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Friday, March 6, 2015

Pressure - C. Science


Pres­sure:

Pres­sure is defined as the force act­ing per unit area of the surface.
math­e­mat­i­cally,
P = \frac{Force}{Area}
from above equa­tion we can say that pres­sure is inversely pro­por­tional to area pro­vided when force is con­stant. That means, a given force (F) will exert more pres­sure if it acts on a smaller area, and less pres­sure if it  acts on a big­ger area.
i.e. P\propto \frac{1}{Area}
Sim­i­larly, it fol­lows that pres­sure is directly pro­por­tional to force pro­vided the on the con­stant area. That means more pres­sure will cre­ate more force and vice versa.
i.e P \propto F

Unit: The SI unit of pres­sure is  N/m^2 or pas­cal (Pa).
Though the force asso­ci­ated with the pres­sure at a point is a vec­tor quan­tity, the pres­sure itself is a scalar i.e. pres­sure has no direction.

Atmos­pheric pressure:
The air that sur­rounds the earth is called atmos­phere. It is extended about 9,600 km above the earth sur­face. The air has weight. Thus, it exerts pres­sure on the sur­face of the earth. The pres­sure exerted by the atmos­pheric air is called atmos­pheric pres­sure.  At sea level, the air pres­sure is said to be one atmos­phere. This is also termed as stan­dard atmos­pheric pres­sure or nor­mal pressure.
The nor­mal pres­sure is the pres­sure exerted by 760 mm long col­umn of mer­cury at O^0C at sea level and at a alti­tude of 45^0 . It is equal to 1.02 * 10^5 N. In short, we can write as, 1 atmosphere = 760 mm of Hg = 1.02 * 10^5 Nm^{-2}or pascal.
1 torr = 1 mm Hg = 103.222 pa

Liq­uid pressure
A liq­uid sub­stance in a con­tainer exerts pres­sure in all direc­tion. For exam­ple liq­uid exerts pres­sure on the bot­tom and to the walls of the con­tainer. The pres­sure of the liq­uid is pro­duced due to its weight.
Pressure
liquid pressure


Derive the for­mula : P = dgh

pressure derivation
Height ‘h’ and area ‘A’
Con­sider a liq­uid of den­sity ‘d’ is con­tained in a beaker of cross — sec­tional area ‘A’. Let ‘h’ be the depth of the liq­uid col­umn from its free sur­face as shown in the figure.
or, V = A * h
the weight of the liq­uid col­umn = Mass * gravity
or, W = m* g
= ( v * d ) * g    [  \because d = \frac{m}{v}, m = d * v ]
\therefore W = A * h * d * g  [ \because V = A * h]

Thus weight of the liq­uid col­umn is the force act­ing on the area ‘A’, nor­mally. Therefore,
P =\frac {F} {A}
P =\frac { A* h * d * g} {A}  [\because W= F ]
P = dgh
Thus pres­sure exerted by a liq­uid of den­sity ‘d’ on the base of the beaker is dgh. There­fore, the pres­sure exerted by liq­uid at rest depends on three factors:
a. depth or height of the liq­uid col­umn (h)
b. den­sity of the liq­uid (d) and
c. accel­er­a­tion due to grav­ity (g)
For a given liq­uid at  a given place, den­sity of the liq­uid and accel­er­a­tion due to grav­ity are con­stant. Hence, pres­sure exerted by a liq­uid is directly pro­por­tional to the depth of the liq­uid col­umn from the free sur­face of the liquid
i.e. P \propto h

Pascal’s law

Blase Pas­cal, born in France in 1623 A.D. was a famous math­e­mati­cian, physi­cist and philoso­pher. He was first to find out the behav­ior of liq­uid pres­sure and put for­ward a law which is pop­u­lar as Pascal’s Law.
Pascal’s law stats that ” When pres­sure is applied on a liq­uid enclosed in a ves­sel, it is trans­mit­ted equally in all the directions.”

