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Force and Pressure Test - 5

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Force and Pressure Test - 5
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  • Question 1
    1 / -0

    Two vessels A and B have the same base area and contain water up to the same height, but the mass of water in vessel A is four times that of water in vessel B. The ratio of pressure on the base of vessel A to that on the base of vessel B is

    Solution

    Pressure due to liquid column is given by:
    P = ρgh, where h is the height and ρ is the density of the liquid.
    Hence, pressure depends only of the height of the liquid column and is independent of the shape of the vessel and the mass of the liquid.
    So, pressure on the base of both the vessels (A and B) is the same. So, the required ratio is 1 : 1.

     

  • Question 2
    1 / -0

    The upthrust on a body immersed in a fluid is equal to the

    Solution

    The upthrust on a body immersed in a fluid is equal to the weight of the fluid displaced.

     

  • Question 3
    1 / -0

    When you press a rubber sucker against a surface, most of the air between its cup and the surface escapes out. To pull the rubber sucker off the surface, the applied force should be

    Solution

    The rubber sucker sticks to the surface because the pressure of atmosphere acts on it. To pull the sucker off the surface, the applied force should be large enough to overcome the atmospheric pressure.

     

  • Question 4
    1 / -0

    Water is flowing through a pipe under constant pressure. At some place, the pipe becomes narrow. The pressure of water at this place

    Solution

    As the area of cross section decreases velocity increases. At the narrow end as the area of cross section decreases, velocity increases.
    As the velocity increases, pressure decreases.

     

  • Question 5
    1 / -0

    What happens when a stream of air is blown through a tube under one of the pans of a physical balance in equilibrium?

    Solution

    As the stream of air is blown through a tube under one of the pans, due to the high air speed, the pressure below the pan decreases as compared to the pressure above the pan. Hence, there will be a net downward force acting on the pan and the pan will go down.

     

  • Question 6
    1 / -0

    For a steel boat floating on a lake, the weight of the water displaced by the boat is

    Solution

    As the boat is floating , the resultant force acting on the boat is zero
    Weight of the body acting downward should be equal to the up thrust acting on the boat.

     

  • Question 7
    1 / -0

    A deep sea diver may hurt his eardrum during diving because of

    Solution

    When a diver dives into the deep sea, there is water on the outside of the ear drum, but air on the inside. The pressure inside will still be equal to atmospheric pressure. However, on the outside, the pressure will be greater. This means that the pressure of air inside the ear will not cancel the pressure of the water outside. There will be net inward force due to this pressure difference, as a result of which the eardrum will stretch. This stretching of the eardrum hurts.

     

  • Question 8
    1 / -0

    What happens to a solid object with density less than that of water when it is put in water?

    Solution

    The object floats on top of the water as the density of water is more than the density of object.

     

  • Question 9
    1 / -0

    A raft is 2.35 m wide, 1.25 m long, and 0.365 m deep. What is the buoyant force on the raft, if only 0.125 m of the raft remains above water? (Take g = 10 m/s2)

    Solution

    Upthrust acting on the object = Weight of the water displaced
    Weight of the water displaced = Volume of water displaced (Volume of the object in the water) × Density × g
    Volume of water displaced = 2.35 × 1.25 × 0.24 = 0.705 m3
    Hence, upthrust = 0.705 × 10× 10 = 7050 N

     

  • Question 10
    1 / -0

    A cube of side 5 cm having density 1600 kg/m3 is submerged in the water with help of a rope. What is the weight of the liquid displaced? (Take g = 10 m/s2)

    Solution

    Weight of the water displaced = Volume of the water displaced (Volume of the object in the water) × Density of water × g
    = (5 × 10-2)× 10× 10 = 1.25 N

     

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