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According to Bernoulli's equation

Pρg+h+12v2g=constant

The terms A, B and C are generally called respectively:





(c) 

A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 kgm-3. The velocity with which gasoline begins to shoot out of the hole is





The velocity of the gasoline shooting out of the hole can be calculated using Bernoulli's equation. The pressure difference between the inside of the tank and the atmospheric pressure provides the driving force for the flow, and the velocity depends on the height of the gasoline column.

A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water





(b) Time taken to be emptied for h height, t=2hg

and for h2 height , t=2h/2g=hgt't=12t'=t2=102=7 minute 

A streamlined body falls through air from a height h on the surface of a liquid. If d and D(D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is





(d) Upthrust – weight of body = apparent weight

VDg-Vdg = Vda,

Where a = retardation of body a=D-ddg

The velocity gained after fall from h height in air,

v=2gh

Hence, time to come in rest,

t=va=2gh×d(D-d)g=2hg×d(D-d)

A large tank of cross-section area A is filled with water to a height H. A small hole of area 'a' is made at the base of the tank. It takes time T1 to decrease the height of water to Hη(η>1) ; and it takes T2 time to take out the rest of water. If T1=T2, then the value of η is





(c) t=Aa2gH1-H2Now, T1=Aa2gH-Hηand T2=Aa2gHη-0

According to problem 

T1=T2H-Hη=Hη-0H=2Hηη=4

 As the temperature of water increases, its viscosity





(b)

As the temperature of a liquid increases, the energy of its molecules increases which increases the movement of molecules.so the liquid becomes more fluid. thus viscosity of liquid decreases.

A small drop of water falls from rest through a large height h in air; the final velocity is





(d) 

Terminal velocity v = 2r2gσ-ρ9η

The rate of flow of liquid in a tube of radius r, length l, whose ends are maintained at a pressure difference P is V=πQPr4ηl where η is coefficient of the viscosity and Q is





(b) In this formula, Q=1/8.

Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2P, the rate of flow becomes





(d) 

V=π(P)r48ηl  V(P)r4     (η and l are constants) V2V1=P2P1r2r14=2×124=18    V2=Q8

Water is flowing in a pipe of diameter 4 cm with a velocity 3 m/s. The water then enters into a tube of diameter 2 cm. The velocity of water in the other pipe is





(c) 

a1v1=a2v2v2v1=a1a2=r1r22v2=3×(2)2=12 m/s

What is the velocity v of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body ? (The densities of metal and of liquid are ρ and σ respectively, and the viscosity of the liquid is η).





(c)

When the acceleration of the ball is g2;Force equation:Mg-B-F=MaVρg-Vσg-6πηrv=Vρg26πηrv=Vρg2-Vσg6πηrv=Vg2ρ-2σ=43πr3g2ρ-2σv=r2g9ηρ-2σ

Consider the following equation of Bernouilli’s theorem.

P+12ρV2+ρgh=K(constant)

The dimensions of K/P are the same as that of which of the following





P+12ρV2+ρgh=K(constant)

According to principle of  homogenecity dimension of K=dimension of ρ

Therefore, KP is dimensionless

out of given options only angle is  dimensionless.

So, the dimension of KP are the same as that of angle.

An incompressible fluid flows steadily through a cylindrical pipe which has radius 2r at point A and radius r at B further along the flow direction. If the velocity at point A is v, its velocity at point B is 






Area×velocity=constantπr2×v=constanti.e., r12v1=r22v2given, r1=2r1   r2=r and v1=vHence, (2r)2×v=r2.v2v2=4v

A block of ice floats on a liquid of density 1.2 g/cm3 in a beaker. Then level of liquid when ice completely melts-







(b) The volume of liquid displaced by floating ice VD=MσL

Volume of water formed by melting ice, VF=MσW

If σL>σW , then, MσL<MσW    i.e. VD<VF

i.e. volume of liquid displaced by floating ice will be lesser than water formed and so the level if liquid will rise.

A vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel having area of cross-section a. The time taken to decrease the level from H1 to H2 will be





At height h the rate of discharge of liquid through hole of bottom = 2gh×a

Let in time dt level of water in tank decreases by dh.

Equating volume

Adh=dt2gh×adt=Aa2gdhh

Now on inegrating

dt=Aa2g H2H1dhht=Aa2gH1-H2