Q 1 :    

Consider a water tank shown in the figure. It has one wall at x=L and can be taken to be very wide in the z-direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ0(θ0<<1) with the x-axis at x=L. If y(x) is the height of the surface, then the equation for y(x) is

[IMAGE 96]

(Take θ(x)=sinθ(x)=tanθ(x)=dydx, g is the acceleration due to gravity)                [2025]

  • d2ydx2=ρgS

     

  • dydx=ρgSx

     

  • d2ydx2=ρgSx

     

  • d2ydx2=ρgSy

     

(4)

[IMAGE 97]

PA=PB=PO

PC=PO-ρgy                        ...(i)

PC=PO-SR                          ...(ii)

So, ρgy=SR                           ...(iii)

Now, R={1+(dydx)2}3/2d2ydx2  (radius of curvature)

As dydx is small, so R=1d2ydx2

Putting in eqn. (iii),

ρgy=S×d2ydx2

d2ydx2=ρgyS



Q 2 :    

A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07Nm-1, then the excess force required to take it away from the surface is           [2024]
 

  •  19.8 mN

     

  • 198 N

     

  • 1.98 mN

     

  • 99 N

     

(1)

Surface tension of water, T=0.07 N m-1

Radius of disc, r=4.5 cm

Required excess pressure, P=2Tr

    Required force to take it away from the surface is F=(P)×Area

or   F=2Tr×πr2=2×T×(πr)

=2×0.07×3.14×4.5×10-2=1.978×10-2=19.8mN



Q 3 :    

Which of the following statement is not true?               [2023]

  • Coefficient of viscosity is a scalar quantity

     

  • Surface tension is a scalar quantity

     

  • Pressure is a vector quantity

     

  • Relative density is a scalar quantity

     

(3)

Pressure is ratio of magnitude of force to the area. Therefore, pressure is a scalar quantity.

 



Q 4 :    

The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03Nm-1)       [2023]
 

  • 3.01×10-4J

     

  • 50.1×10-4J

     

  • 30.16×10-4J

     

  • 5.06×10-4J

     

(1)

We know that surface tension,

S=Surface energySurface areaS=UAU=SA

=(0.03)×2×(4πr2)=(0.03)×2×(4×3.14)×(2×10-2)2

=3.014×10-4J.



Q 5 :    

If a soap bubble expands, the pressure inside the bubble            [2022]

  • decreases

     

  • increases

     

  • remains the same

     

  • is equal to the atmospheric pressure

     

(1)

The excess pressure inside a soap bubble is

Pi=4TR+Po

where, Piinside pressure, Pooutside pressure

T=surface tension and R is radius.

Here, Po,T is constant, so when R increases, Pi decreases.

So, the pressure inside the bubble decreases.



Q 6 :    

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is              [2020]

  • 2.5 g

     

  • 5.0 g

     

  • 10.0 g

     

  • 20.0 g

     

(3)

Force of surface tension balances the weight of water in capillary tube.

          Fs=Tcosθ(2πr)=mg

mr

Hence, m'm=r'r  m'5g=2rr  m'=10g



Q 7 :    

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5×10-2N/m. The pressure inside the bubble equals that at a point Z0 below the free surface of water in a container. Taking g=10m/s2, density of water =103kg/m3, the value of Z0 is          [2019]

  • 0.5 cm

     

  • 100 cm

     

  • 10 cm

     

  • 1 cm

     

(4)

The pressure at a point Z0 below the surface of water, PZ0=P0+ρgZ0

Also, pressure inside a soap bubble, P=P0+4TR

As per question, PZ0=P         P0+4TR=P0+ρgZ0

Z0=4TRρg=4×2.5×10-21×10-3×103×10=1×10-2m=1 cm



Q 8 :    

A rectangular film of liquid is extended from (4cm×2cm) to (5cm×4cm). If the work done is 3×10-4J, the value of the surface tension of the liquid is         [2016]
 

  • 0.250 N m-1

     

  • 0.125 N m-1

     

  • 0.2 N m-1

     

  • 8.0 N m-1

     

(2)

Work done = Surface tension of film×Change in area of the film

or,    W=T×ΔA

Here,  A1=4cm×2cm=8cm2,  A2=5cm×4cm=20cm2

        ΔA=2(A2-A1)=24cm2=24×10-4m2

          W=3×10-4J,  T=?

   T=WΔA=3×10-424×10-4=18=0.125 N m-1



Q 9 :    

Three liquids of densities ρ1,ρ2 and ρ3 (with ρ1>ρ2>ρ3), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ1,θ2 and θ3 obey               [2016]
 

  • π2>θ1>θ2>θ30

     

  • 0θ1<θ2<θ3<π2

     

  • π2<θ1<θ2<θ3<π

     

  • π>θ1>θ2>θ3>π2

     

(2)

Capillary rise, h=2Tcosθrρg

For a given value of T and r, hcosθρ

Also, h1=h2=h3   or   cosθ1ρ1=cosθ2ρ2=cosθ3ρ3

Since,  ρ1>ρ2>ρ3, so  cosθ1>cosθ2>cosθ3

For 0θ<π2, θ1<θ2<θ3

Hence, 0θ1<θ2<θ3<π2



Q 10 :    

Water rises to a height h in capillary tube. If the length of capillary tube above the surface of water is made less than h, then              [2015]

  • water rises up to a point a little below the top and stays there

     

  • water does not rise at all

     

  • water rises up to the tip of capillary tube and then starts overflowing like a fountain

     

  • water rises up to the top of capillary tube and stays there without overflowing.

     

(4)

Water will not overflow but will change its radius of curvature.