Upthrust in Fluids, Archimedes Principle And Floatation Numericals Class 9

 Upthrust in Fluids, Archimedes Principle And Floatation 

Numericals:

1: A body of volume 100 cm³ weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 *10³ 

kg m-3. Find :

i) the upthrust due to liquid and

ii) the weight of the body in liquid.

Ans. : Weight of the body in air = 5 kgf

(i) Upthrust due to the liquid = weight of the liquid displaced

                                               = Volume of liquid displaced * density of liquid * g

                                               = 100 * 10-6 m3* 1.8 * 10³  kg m-3 *9.8 ms -2

                                                          = 0.18 * 9.8 N = 0.18 * 9.8 / 9.8 kgf = 0.18 kgf

(ii) Weight of the body in liquid = Weight of the body in air - upthrust 

                                                    = ( 5 - 0.18) kgf = 4.82 kgf

2. A body weighs 450 gf in air and 310 gf when completely immersed in water. Find : 

(i) the volume of the body.

(ii) the loss in weight of the body , and 

(iii) the upthrust on the body,

State the assumption made in part (i).

Ans : Weight of the body in air = 450 gf

          Weight of the body in water = 310 gf

(i) Volume of the body = Volume of water displaced

                                     = Loss of the weight of the body 

                                     = 450 - 310 = 140 cm3

(ii) The loss in the weight of body = 450 - 310 = 140 gf

(iii) The upthrust on the body = Loss in weight of body  = 410 gf

Assumption made in (i) is that density of water is 1.0 g cm^3

3. You are provided with a hollow iron ball A of volume 15 cm3 and mass 12 g and solid iron ball B of mass 12 g . Both are placed on the surface of water contained in a large tub.

(a) Find upthrust on each ball.

(b) Which ball will sink ? Give reason for your answer. ( Density of iron = 8.0 g cm^3)

Ans : (a) Ball - A

 v1  = 15 cm^3 , m1 = 12 g

weight of hollow ball - A in downward direction

F1 = mass * acceleration due to gravity = m1g

F2 = 12 gf

As the ball A floats on the surface of water

It floats with that much part submerged for which upthrust become 12 gf

Upthrust on ball A = 12 gf

When ball is fully submerged upthrust acting will be F2 = vdg

F2 = 15 * 1 *g 

F2 = 15 gf 

15 gf > 12 gf

(b) since F2 > F1

The ball will float.

For Ball B 

m2 = 12 g, d2 = 8 g cm^3

v = mass / density = 12 / 8 = 1.5 cm^3

weight of ball F1 = m2 

g = 12 gf

upthrust on ball B 

 F2 = v d2g = 1.5 * 1g = 1.5 gf

F2 < F1 

Ball B will Sink.

4. A solid of density 5000 kg m^3 weighs 0.5 kgf in ait. It is completely immersed in water of density 1000 kg m^3. Calculate the apparent weight of the solid in water.

Ans : Density of solid, d(s) = 5000 kg m^-3. 

          Weight of the body in air = 0.5 kgf

          mg = 0.5 kgf

           m = 0.5 kg

          Volume of solid V = m / d(g) = 0.5/5000 = 1/10000 m^3

          Volume of water displaced = 1/ 10000 m

          density of water = 1000 kg m^3

          mass of water displaced = V * D

         = 1 / 10000 * 1000 = 1/10 kg

      weight of water = 0.1 kg

(i) Apparent weight of the solid in water = 0.5 - 0.1 = 0.4 kgf

(ii) Apparent weight of body in liquid of density 8000 kg m^3 is zero.

Density of solid is less than density of liquid  i.e.  upthrust is more than weight of body.

5. Two spheres A and B, each of volume 100 cm^3 are placed on water (density = 1.0 g cm^-3). The sphere A is made of wood of density 0.3 g cm^-3 and the sphere B is made of iron of density 

8.9 g cm ^-3.

(a) Find : (i) The weight of each sphere  and

                (ii) the upthrust on each sphere.

(b) Which sphere will float ? Give reason.

