A capillary tube of length L and radius r is connected with

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

111.

The speed of air flows on the upper and lower surfaces of a wing of an aeroplane are v1 and v2 respectively. If A is the cross section area of the wing and ρ is the density of air, then the upward life is

  • 12ρ Av2 - v1

  • 12ρ Av1 + v2

  • 12 ρ A  v12 - v22 

  • 12ρ A v12  +  v22 


112.

The work done in increasing the size of a soap film for 10cm × 6cm  to 10cm × 11cm is 3  × 10-4 J. The surface tension of the film is

  • 1.0  × 10-2 N/m

  • 6.0  × 10-2 N/m

  • 3.0  × 10-2 N/m

  • 1.5  × 10-2 N/m


113.

A balloon contains 500 m3 of helium at 27o C and 1 atmosphere pressure. The volume of the helium at - 3° C temperature and 0.5 atmosphere pressure will be

  • 1000 m3

  • 900 m3

  • 700 m3

  • 500 m3


114.

An ice-cube of density 900 kg/m3 is floating in water of density 1000 kg/m3. The percentage of volume ofice-cube outside the water is

  • 20%

  • 35%

  • 10%

  • 25%


Advertisement
115.

The lower end of capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, upto what height will the water rise in the capillary?

  • 5.9

  • 4.9

  • 2.9

  • 1.9


Advertisement

116.

A capillary tube of length L and radius r is connected with another capillary tube of the same length but half the radius in series. The rate of steady volume flow of water through first capillary tube under a pressure difference of p is V. The rate of steady volume flow
through the combination will be (the pressure difference across the combination is p)

  • 17 V

  • 1617 V

  • V17

  • 1716 V 


C.

V17

Problems of series and parallel combination of pipes can be solved in the similar manner as is done in case of an electrical circuit. The only difference is potential difference is replaced by Δp and electrical resistance

        R18 η Lπ r4 = R

and    R28 ηLπ r24

               = 16 R

Electric current is replaced by rate of volume flow V'

       p = VR1 = VR                      .....(i)

        p' = p = V' Req

                  = V' (R1 + R2 )  

                   = 17 V'R

      From Eqs. (i) and (ii), we get

         VR = 17 V'R

⇒         V' = V17    


Advertisement
117.

A small spherical drop fall from rest in viscous liquid. Due to friction, heat is produced. The correct relation between the rate of production of heat and the radius of the spherical drop at terminal velocity will be

  • dHdt  1r5

  • dHdt  r4

  • dHdt  1r4

  • dHdt  r5


118.

A solid sphere of volume V and density p floats at the interface of two immiscible liquids of densities and p, respectively. If ρ1 < ρ < ρ2, then the ratio of volume of the parts of the sphere in upper and lower liquid is

  • ρ - ρ1ρ2 - ρ

  • ρ2 - ρρ - ρ1

  • ρ + ρ1ρ + ρ2

  • ρ + ρ2ρ + ρ1


Advertisement
119.

Assertion:  A needle placed carefully on the surface of water may float, whereas the ball of the same material will always sink. 

Reason:  The buoyancy of an object depends both on the materials and shape of the object.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


120.

A cylindrical tank is filled with water to level of 3 m. A hole is opened at height of 52.5 cm from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is 0.1. The square of the speed with which water is coming out from the orifice is (Take g = 10 m s-2)

  • 50 m2 s-2

  • 40 m2 s-2

  • 51.5 m2 s-2

  • 50.5 m2 s-2


Advertisement