Appli­ca­tion of pascal’s law

Pascal’s law tells that if pres­sure at any point in a liq­uid is changed, there will be equal change in pres­sure at any point in a liq­uid is changed, there will be equal change in pres­sure at other points in the liq­uid. This fact is used in hydraulic machines such as hydraulic press, hydraulic brake, hydraulic garage lift, hydraulic cranes etc.

Prin­ci­ple of hydraulic machine

If a force F_1 is applied at first pis­ton of area A_1, the pres­sure exerted at any point on it is given by P_1= \frac {F_1}{A_1}
This pres­sure is trans­mit­ted unchanged to sec­ond pis­ton of area A_2. There­fore, the upward force F_2 exerted on sec­ond pis­ton is given by P_2 = \frac {F_2}{A_2}

From pascal’s law hydraulic machine

P_1 = P_2
\frac {F_1}{A_1} = \frac {F_2}{A_2}
\frac {F_2}{F_1} = \frac {A_2}{A_1}
If A_2 > A_1, then F_2 > F_1. Thus a small force can be used to exert a much larger force. Thus, a hydraulic machine is a force multiplier.
The prin­ci­ple of hydraulic machine can be summed up as given below.
~ Any liq­uid can­not be compressed.
~ Pres­sure applied on an enclosed liq­uid is trans­mit­ted at each part of it (Pascal’s law)
~ The ratio of cross-sectional area of big cylin­der A_2 to that of smaller cylin­der A_1 is equal to the ratio of load over­come F_2 to effort applied F_1.
i.e. \frac {A_2}{A_1} = \frac {F_2}{F_1}
This prin­ci­ple is applic­a­ble in dif­fer­ent hydraulic machines.

Den­sity:

