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Fluids Test - 2

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Fluids Test - 2
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Weekly Quiz Competition
  • Question 1
    1 / -0

    A diagram shows a venturimeter, through which water is flowing. The speed of water at X is 2 cms-1. The speed of water at Y (g = 1000 cms-2) is

    Solution

    v= 32 cms-1 

     

  • Question 2
    1 / -0

    A piece of paraffin wax of density 0.9 gm/cc floats on water. A layer of turpentine of density 0.87 gm/cc is added on top of water until the wax is cntircly submerged. The ratio of the volume of wax immersed in water to that in turpentine is

    Solution

    Let V be volume of paraffin wax and V1 be the volume immersed in water and the rest (V - V1) immersed in turpentine, then by condition of floatation weight of wax = weight of water displaced + welght of turpentine displaced.

    ∴  V x 0.90 x g = V1 x 1 x g + (V - V1) x 0.87 g

    V x 90/100 g = V1 x 1 x g + (V - V1) 87/100. g

    90 V = 100 V+ 87 - 87 V1

     

  • Question 3
    1 / -0

    Bernoulli's principle is based on the law of conservation of

    Solution

    As according to Bernoulli's principle, an ideal fluid has streamline flow in a tube of non uniform cross-section, then the sum of pressure energy, kinetic energy and potential energy at any cross-section per unit volume is constant.

     

  • Question 4
    1 / -0

    By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density = 13.6 gm/cm3). Using the straw, he can drink water from a glass up to a maximum depth of

    Solution

    Pressure difference between lungs and atmosphere = 760 - 750 = 10 mm of Hg

     

  • Question 5
    1 / -0

    A soap bubble of diameter 8 mm is formed in air. The surface tension of liquid is 30 dyne/cm. The excess pressure inside the soap bubble is

    Solution

    Excess pressure inside the soap bubble i

     

  • Question 6
    1 / -0

    Two capillary tubes of the same length but different radii r1 and r2 are fitted in parallel to the bottom of a vessel. The pressure head is p. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as before

    Solution

    V = V+ V2

     

  • Question 7
    1 / -0

    A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance ℓ and h are shown there. After some time the coin fails into the water, then

    Solution

    When the coin falls into the water the block rises as it needs less amount of water displaced to satisfy the condition of floatation. As a result ℓ decreases.

    When the coin falls inside water, h also decreases as when the coin is inside water, it displaces water equal to its own volume. which is less than the water being displaced when it was floating along with the block. 

     

  • Question 8
    1 / -0

    A solid ball of density half of that of water falls freely under gravity from a height of 19.6 m and then enters into water. Upto what depth will the ball go ?

    Solution

    Velocity of body when it strikes water surface after having fallen from height h = 19.6 m is v=

    h = 19.6

     

  • Question 9
    1 / -0

    Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by 75 x 10-4 N force due to the weight of the liquid. If the surface tension of water is 6 x 10-2 N/m, the inner circumference of the capillary must be

    Solution

    6 x 10-2 x circumference = Force

    = 12.5 x 10-2 m

     

  • Question 10
    1 / -0

    A cubical block of wood 10 cm on a side floats at the interface between oil and water as shown in figure with its lower face 2 cm below the interface. The density of oil is 0.6 gm/cm3. The mass of the block is

    Solution

    area of base of cube = 100 cm2

    volume of water displaced = 100 x 2 = 200 cm2

    volume of oil displaced = 100 x 8= 800 cm2

    By law of floatation = wt of cube

    = weights of liquids displaced

    Mg = 200 x 1g + 800 x 0.6 x g

    M  = 680 gm.

     

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