NEET Physics Questions: Mechanical Properties of Fluids

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To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.1% $( Take : density of sea water = 1000 kg /m^3 , bulk modulus of rubber = 9 \times 10^8 N/m^2 , acceleration due to gravity = 10 m/s^2 ) $




$ k = { P \over { -\triangle v \over v } } $
The compressibility of water $4 \times 10 ^{-5} $ per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be................




$ C = { \triangle v \over v } \triangle P = 0.4 CC $
Writing on black board with a piece of chalk is possible by the property of




Writing on a blackboard with a piece of chalk is possible due to adhesive forces. Adhesive forces are the attractive forces between unlike substances, such as chalk and the blackboard. These forces enable the chalk particles to stick to the surface of the blackboard, allowing writing to occur. Therefore, option o1 is correct.
When there is no external force, the shape of liquid drop is determined by




When there is no external force acting on a liquid drop, its shape is determined by surface tension. Surface tension is the property of the liquid that causes it to acquire the least surface area possible, resulting in a spherical shape for the drop. Therefore, option o1 is correct.
Soap helps in cleaning because




Soap helps in cleaning by lowering the surface tension of water. This allows the water to spread more easily and interact with oils and dirt, emulsifying them and making them easier to rinse away.
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A beaker of radius 15 cm is filled with liquid of surface tension 0.075 N/m. Force across an imaginary diameter on the surface of liquid is




Soap helps to lower the surface tension of solution thus soap get stick to the dist partcles and grease and these are removed by action of water.
A square frame of side L is dipped in a liquid on taking out a membrane is formed if the surface tension of the liquid is T, the force acting on the frame will be.




When a square frame of side L is dipped in a liquid and a membrane is formed, the force due to surface tension acting on the frame can be calculated using the formula: \(F = 4TL\), where T is the surface tension and L is the length of one side of the square. However, since there are two surfaces (top and bottom of the membrane), the total force is \(8TL\).
The force required to separate two glass plates of area $ 10 ^ {-2} m^2 $ with a film of water 0.05 mm thick between them is (surface tension of water is $ 70 \times 10 ^ {-3} N/m $ )




Force required to separate the plates $ = { 2TA \over t } $
Surface tension of a liquid is found to be influenced by




No explanation available.
A thmmetal disc of radius r floats on water surface B and bends the surface down wards along the perimeter making an angle Q with vertical edge of the disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight at metal dis




When a metal disc floats on the water surface, the surface tension acts along the perimeter of the disc. The vertical component of this force balances the weight of the disc. Hence, the weight of the metal disc is given by the sum of the vertical component of the surface tension force and the weight of the displaced water. Therefore, the correct expression is $2 \\pi r T \\cos heta + W$.
Radius of a soap bubble is 'r', surface tension of soap solution is T. Then without incresing the temperature how much energy will be needed to double its radius.




$ w = 8 \pi T ( R_2^2 - R_1 ^2 ) = 8 \pi T ( (2r)^2 - (r)^2 ) = 24 \pi r^2 T $
A soap bubble of radius r is blown up to forma bubble of radius 2r under isothermal conditions if the T is the surface tension of soap solution the energy spent in the slowing is.




$ Energy spent = T \times increase in surface ared$ $ = T \times 2 ( 4 \pi ( 2r)^2 - 4 \pi r^2 ) = ( 24 \pi T r^2) joule $
The surface tension of a liquid is 5 N/m. If a thin film of the area 0.02 m2 is formed on a loop, then its surface energy will be




$ w = T \times \triangle A $
A frame made of a metalic wire enclosing O surface area Ais covered with a soap film. If the area of the frame metalic wire is reduced by 50% the energy of the soap film will be changed by




$ Surface energy = Surface tension \times surface ared $ $ E = T \times 2 A $ $ New surface energy F_1 = T \times 2 \left( { A \over 2} \right) $ $ \% decrase in surface energy = { E - E_i \over E } \times 100 $
Two small drops mercury, each of radius R, coaless the form a single large drop. The ratio of the total surface energies before and after the change is.




The ration of the total surface energies before and after the change $ = n^{1 \over 2 } : 1 = 2 ^ {1 \over 3} : 1 $