The den­sity of a sub­stance is defined as the mass per unit volume
Math­e­mat­i­cally,
Density = \frac {Mass} {Volume}
i.e. d = \frac {m}{v}
In SI sys­tem den­sity is mea­sured in kg/m^3 and in CGS sys­tem it is mea­sure in gm/cm^3.
1000kg/m^3 = 1gm/cm^3
The sub­stance hav­ing lesser den­sity is said to be lighter one while the sub­stances hav­ing greater den­sity so is said to be heav­ier one.
Rel­a­tive Den­sity:
 When the den­sity of any sub­stance is com­pared with water at 4^0C, it is called rel­a­tive den­sity of that sub­stance. There­fore, When the mass of a par­tic­u­lar vol­ume of sub­stance is divided with the mass of the same vol­ume of water at 4^0C, it is called rel­a­tive den­sity of that substance.
Math­e­mat­i­cally
relative \: density = \frac {mass\: of \:any\: volume\: of \:substance}{mass \:of \:same \:volume\: of\: water\:at \:4^0C}
relative \: density = \frac {Density \:of \:substance}{Density\: of \:water\: at \:4^0}
relative\: density = \frac {weight \:of\: body \:in \:air}{wt \:of\: body \:in \:air - weight\: of \:body\: in \:water}
Rel­a­tive den­sity is a sim­ple ratio that’s why it does not have any unit.
Achimedes 
Other Important Questions
1. The bot­tom of dam is wider and stronger than upper part.
OR
      Walls of a dam are made thicker at the bot­tom. Why?
Liq­uid pres­sure is directly pro­por­tional to the depth of liq­uid from free sur­face. At the bot­tom of dam, there is high water pres­sure. To pro­tect the dam from the high water pres­sure, the bot­tom is made wide and stronger than the upper part.
2. The pres­sure exerted by water at the bot­tom of the drum decreases slightly at the Himalayan region than that in the Terai region.
Since pres­sure of a liq­uid is directly pro­por­tional to the value of g and the value of g is more at Terai region than that in the Himalayan region. So, the pres­sure exerted by water at the bot­tom of the drum decreases slightly at the Himalayan region than that in the Terai region.
3. The bucket filled faster in the down­stairs tap than in upstairs tap.
Pres­sure of a liq­uid is directly pro­por­tional to its depth or height  from the free sur­face. The tap at the down­stairs is at greater height than the tap at the upstairs. Water comes out with greater pres­sure from the down­stairs tap. So, bucket is filled faster in the down­stairs tap than in upstairs tap.
4. The speed of flow of water out of a tap of upper floor is less than that of the down floor.
Since pres­sure of a liq­uid is directly pro­por­tional to its depth or height from the free sur­face. The height of the liq­uid col­umn on the down floor is more than that on the upper floor. So the speed flow of water out of a tap of upper floor is less than that of the down floor.
5. A deep sea-diver needs to wear div­ing suit.
The pres­sure of a liq­uid is directly pro­por­tional to the height of liq­uid col­umn from free sur­face. When a per­son dives deep into sea, he or she has to bear high water pres­sure. To with­stand with high water pres­sure, it is nec­es­sary to wear div­ing suit.
6.Weight of stone decreases in the salt solution.
When salt is dis­solved in water, the upthrust increases due to the increases in the den­sity of water  (upthrust \propto density\: of \: liquid). As a result, the weight of a stone decreases in the salt solution.
7. It is eas­ier to lift a heavy stone inside the water.
The weight of the stone inside the water decreases due to the upthrust. So it is eas­ier to lift a heavy stone inside the water.
8. The weight of a body is lighter in water than in air.
Upthrust due to water is greater than that of air. Upthrust decreases weight of body. Hence, the weight of a body is lighter in water than in air.
9. The weight of a body is more in vac­uum than in air.
There is no upthrust in vac­uum. Upthrust decreases the weight of body. Hence, the weight of body is less in air due to its upthrust.
10. It is eas­ier to pull a bucket of water from the well until it is inside the water but dif­fi­cult when it is out of water.
Upthrust  due to water is greater than that of air. Upthrust exerted by water decreases weight of body. Hence, it is eas­ier to pull a bucket of water from the well until it is inside the water but dif­fi­cult when it is out of water.
11. Studs are made on the sole of foot­ball player’s boot.
Studs are made on the sole of foot­ball player’s boot to increase the pres­sure on the ground that pre­vents the player from falling or slid­ing dur­ing sud­den run and turning.
12. It is eas­ier to cut veg­eta­bles with the sharp-edged knife than the blunt one.
The area of the edge of sharp knife is less than that of the blunt one. So sharp-edged knife can exert more pres­sure while apply­ing force than that of the blunt edged knife as the pres­sure is inversely pro­por­tional to the area. Due to more pres­sure, it is eas­ier to cut veg­eta­bles with the sharp-edged knife than the blunt one.
13. Buses and trucks have broad and dou­ble wheeled tyres.
The buses and trucks have to carry greater weight which exerts greater pres­sure on the tyres. The pres­sure on them is reduces by increas­ing the area of the tyres as the pres­sure is inversely pro­por­tional to the area.