Ans : (a) (i) The weight of sphere A = Vρg = 100 * 0.3 * g = 30 gf    

6. The mass of a block made of certain material is 13.5 kg and its volume is 15 * 10^-3 m^3.

(a) Calculate upthrust on the block if it held fully immersed in water.

(b) Will the block float or sink in water when released ? Give reason for your answer.

(c) What will be the upthrust on block while floating ?

Take density of water = 1000 kg m^-3.

Ans : (a) Mass of block m = 13.5 kg

               Volume of block = 15 * 10^-3 m^3

               Upthrust on block when fully immersed in water = vdg = wt. of liquid displaced

                                                                                              = 15 / 1000 * 1000 * g = 15 kgf

          (b) Density of block S = m/v = 13.5 / (15 * 10^-3)

                                                         = (135 * 1000) / (15 * 10)

                                                         = 900 kg m^-3

            Density of block (900) < density of water ( 1000 kg m^3)

             block will float since density of block is less than density of water.

           (c) Upthrust on block while floating =13.5 kgf

                 One kgf is the force due to gravity on a mass of 1 kg

                 Force due to gravity on 13.5 kg = 13.5 kgf

7. A piece of brass weights 175 gf in air and 150 gf when fully immersed in water. The density of water is 1.0 g cm^-3

(i) What is the volume of the brass piece?              

(ii) Why does the brass piece weigh less in water ?

Ans : Weight of brass piece in air = 175 gf

          Weight of brass piece in water = 150 gf

          Loss of weight of solid in water = ( 175 - 150) = 25 gf

     (i) Volume of brass piece = 25 cm^3

          Density of water = 1.0 g cm^-3

     (ii) Brass piece weights less in water as upthrust acts on brass piece.

8. A metal cube of edge 5 cm and density 9.0 g cm^-3 is suspended by a thread so as to be completely immersed in a liquid of density 1.2 g cm^-3. Find the tension in the  thread . ( Take g = 10 m s^-2)

Ans : Volume of metal cube = side ^3 = 5^3 = 125 cm^3

         Density of cube = 9 g cm^-3

        weight of cube acting downward = mg = v * d

         F1 = 125 * 9 = 1125 gf

        Density of liquid d(L) = 1.2 g cm^-3

        Upthrust due to liquid in the upward direction 

        F2 = v d(L) = 125 * 1.2 = 150.0gf

     Tension in the string = Net downward force

                                      = F1 - F2 = 1125 - 150 = 975 gf = 9.75 N

9. A block of wood is floatinf on water with its dimensions 50 cm * 50 cm * 50 cm inside water. Calculate the buoyant force acting on the block. Take = 9.8 N kg^-1.

Ans : Volume of water displaced by the block = 50 cm * 50 cm * 50 cm 

                                                                          = 50 * 50 * 50 cm^3

                                                                          = 50 * 50 * 50 * 10^-6 m^3 

                                                                          = 125 * 10^-3m^3

              Density of water = 10^3 kg m^-3

                                      g = 9.8 N kg^-1

              Buoyant force acting on the block = weight of water displaced by the block

                                                                     =Vol of water displaced by the block * density of water * g

                                                                     =125 * 10^-3 * 10^3 * 9.8 N = 1225 N

10. A body of mass 3.5 kg displaces 1000 cm^3 of water when fully immersed inside it. Calculate:

(i) the volume of body

(ii) the upthrust on body and

(iii) the apparent weight of body in water.

Ans : Mass of body = 3.5 kg g = 10 N kg^-1

(i) Volume of the body = Volume of water displaced by it = 1000cm^3

(ii) Upthrust on the body = Weight of water displaced by the body

                                         = Volume of water displaced by the body * density of the body * g

                                         = 1000 * 10^-6 m^3 * 10^3 kg m^-3 * 10 N kg^-1

                                         = 10N = 10/10 kgf = 1 kgf'

(iii) Apparent weight of body in water = Weight of body in air - upthrust

                                                              = 3.5 kgf - 1 kgf

                                                              = 2.5 kgf