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A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. It is stretched by an amount x. The work done is -





(c)

W = 12×AYL×x2Where x is deformation

A stretched rubber has





(b)

Due to stretching, potential energy is stored in the rubber band which is released on releasing the rubber. Thus, potential energy is increased.

When load of 5kg is hung on a wire then extension of 3m takes place, then work done will be





(a) W=12Fl=12×Mg×l=12×5×10×3=75 J

The work per unit volume to stretch the length by 1% of a wire with constant cross sectional area   will be. Y=9×1011N/m2





(b) U=12×Y×Strain2=12×9×1011×11002                                     =4.5×107 J

A tube of length L is filled completely with an incompressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by liquid at the other end is





Force on element of width dx = MLdxxω2Total force F = 0LML2 dx= MLω22

Choice A is correct.

 

A cubical vessel of height 1 m is full of water. Find the work done in pumping out whole water.





Work done=PE of the water=mgh2=4900J

 

The flow rate from a tap of diameter 1.25 cm is 3 lit/min. The coefficient of viscosity of water is 10-3 Pas. The nature of flow is :





Let the speed of the flow be v. The diameter of the tap = d = 1.25 cm = 1.25 ×10-2 m Density of water = ρ = 103 kg m-3 Viscosity = η = 10-3PasThe volume of the water flowing out per second is  Q = v × πd2/4v = 4Q/d2πReynolds number is given by:R = ρvd η= 4ρQπdη=4 ×103 × Q  (3.14 × 1.25 ×10-2 × 10-3)= 1.019 × 108QQ = 3 L / min = 3 × 10-3 60=5 × 10-5m3/secR = 5095 The flow will be turbulent.

Water flowing from a hose pipe fills a 15-liter container in one minute. The speed of water from the free opening of radius 1 cm is (in ms-1) :





Volume of the water=15 litre=15×10-3m3Volume=Area×length=Area×velocity×timeVelocity=15×10-3π×1×10-22×60=2.5πm/s

Work of 6.0 x 10-4 Joule is required to be done in increasing the size of a soap film from 10cm x 6cm to 10cm x 11cm. The surface tension of the film is :





Surface Tension=Surface EnergyAreaSurface Energy=Surface Tension×AreaWork done= change in surface energy=Surface Tension×Change in areaSurface Tension=6×10-42×10×5×10-4=6100=6×10-2N/m

When a cylindrical tube is dipped vertically into a liquid the angle of contact is 140o. When the tube is dipped with an inclination of 40o, the angle of contact is- 

 





The angle of contact is independent of the tilting angle.

Two liquid drops have their diameters as 1 mm and 2 mm. The ratio of excess pressure in them is :





2P = 2TrP1P2=r2r1=21

If a soap bubble of radius 3 cm coalesce with another soap bubble of radius 4 cm under isothermal conditions, the radius of the resultant bubble formed is in cm-





P1=4Tr1 & V1=43πr13P2=4Tr2 & V2=43πr23Two bubbles coelesce isothermally:PV=P1V1+P2V24Tr×43πr3=4Tr1×43πr13+4Tr2×43πr23r2=r12+r22r=r12+r22=5cm

Water flows through a non-uniform tube of area of cross sections A, B, and C whose values are 25, 15, and 35 cm2 respectively. The ratio of the velocities of water at the sections A, B, and C is-





By equation of continuityAv = const. A1v1 = A2v2 = A3v3Since A1:A2:A3 = 25:15:35v1:v2:v3 = 21:35:15

W is the work done in forming a bubble of radius r, the work is done in forming a bubble of radius 2r will be:-





1W=T×A=T×4πr2W  r2

Equation of continuity based on :





It is based on the conservation of mass