Q 1 :    

Given below are two statements:

 

Statement (I):Viscosity of gases is greater than that of liquids.

 

Statement (II): Surface tension of a liquid decreases due to the presence of insoluble impurities.

 

In the light of the above statements, choose the most appropriate answer from the options given below:                   [2024]

  • Statement I is correct but statement II is incorrect

     

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I and Statement II are incorrect

     

  • Both Statement I and Statement II are correct

     

(2)   

        Gases have less viscosity.

         Due to insoluble impurities like detergent, surface tension decreases.

 



Q 2 :    

A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be               [2024]

  • 4πR2T

     

  • 8πR2T

     

  • 3πR2T

     

  • 1/8πR2T

     

(2)   

       Volume of liquid of drops should be constant,

        43πR3=27×43×πr3

        R3=27r3R=3r

        r=R3r2=R29

       Work done = T·A

       =27T(4πr2)-T4πR2

        =27T4πR29-4πR2T=8πR2T

 



Q 3 :    

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become                              [2024]

  • 100 times

     

  • 10 times

     

  • 1100th

     

  • 110th

     

(4)   

         Lets say radius of small droplets is r and that of big drop is R,

         By conservation of volume:

         43πR3=100043πr3R=10r

           Surface energy of 1000 droplets

         Ui=1000(4πr2S)

         Uf=4πR2=100(4πr2S)

         Uf=110Ui

 



Q 4 :    

A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap solution is 40 dyne/cm then the new radius is _______ cm. Take (π=227)                          [2024]



(7)         ΔW=2T·4πr22-2T·4πr12

             r22-r12=ΔW8Tπ=36960×78×40×22=36.75

             r22=36.75+494=49r2=7cm

 



Q 5 :    

A big drop is formed by coalescing 1000 small identical drops of water. If E1 be the total surface energy of 1000 small drops of water and E2 be the surface energy of single big drop of water, the E1:E0 is x:1 where x= __________ .                 [2024]



(10)        Let radius of small drops = r

              Let radius of big drops = R

              Here volume will remain conserve (volume conservation)

              (1000)Vs=(1)Vbvolume of big drop

              Volume of small drop103(43πr3)=43πR3

              R=10r

              We know, surface tension =Surface energyArea

              Surface energy=(Surface tension)(Area)

              ε1=T×4πr2×103

              ε2=T×4πR2

              ε1ε2=(103)(r2)R2=(103)r2(10r)2=101

               ε1ε2=101=X1,

              x=10



Q 6 :    

A big drop is formed by coalescing 1000 small droplets of water. The ratio of surface energy of 1000 droplets to that of the energy of the big drop is 10x. The value of x is ____.            [2024]



(1)

100043πr3=43πR3    R=10r

Surface energy of 1000 drops =1000×T×4πr2

Surface energy of bigger drop =T×4πR2

S.E. of 1000 dropsS.E. of Big drop=1000(4πr2)T4πR2T=1000×r2(10r)2

S.E. of 1000 dropsS.E. of Big drop=10=10x

 x=1



Q 7 :    

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m3. If the pressure inside the bubble is 2100 N/m2 greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 m/s2)          [2025]

  • 0.02

     

  • 0.1

     

  • 0.25

     

  • 0.05

     

(4)

P1=P0+pgh=P0+1000×10×20100

 P1=P0+2000

So, P2P0=2TR=(2T1×103)

 P2=P0+2100   (given)

So, P0+2100P02000=2T×103

100=2T×103

 T=(120)=0.05 N/m



Q 8 :    

The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be          [2025]

  • 15 J

     

  • 10 J

     

  • 20 J

     

  • 5 J

     

(1)

The amount of work done to break a big water drop into small drops,

W=U=TA

One drop to n drops

43πR3=n43πr3  r= Rn13

So, W=T(n4πr24πR2)=T4πR2(n1/31)

For one drop to 27 drops

W=T4πR2(271/31)                         ...(i)

For one drop to 64 drops

W'=T4πR2(641/31)                       ...(ii)

(ii)/(i)

W'W=4131=32  W'=32W= 15 J



Q 9 :    

Consider following statements:

A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid.

B. As the temperature of liquid rises, the coefficient of viscosity increases.

C. As the temperature of gas increases, the coefficient of viscosity increases.

D. The onset of turbulence is determined by Reynold's number.

E. In a steady flow two stream lines never intersect.

Choose the correct answer from the options given below:          [2025]

  • A, D, E only

     

  • C, D, E only

     

  • B, C, D only

     

  • A, B, C only

     

(2)

Surface tension arises due to extra energy of the molecules at the surface as compared to the interior of a liquid. The coefficient of viscosity for a liquid decreases with rise in temperature where as it increases for gases with increase in temperature. The flow is turbulent for a Reynold's number greater than 2000. Stream lines never intersect in a steady flow.



Q 10 :    

Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is:          [2025]

  • 4πr2T[2223]

     

  • 4πr2T[2213]

     

  • 4πr2T[1+2]

     

  • 4πr2T[21]

     

(1)

2×43πr3=43πR3  R=21/3r

Ei=2×4πr2T

Ef=4πR2T=4πr222/3T

Heat lost Q=EiEf=4πr2T[222/3]