14. Camel can walk eas­ily on sand.
Camel has flat-soled feet which exert less pres­sure due to the body weight of camel. As there is less pres­sure exerted by flat-soled feet of camel, it can eas­ily walk on the sand.
15. The wheel of trac­tor is wider than that of a car.
The rear wheels of a trac­tor are made large and flat to reduce the pres­sure and pre­vent the tyres from descend­ing into the muddy land. As trac­tor is used for plough­ing and thresh­ing in crop fields, small and nar­row real wheels may descent into the muddy and marshy load due to more pres­sure. (i.e. P\propto \frac {1}{A})
16. The water sup­ply tanks are placed higher than any other build­ing in a town.
The water sup­ply tanks are placed higher than any other build­ing in a town so that the pres­sure of water is more. Due to this, water can flow to the build­ings located at higher level. As a result, water can be sup­plied eas­ily on the taps of upper­most floors and houses built at higher level of ground.
17. The grav­ity bulb of hydrom­e­ter is made heavier.
The grav­ity bulb of hydrom­e­ter is made heav­ier so that its cen­tre of grav­ity remains very low, when it is float­ing in denser liq­uids. It helps the hydrom­e­ter to float upright in sta­ble equilibrium.
18. Hydrom­e­ter  has a heavy bulb and nar­row stem.
Hydrom­e­ter has a heavy bulb to float in a liq­uid ver­ti­cally. The stem of hydrom­e­ter is made long and nar­row to find the rel­a­tive den­sity of less dense liq­uid i.e. it increase its sensitivity.
19. The floata­tion bulb (buoy­ancy bulb) of hydrom­e­ter is made small.
The floata­tion bulb of hydrom­e­ter is made small because denser liq­uids pro­vide more upthrust and hence the whole hydrom­e­ter does not float on the dense liquid.
20. An egg sinks in fresh water but floats in a strong solu­tion of salt (or in a sat­u­rated salt solution).
Fresh water has less den­sity than that of the egg so the egg can­not dis­place water equal to its weight and it sinks. But salt on mix­ing with water increases the den­sity of water which increase upthrust and hence the egg can dis­place water equal to its weight and floats.
21. Ice­berg made up of water floats in water.
The den­sity of ice is less than that of water. Due to this, ice­berg is able to dis­place the water equal to its weight. Hence, Ice­berg made up of water floats in water.
22. An iron nail sinks in water but a ship made up of iron floats.
Since the den­sity of iron is greater than that of water, the iron nail sinks as the weight of the nail is greater than the weight of the dis­placed water. But the ship made of up iron is designed in such a way that the rel­a­tive den­sity of ship becomes always less than that of water. Hence, the weight of water dis­placed becomes equal to the total weight of the ship and the ship floats.
23. It is eas­ier for a man to swim in sea water than in river water.
The den­sity of sea water is more than that of river water. Thus for the same vol­ume,  the sea water pro­vides more upthrust as com­pared to the river water. So it is eas­ier for a man to swim in sea water than in river water.
24. A hole at the bot­tom of a boat is more dan­ger­ous than that at the side.
The bot­tom of the boat exerts more pres­sure, as the force acts per­pen­dic­u­larly down­wards, than that at the sides. If there is hole at the bot­tom of the boat, more water enters inside due to more upthrust and the boat may sink. So, a hole at the bot­tom of a boat is more dan­ger­ous than that at the side.
25. An iron nail sinks in water but floats in mercury.
The den­sity of iron is more than that of the water. The iron nail can­not dis­place the eater equal to its weight and it sink. But the den­sity of mer­cury is much more than that of the water. It gives more upthrust which helps the iron nail to float on mercury.
26. A girl with heeled shoe exerts a higher pres­sure on the ground than an elephant.
The area of the heels of the girl’s shoes is much smaller in com­par­i­son with the area cov­ered by the flat feet of the ele­phant. As  P\propto \frac {1}{A} a girl with heeled shoe exerts a higher pres­sure on the ground than an elephant.
27. A man exerts more pres­sure under his foot when he stands with one foot than when he stands on two feet.
The area of one foot is less than that of two feet. As  P\propto \frac {1}{A} so a man exerts more pres­sure under his foot when he stands with one foot than when he stands on two feet.
28. The base of the house is made wider than its wall.
The walls and roof of house exerts pres­sure on the base of a house. The pres­sure exerted by the build­ing is reduced by mak­ing the base of the house wider as pres­sure is inversely pro­por­tional to the area.
29. The loaded ship sinks more than empty ship.
The loaded ship has larger mass. Hence, the weight (or vol­ume) of dis­placed water is more for loaded ship. But the empty ship has lesser mass. Hence, the weight (or vol­ume) of dis­placed water is less for empty ship. So the loaded ship sinks more than empty ship.

UPTHRUST, ARCHIMEDES PRINCIPLE AND FLOATATION


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