Q 1 :

An open-ended U-tube of uniform cross-sectional area contains water (density 103 kgm-3). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density 800 kgm-3 is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio (h1h2) of the heights of the liquid in the two arms is                           [2020] 

[IMAGE 378]

  • 1514

     

  • 3533

     

  • 76

     

  • 54

     

(2)

[IMAGE 379]

Pressure will be same at the same horizontal level.

  PA=PB

P0+ρkg×0.1+ρwg×(h1-0.1)=ρwgh2+p0

80+1000(h1-0.1)=1000h2

80+1000h1-100=1000h2

h1-h2=201000

h1-h2=0.02        (i)

Also, h1-0.1+h2=2×0.29

h1+h2=0.68             (ii)

Solving (i) & (ii), we get

      2h1=0.70

   h1=0.35 m

and h2=0.33 m

So, h1h2=3533



Q 2 :

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density σ at equilibrium position. The extension x0 of the spring when it is in equilibrium is:                     [2012]

  • Mgk

     

  • Mgk(1-LAσM)

     

  • Mgk(1-LAσ2M)

     

  • Mgk(1+LAσM)

     

(3)

[IMAGE 380]

From figure, kx0+FB=Mg

kx0+σL2Ag=Mg

[ mass = density × volume]

kx0=Mg-σL2Ag

x0=Mg-σLAg2k=Mgk(1-LAσ2M)

Hence, extension of the spring when it is in equilibrium is,

x0=Mgk(1-LAσ2M)



Q 3 :

A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance  and h are shown here. After some time the coin falls into the water. Then         [2002]

[IMAGE 381]

  •  decreases and h increases

     

  •  increases and h decreases

     

  • both  and h increase

     

  • both  and h decrease

     

(4)

When the coin falls from the top of block into water the block moves upwards because the weight of floating body becomes less and hence l decreases. When the coin was floating, it displaces water equal to its own weight. When the coin is inside the water, it displaces water equal to its own volume. As its density is greater than that of water, it displaces more water in first case. Hence, h decreases when coin fall into the water.



Q 4 :

A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass M. It is suspended by a string in a liquid of density ρ where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is      [2001]

[IMAGE 382]

  • Mg

     

  • Mg-Vρg

     

  • Mg+πR2hρg

     

  • ρg(V+πR2h)

     

(4)

[IMAGE 383]

From Archimedes principle

Upthrust=wt. of fluid displaced

Now, Fbottom-Ftop=Vρg

Fbottom=Ftop+Vρg

=P1×A+Vρg

=(hρg)×(πR2)+Vρg

=ρg[πR2h+V]



Q 5 :

A cubical solid aluminium (bulk modulus =-VdPdV=70 GPa) block has an edge length of 1 m on the surface of the earth. It is kept on the floor of a 5 km deep ocean. Taking the average density of water and the acceleration due to gravity to be 103 kg m-3 and 10 m s-2, respectively, the change in the edge length of the block in mm is ______.             [2020]



(0.24)

As we know, bulk modulus B=P(-dVV)

dVV=-PB3(Δ)=-PB

Δ=(-PB)3=(ρghB)3

=103×10×5×10370×109×13=0.238×10-3

  Δ=0.238 mm0.24 mm



Q 6 :

A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration a along a fixed inclined plane with angle θ=45°. P1 and P2 are pressures at points 1 and 2, respectively, located at the base of the tube. Let  β=(P1-P2)ρgd, where ρ is density of water, d is the inner diameter of the tube and g is the acceleration due to gravity. Which of the following statement(s) is(are) correct?                        [2021]

[IMAGE 384]

  • β=0 when a=g2

     

  • β>0 when a=g2

     

  • β=2-12 when a=g2

     

  • β=12 when a=g2

     

Select one or more options

(1, 3)

[IMAGE 385]

P1=P2-ρ a cos45°d+ρ(g-asin45°)d

P1-P2ρgd=1-2ag

  β=0  for  a=g2

and    β=2-12  for  a=g2



Q 7 :

A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r(r<R), then the correct option(s) is(are)                             [2015]

  • P(r=0)=0

     

  • P(r=3R/4)P(r=2R/3)=6380  

     

  • P(r=3R/5)P(r=2R/5)=1621  

     

  • P(r=R/2)P(r=R/3)=2027  

     

Select one or more options

(2, 3)

[IMAGE 386]

Let us consider an elemental mass dm shown in the shaded portion of the figure.

