moves down with terminal velocity.
  • Question 9
    1 / -0
    A ball is dropped into coaltar. Its velocity-time curve will be
    Solution
    Initially the velocity keeps increasing, but there is an opposing force the viscous force which decelerates the body. the two opposing force act on each other until they attain an equilibrium & velocity becomes constant.
  • Question 10
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    The reading of a pressure meter attached with a closed water pipe is 3.5 x 10$$^{5}$$ N m$$^{-2}$$. On opening the valve of the pipe, the reading of pressure meter is reduced to 3 x 10$$^{5}$$ N m$$^{-2}$$. Calculate the speed of water flowing in the pipe.
    Solution


    Applying Bernoulli's principle when the pipe was closed and after it was opened,
    $$\Rightarrow P_{1}=P_{2}+\dfrac{1}{2}\rho V^{2}$$
    $$\Rightarrow 3.5 \times 10^{5}= 3 \times 10^{5} + \dfrac{1}{2}\rho V^{2}$$
    $$\Rightarrow \dfrac{0.5 \times 10^{5} \times 2}{10^{3}}=V^{2}$$
    $$\Rightarrow 10^{2}= V^{2}\Rightarrow V=10 m/s$$

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