Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. A ball whose density is 0.4×103 kg m(-3) falls into water from a height of 9 cm. To what depth does the ball sink?
  •  9cm
  •  6cm
  •  4.5cm
  •  2.25cm
The velocity of ball before entering the water surface v=√2gh=√(2g×9) When a ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded) The retardation, a=(apparent weight)/(mass of ball) a=V(ρ-σ)g/Vρ=(ρ-σ)g/ρ ((0.4-1)/0.4)g=-3/2 g If h be the depth upto which ball sink, then 0-v2=2×((-3)/2 g)×h ⟹ 2g×9=3gh ∴ h=6 cm

Q2.Choose the correct statement(s) for a cricket ball that is spinning clockwise through air S1 : Streamlines of air are symmetric around the ball S2 : The velocity of air above the ball relative to it is larger than that below the ball S3 : The velocity of air above the ball relative to it is smaller than that below the ball S4 : There is a net upward force on the ball
  •  S1, S2 and S4
  •  S2 and S4
  •  S4 only
  •  S3 only
The streamlines of air for a ball which is moving and spinning at the same time is as shown in figure below. The ball is moving forward and relative to it the air is moving backwards. Therefore, the velocity of air above the ball relative to it is larger and below it is smaller. The streamlines thus get crowded above and rarified below. This difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is known as magnus effect

Q3.  Horizontal tube of non-uniform cross-section has radii of 0.1 m and 0.05 m respectively at M and N. For a streamline flow of liquid the rate of liquid flow is
  •   Changing continuously with time
  •  Greater at M than N
  •  Greater at N than at M
  •  Same at M and N
The velocity of flow will increases if cross-section decreases and vice-versa
 ie, A1 v1=A2 v2 
 or Av=constant 
 Therefore, the rate of liquid flow will be greater at N than at M.

Q4. A piece of solid weighs 120 g in air, 80 g in water and 60 g in a liquid. The relative density of the solid and that of the liquid are respectively
  •  3,2
  •  2,3/4
  •  3/2,2
  •  4,3
Relative density of solid =(weight in air)/(weight in air-weight in water) ⟹ Relative density of solid =120/(120-80)=120/40=3 Relative density of liquid =(weight in air-weight in liquid )/(weight in air-weight in water) ⟹ Relative density of liquid =(120-60)/(120-80)=60/40=3/2

Q5.An aeroplane gets its upward lift due to phenomenon described by the
  •  Archimedes’ principle
  •  Bernoulli’s principle
  •  Buoyancy principle
  •  Pascal law
 Bernoulli’s principle

Q6. When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is
  •  H
  •  2H
  • 7H
  •  8H
P1 V1=P2 V2⇒(P0+hρg)×4/3 πr3=P0×4/3 π(2r)3 Where, h= depth of lake ⇒hρg=7P0⇒h=7×Hρg/ρg=7H

Q7. The velocity of a small ball of mass M and density d_1 when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d_2, the viscous force acting on the ball is
  •  Mg(1 d2/d1 )
  •  Mg d1/d2
  •  Mg(d1-d2)
  •  Mgd1 d2
Viscous force upwards = Apparent weight in liquid ∴ Viscous force=Mg-(Mgd2)/d1 Viscous force=Mg(1-d2/d1 )

Q8.In the following fig. is shown the flow of liquid through a horizontal pipe. Three tubes A,B and C are connected to the pipe. The radii of the tubes A,B and C at the junction are respectively 2 cm,1 cm and 2 cm. It can be said that the
  •  Height of the liquid in the tube A is maximum
  •  THeight of the liquid in the tubes A and B is the same
  •  Height of the liquid in all the three tubes is the same
  •  Height of the liquid in the tubes A and C is the same
As cross-section areas of both the tubes A and C are same and tube is horizontal. Hence according to equation of continuity vA=vC and therefore according to Bernoulli’s theorem PA=PCi.e.height of liquid is same in both the tubes A and C

Q9.A drop of oil is placed on the surface of water then it will spread as a thin layer because
  •  surface tension tends to give the oil a spherical surface
  •  surface tension of water is greater than that of oil
  •  both oil and water have nearly equal surface tension
  •  oil is lighter than water
An oil drop spreads as thin layer, on the surface of water because the cohesive force between water molecules is greater than the adhesive force between water-oil molecules, hence the surface tension of water is greater than that oil.

Q10. The excess pressure in a bubble of radius R of a gas in a liquid of surface tension S is
  •  2S/R
  •  2R/S
  •  2S/R2
  • 2R2/S
Excess pressure inside a bubble of radius R=2S/R where S is the surface tension of the liquid.

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