Here,  P(4πr2)-(P+dP)(4πr2)=GMrR3ρ(4πr2)dr

   -0PdP=GMρR3rRrdr

  P=GMρ2R3[R2-r2]

  P(r=3R/4)P(r=2R/3)=[R2-9R216][R2-4R29]=7R2165R29=6380

and    P(r=3R/5)P(r=2R/5)=[R2-9R225][R2-4R225]=1621



Q 8 :

A solid sphere of radius R and density ρ is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s) is(are)           [2013]

  • The net elongation of the spring is 4πR3ρg3k  

     

  • The net elongation of the spring is 8πR3ρg3k  

     

  • The light sphere is partially submerged

     

  • The light sphere is completely submerged

     

Select one or more options

(1, 4)

[IMAGE 387]

The complete system is as shown in figure.

Let x be the net elongation of the spring.

At equilibrium, for upper sphere

W+FS=FB

43πR3ρg+kx=43πR3(2ρ)g

kx=43πR3(2ρ)g-43πR3ρg

kx=4πR3ρg3   or   x=4πR3ρg3k



Q 9 :

A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered by a wooden cylindrical block of diameter 4r, height h and density ρ/3.

[IMAGE 388]

Situation I: Initially, the tank is filled with water of density ρ to a height such that the height of water above the top of the block is h1 (measured from the top of the block).

Situation II: The water is removed from the tank to a height h2 (measured from the bottom of the block), as shown in the figure. The height h2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere.

Q.     Find the minimum value of height h1 (in situation 1), for which the block just starts to move up?                [2006]

  • 2h3

     

  • 5h4

     

  • 5h3

     

  • 5h2

     

(3)

[IMAGE 389]

Consider the equilibrium of wooden cylindrical block.

Forces acting in the downward direction are

(i)  Weight of wooden cylinder

        W=π(2r)2×h×ρ3×g

            =π×4r2×hρ3g

(ii)  Force due to pressure (P1) created by liquid of height h1 above the wooden block

        F1=P1×π(2r)2=[P0+h1ρg]×π(2r)2

(iii)  Force acting on the upward direction due to pressure P2 exerted from below the wooden block and atmospheric pressure

          F2=P2×π[(2r)2-r2]+P0×π(r)2

          =[P0+(h1+h)ρg]×π×3r2+P0πr2

     In situation 1, at the verge of rising the block, F2=F1+W

           [P0+(h1+h)ρg]×(π×3r2)+πr2P0

           =[P0+h1ρg]×4πr2+π×4r2hρg3     or,   h1=5h3



Q 10 :

A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered by a wooden cylindrical block of diameter 4r, height h and density ρ/3.

[IMAGE 390]

Situation I: Initially, the tank is filled with water of density ρ to a height such that the height of water above the top of the block is h1 (measured from the top of the block).

Situation II: The water is removed from the tank to a height h2 (measured from the bottom of the block), as shown in the figure. The height h2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere.

Q.    Find the height of the water level h2 (in situation II), for which the block remains in its original position without the application of any external force.           [2006]

  • h3

     

  • 4h9

     

  • 2h3

     

  • h

     

(2)

Again considering equilibrium of wooden block.

Total downward force=total force upwards

Wt. of block+force due to atmospheric pressure=force due to pressure of liquid+Force due to atmospheric pressure

π(16r2)ρ3×g+P0π×16r2=[h2ρg+P0]π[(16-4)r2]+P0×4r2

where h2=height of the water level in situation-2 for which the block remains in its original position.

    h2=49h



Q 11 :

A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered by a wooden cylindrical block of diameter 4r, height h and density ρ/3.

[IMAGE 391]

Situation I: Initially, the tank is filled with water of density ρ to a height such that the height of water above the top of the block is h1 (measured from the top of the block).

Situation II: The water is removed from the tank to a height h2 (measured from the bottom of the block), as shown in the figure. The height h2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere.

Q.     In situation 2, if h2 is further decreased, then                       [2006]

  • cylinder will not move up and remains at its original position

     

  • for h2=h3, cylinder again starts moving up

     

  • for h2=h4, cylinder again starts moving up

     

  • for h2=h5, cylinder again starts moving up

     

(1)

In situation - 2 when the height h2 of water level is further decreased, then the upward force acting on the wooden block decreases. The total force downward remains the same. This difference will be compensated by the normal reaction by the tank wall on the wooden block. Hence the block does not moves up and remains at its original position.