A force of 2x102 N is applied on one of the piston of area of cross section 100cm2of hydraulic lift to support a car placed on the second piston of lift of area of cross section 1960cm2. Find the mass of the car.

Given,
Force, F = 2×102
Area of cross-section of the first piston = 100 cm2 
Area of cross-section of the second piston = 1960 cm

Weight of the car = ?

Let, M be the mass of car.

The pressure on the piston on which force is applied is, 

              P1=Fa=2×102100=2 N/cm   ...(1)

The pressure on the second piston on which car is placed is, 

           P2=MgA=M×9.81960=M200N/cm    ...(2)

Now, According to Pascal law we have

                       P=  P2    

 we get from equation (1) and (2),  

Mass of the car, M=400  kg 
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What is manometer?

It is U tube open at both the ends having some mercury in it. It is used to measure the gauge pressure or pressure difference of gases on the two sides of manometer tube by measuring the height difference of mercury column in two limbs


It is U tube open at both the ends having some mercury in it. It is u

of manometer tube. If the difference in the height column of mercury in the two limbs of tube is h, then the gauge pressure is ρgh and absolute pressure is Pa + ρgh.
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The atmospheric pressure at sea level is 760mm of mercury. At what height it will be 740mm of mercury? Take the density of air 1.293kg/m3 constant upto height about 1 km.


Given, atmospheric pressure, P1= 760 mm of Hg
Atmospheric pressure, P2 = 740 mm of Hg
Density of air, ρ = 1.293 kg/m3

Let the height at which the atmospheric pressure is decreased be 'h'. 

Now, Pressure due to h height of air = Pressure due to 2 cm or 0.02 m of mercury. 

Threfore,

1.293 x g x h = 13600 x g x 0.02 

              h = 210 m, is the height at which the atmospheric pressure will decrease. 
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A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?


Given, 
Weight of the girl = 50 kg = 490 N
Circular diameter of the heel = 1.0 cm = 10-2

Area of the heel is given by, 

           A=14πD2=14π×10-4m2 

Threfore, the pressure exerted by the heel on the ground is given by, 

       P = WA=49014π×10-4=6.2×106N/m2

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What height of water column produces the same pressure as 50 cm of Hg column?

The pressure due to height column, P = ρgh

If the pressure due to height column of different liquids is same, then the relation between the height coulumn is given by,

              ρ1gh1=ρ2gh2 h2=ρ1ρ2h1=13.61×50                         = 680cm                         =6.80m , 
is the required height of the water column which will produce the same pressure as 50 cm of Hg column. 

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