Pressure In Fluids And Atmospheric Pressure Physics Numericals Class 9

 Pressure In Fluids And Atmospheric Pressure

Q.1  A hammer exerts a force of 1.5 N on each of the two nails A & B. The area of cross section of tip of nail A is 2 mm2 while that the nail B is 6 mm2 . Calculate pressure on each nail in Pascal.

Ans.

 Here F = 1.5 N 

          P = ?

  we know that, P = F / A

   In the case of nail A;

    A = 2 mm2 = 2 * 10-6 m2 

    P = 1.5 N / 2 * 10-6 m2 

 

        = 0.75 * 106 N m-2 

        =  7.5 * 105 N m-2 or pascal

In the case of nail B;

    A = 6 mm2 = 6 * 10-6 m2 

    P = 1.5 N / 6 * 10-6 m2 

 

        = 0.25 * 106 N m-2 

        =  2.5 * 105 N m-2 or pascal

Q. 2) A block of iron of mass 7.5 kg and of dimensions 12 cm *  8 cm * 10 cm is kept on the table top on its base of side 12 cm * 8cm. Calculate (a) thrust And (b) pressure exert on the table top. 

Take 1 kgf = 10 N.

Ans. Mass of iron block = 7.5 kg

 Given, 1 kgf = 10 N 

 Trust (f) = m  * g = 7.5 * 10 = 75 N 

 Now, Area of table (A) = 12 * 8 = 96 cm2 

 

                                       =  96 * 10-4 m2 

Pressure = Thrust / Area = F / A = 75 / 96 * 10-4

                                                    = 7812.5 pa

Q 3) A vessel contains water up to height of 1.5 m. Taking the density of water 103 kg m-3   , acceleration due to gravity 9.8 m s-2 ana area of the base of vessel 100 cm2 . 

Calculate : 

(a) The pressure and

(b) The thrust at the base of vessel. 

Ans : Here, h = 1.5m, 

                    ρ = 103 kg m-3               

                    g =  9.8 m s-2

(a) P = h ρ g = 1.5 * 103  * 9.8 N m-2 = 1.47 * 104 N m-2 

(b) Area of the vessel, A = 100 cm2 

                                        = 100 * 10-4 m2

Let thrust on the base of the vessel = F

Using P = F / A,

we have 

          F = P * A =     1.47 * 104 * 100 * 10-4 N = 147 

Q 4) The area of base of a cylindrical vessel is 300 cm2 . Water ( 103 kg m-3 ) is poured into it upto a depth of 6 cm.  Calculate : ( a) the pressure and (b) the thrust of water on the base. (g =  9.8 m s-2).

Ans . (a) Height of water column = 6 cm = 6/100 m

               Density of water d =  103 kg m-3 

               g =  9.8 m s-2

               A =   300 cm2  =  300/10000 = 3/ 100 m2

               P = h ρ g = 6 /100 *1000 * 10 = 600 Pa

          ( b) Thrust = P * A = 600 * 3 /100 = 18 N

Q 5 ) ( a ). Calculate the height of  a water column which will exert on its base the same pressure as the 70cm column of mercury. Density of mercury is 13.6 g  cm-3

          ( b ) Will the height of the water column in part (a) change if the cross section of the water column is made wider ?

Ans : ( a) h = ? 

                P = 70 / 100 dg of Hg column

                   = 70 / 100 * 13600 * g

                Pressure of water = 70 / 100 * 13600 g

                     h * 1000 * g    =  70 * 136 g

                                      h     =  70 * 136 /1000

                                             =   9.52 m

           (b) No - pressure does not depend on width of vessel, it depend on depth.

Q. 6)  The Pressure of water on the ground floor is 40000 Pa and on the first floor is 10,000 Pa. Find the height of the first floor. ( Take : density of water = 1000  kg m-3 , g = 10 m s-2)     

Ans : Difference of pressure = 40000 - 10000 = 30000 Pa

          Density of water = 1000  kg m-3 , g = 10 m s-2

                         hdg = 30000

                          h * 1000 * 10 = 30000

                        h height of 1st floor = 30000 / 1000 * 10 = 3m

Q 7 ) A simple U tube contains mercury to the same level in both of its arms. If water is poured to a height of 13.6 cm in one arm, how much will be the rise in mercury level in the other arm ?

Given : density of mercury = 13.6 * 103 kg m-3  and density of water =  103 m s-2

Ans : Press due to mercury column = Pressure due to water column

          h * 13.6 * 103 * g = 13.6 / 100 *  103 *  g

    

                  h = 1/ 100 m = 1/100 * 100 = 1 cm

Q 8) In a hydraulic machine, a force of 2 N is applied on the piston of area if cross section 10 cm2. What force is obtained on its piston of area of cross section 100  cm2 ?

Ans : Area of cross section of piston 1, A1 = 10 cm2.

         Force applied on it , F1 = 2N

                                           A2 = 100  cm2.

                                           F2 = ?

                       By the principle of hydraulic machine :

                                        F1/A1 = F2/A2

                                        2/10 = F2/100

                                         F2 = 2 * 100/10 = 20 N

                        Force on piston of area of cross-section 100 cm2 is 20 N

Q 9) What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of its brake shoe by exering a force of 0.5 N on the pedal ?

Ans : Here: F1 = 0.5 N, F2 = 15 N , A1/A2 = ?

           Using F2/F1 = A2 / A1, we have

                      

             15/0.5 = A2 / A1 

  or

            30 / 1 = A2 / A1

  or

              A1/A2 = 1/30

  i.e. Ratio of area of cross section of master cylinder to that of wheel = 1 : 30

Q.10) The area of piston in a hydraulic machine are 5 cm2 and 625 cm2. What force on the smaller piston will support a load of 1250 N on the larger piston ? State any assumption which you make calculation .

Ans : A1 = 5 cm2,  A2 = 625 cm2,

          F1 = ?           F2 = 1250 N

          F1 / A1 = F2 / A2

                 F1 = F2 /A2 * A1

                

                 F1 = 1250 * 5 / 625

                      =  10 N

Assumption Made  : No friction and no leakage of liquid.