Practice free Mechanical Properties of Fluids (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
When a large bubble rises drom the bottom of a lake to the surface. Its radius double S. It atmospheric pressure in euqal to that of colurnn of colurnn of water height H then the depth of Lake is
$ p_1 v_1 = p_2 v_2 $ $ ( p + h \rho g ) {4 \over 3} \pi r^3 = p_0 \times {4 \over 3} (2 r)^3 $
A triangular lamind of area A and, height h is immersed in a liquid of density S in a vertical plane with its base on the surface of the liquid. The thrust on lamina is
$ Thrust on lamina = pressure at centroid \times area $ $ = { h \rho g \over 3 } \times A = {1 \over 3 } A \rho gh $
The density S of coater of bulk modulus B at a depth y in the ocean is related to the density at surface so by the relation.
Bulk modulas $ B = -V_o { \triangle p \over \triangle v } \Rightarrow \triangle v = v_o { \triangle p \over B } \Rightarrow v = v_o [ 1 - { \triangle p \over B } ]$ $ density \rho = \rho_o [ 1 - { \triangle p \over B } ]^{-1} = \rho_o [1 + { \triangle p \over B } ] $
By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg ($ density = 13.6 gm / cm^3 $) using the straw, he can drink water from a glass up to a maximum depth of
= 760 - 750 = mn at Hg = $ 4 \rho g $ K2086
The pressure on a swimmer 20 m below the surface of coater at sea level is
Here, h = 20 m $ Density of water \rho = 10 ^ 3 kg /m^3 $ Atmospheric pressure $ Pa = 1.01 \times 10^5 pa $ $ p = pa + \rho g h$
A spherical solid ball of volume V is made of a material of density S. It is falling through a liquid of density $ S_2 (S_2 < S_1)$ . Assume that the liquid opplies a viscous force on the ball that is proportional to square of its speed V. i.e. $Fviscous = - KV^2 (K > 0) $ The terminal speed of the ball is
Weight of the ball = Buyoant force + viscous force
The fraction of floating object of volume $V_O$ and density do above the surface of a Liquid as density d will be
For the floatation Vo dog = Vin dg $ Vin = V_o { do \over d } $ $Vout = V_o - Vin \Rightarrow V_o - V_o {d_o \over d } $
Abody floats in water with one-thired od its volume above the surface of water. It is placed in oil it floats with half of : Its volume above the surface of the oil. The specific gravity od the oil is.
Weight of body Weight of water displaced Weight of oil displaced Specitic grvity of oil = $ { \rho_0 \over \rho_w } = {4 \over 3 } $
If there were no gravity which of the following will not be there for a fluid.
A schemedies principal explains buoyant force and bouant force depends on acceleration due to gravity
A piece of solid weighs 120 g in air, 80 g in water and 60 g in liquid the relative density of the solid and that of the solid and that of the liquid are respectively.
To find the relative densities, we use Archimedes' principle. The relative density of the solid is given by $ rac{ ext{Weight in air}}{ ext{Loss of weight in water}} = rac{120}{120 - 80} = 3$. For the liquid, it's relative density compared to water is $ rac{ ext{Loss of weight in water}}{ ext{Loss of weight in liquid}} = rac{120 - 80}{120 - 60} = rac{40}{60} = rac{2}{3}$, which means the relative density of the liquid is $ rac{3}{2}$. So, the relative densities are 3 and $ rac{3}{2}$. Thus, option o4 is correct.
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