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A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8. The relative density of silver is 10.5. The tension in the string in kg-wt is





(b) Apparent weight 

=V(ρ-σ)g=Mρ(ρ-σ)g

=M1-σρg=2.11-0.810.5g=1.94g N=1.94 kg-wt

Two solids A and B float in water. It is observed that A floats with half its volume immersed and B floats with 2/3 of its volume immersed. Compare the densities of A and B





(c) If two different bodies A and B are floating in the same liquid then

ρAρB=(fin)A(fin)B=1/22/3=34

The fraction of a floating object of volume V0 and density d0 above the surface of a liquid of density d will be





(c) For the floatation  V0d0g=Vindg

Vin=V0d0d

 Vout=V0-Vin=V0-V0d0d=V0d-d0dVoutV0=d-d0d

Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. This law was first formulated by:






(d)

According to Pascal's principle, the pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid with the walls of the containing vessel too. A common application of this principle is in hydraulic lift, to lift the car. Here, a small amount of force applied to the small-area piston, which will be transformed into a large force at a large-area piston.

A cork is submerged in water by a spring attached to the bottom of a bowl. When the bowl is kept in an elevator moving with acceleration downwards, the length of spring





(b)

Upthrust reduces from Vdg to Vdg-a which is responsible for the extension in the spring.

A hollow sphere of volume V is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water ?





(a) When body (sphere) is half immersed, then
upthrust = weight of sphere 

V2×ρliq×g=V×ρ×g     ρ=ρliq2

When body (sphere) is fully immersed then,

Upthrust = wt. of sphere + wt. of water poured in sphere

V×ρliq×g=V×ρ×g + V'×ρliq×gV×ρliq=V×ρliq2+V'×ρliqV'=V2

A ball whose density is 0.4×103 kg/m3 falls into water from a height of 9 cm . To what depth does the ball sink





(b) The velocity of ball before entering the water surface

v=2gh=2g×9

When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded)

The retardation, a=apparent weightmass of ball

=V(ρ-σ)g=ρ-σρg=0.4-10.4×g=-32g

If h be the depth upto which ball sink, then,

0-v2=2×-32g×h2g×9=3gh h=6 cm

 

A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water ?





(c)

ρball × Vball × g = ρwater × Vdisplaced × gAs ρball  >  ρwaterVdisplaced > VballSo water level falls

Two pieces of metal when immersed in a liquid have equal upthrust on them; then





(c) Since, upthrust (F) =Vσg   i.e. F∝V

Construction of submarines is based on 





Submarine has a large volume ,so buoyant force  is more than the weight. Hence it floats on water surface. When it has to go inside water, it is filled with water so that weight is more than the buoyant force acting on submarine. So it sinks.

In which one of the following cases will the liquid flow in a pipe be most streamlined  ?





(b) For streamline flow, Reynold's number NRη should be less. For less value of NR, radius and density should be small and viscosity should be high.

Two water pipes of diameters 2 cm and 4 cm are connected with the main supply line. The velocity of flow of water in the pipe of 2 cm diameter is 





(a) dA=2cm and dB=4 cm                rA=1cm and rB=2 cm

From equation of continuity , av= constant

vAvB=aBaA=π(rB)2π(rA)2=212vA=4vB

The velocity of kerosene oil in a horizontal pipe is 5 m/s. If g=10m/s2 then the velocity head of oil will be





(a) Velocity head h=v22g=(5)22×10=1.25m

A manometer connected to a closed tap reads 3.5×105 N/m2. When the valve is opened, the reading of manometer falls to 3.0×105 N/m2 , then velocity of flow of water is





(b) Bernoulli's theorem for unit mass of liquid

Pρ+12v2= constant

As the liquid starts flowing, it pressure energy decreases 

12v2=P1-P2ρ12v2=3.5×10-5-3×105103=2×0.5×105103v2=100v=10 m/s

A large tank filled with water to a height ‘h’ is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to h2 and from h2 to zero is





(c)

When the tank is full upto height h, velocity at the hole, v=2ghLet the rate of decrease height of water=-dhdtSo, using continuity equation, -Adhdt=A0v=A02gh0tdt=-H1H2AA0dh2ght=-AA0×22ghH1H2=AA0×2gH1-H2

Time taken for the level to fall from H to H'

t1=AA02gH-H'

According to problem-the time taken for the level to fall from h to h2

t=AA02gh-h2

and similarly time taken for the level to fall from h2 to zero

t2=AA02gh2-0

t1t2=1-1212-0=2-1