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Physics MCQs for NEET — Practice Questions with Answers

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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.

A. $ \rho = \rho_0 \left[ 1 - { \rho_0gy \over B } \right] $ B. $ \rho = \rho_0 \left[ 1 + { \rho_0gy \over B } \right] $ C. $ \rho = \rho_0 \left[ 1 + { \rho_0gyh \over B } \right] $ D. $ \rho = \rho_0 \left[ 1 - { B \over \rho_0gy } \right] $

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Explanation

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 } ] $

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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

A. 10 cm B. 75 cm C. 13.6 cm D. 1.36 cm

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Explanation

= 760 - 750 = mn at Hg = $ 4 \rho g $ K2086

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The pressure on a swimmer 20 m below the surface of coater at sea level is

A. 1.0 atm B. 2.0 atm C. 2.5 atm D. 3.0 atm

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Explanation

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$

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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

A. $ { vg \rho_1 \over K } $ B. $ \sqrt { vg \rho_1 \over K } $ C. $ { vg (\rho_1 - \rho_2 ) \over K } $ D. $ \sqrt { vg (\rho_1 - \rho_2 ) \over K } $

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Explanation

Weight of the ball = Buyoant force + viscous force

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The fraction of floating object of volume $V_O$ and density do above the surface of a Liquid as density d will be

A. $ { d_0 \over d } $ B. $ { d d_0 \over d + d_0 } $ C. $ { d - d_0 \over d } $ D. $ { d d_0 \over d - d_0 } $

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Explanation

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 } $

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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.

A. 5/3 B. 4/3 C. 3/2 D. 1

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Explanation

Weight of body Weight of water displaced Weight of oil displaced Specitic grvity of oil = $ { \rho_0 \over \rho_w } = {4 \over 3 } $

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If there were no gravity which of the following will not be there for a fluid.

A. Viscosity B. Surface tension C. pressure D. Archime des's upward thrust

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Explanation

A schemedies principal explains buoyant force and bouant force depends on acceleration due to gravity

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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.

A. 3,2 B. $ 2 , {3 \over 4 } $ C. $ {3 \over 4 } ,2 $ D. $ 3 , { 3 \over 2 } $

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Explanation

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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Ice pieces are floating in a beaker A containing watre and also in a beakre B containing miscible liquid of specific gravity 1.2 Ice melts the level of

A. water increases in A B. water decreases in A C. Liquid in B decrease B D. Liquid in B increase

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An engine pumps water continuously through a hose water leares the hose with a velocity V and m is the mass per unit length of the water Jet what is the rate at which kinetic energy is imperted to water.

A. $ {1 \over 2 } mv^3 $ B. $ mv^3 $ C. $ {1 \over 2 } mv^2 $ D. $ {1 \over 2 } mv^{alpha} v^2 $

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