Tag: pressure in fluids and atmospheric pressure

Questions Related to pressure in fluids and atmospheric pressure

The radii of the press plunger and the pump plunger are in the ratio 30 : 4. If an effort of 32 kgf acts on the pump plunger. Find the maximum effort the press plunger can overcome.

physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

  1. 1600 kgf

  2. 1700 kgf

  3. 1800 kgf

  4. 1900 kgf

Show answer
Correct Option: C
Explanation:

$\dfrac {\text {Radius of the press plunger}}{\text {Radius of pump plunger}}=\dfrac {R}{r}$


$=\dfrac {30}{4}$

According to the principal of hydraulic machine,

$\dfrac {L}{E}=\dfrac {\pi R^2}{\pi r^2}$

$\Rightarrow \dfrac {L}{32 kgf}=(\dfrac {30}{4})^2$


$\Rightarrow \dfrac {L}{32}=\dfrac {900}{16}$

$\Rightarrow L=\dfrac {900\times 32}{16}=1800 kgf$

Water is poured from a height of 10 m into an empty barrel at the rate of 1 litre per second. If the weight of the barrel is 10 kg, the weight indicated at time t = 60 s will be 

physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

  1. $71.4 \mathrm { kg }$

  2. $68.6 \mathrm { kg }$

  3. $70.0 \mathrm { kg }$

  4. $84.0 \mathrm { kg }$

Show answer
Correct Option: B
Explanation:

Time taken to reach $10m$ below $=\sqrt {\dfrac{{2 \times 10}}{{10}}}  = \sqrt 2 \,\sec $

$\therefore $ only $\left( {60 - \sqrt 2 } \right)\sec $ is effective for pouring of water
$\therefore $ total weight $=10lkg+\left( {60 - \sqrt 2 } \right)l \times \dfrac{{1\,kg}}{l}$
$=58.6kg+10kg$
$=68.6kg$
Hence,
option $(B)$ is correct answer.

The residual pressure of a vessel at ${27}^{o}C$ is $1\times{10}^{-11}N/{m}^{2}$. The number of molecules per cc in this vessel is nearly

physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

  1. $2400$

  2. $2.4\times{!0}^{8}$

  3. ${10}^{-11}\times 6\times{10}^{23}$

  4. $2.68\times{10}^{-19}\times {10}^{-11}$

Show answer
Correct Option: C

Two identical holes of each of cross sectional area $A$ are opened on the opposite sides of a wide vertical vessel filled with water.The difference in vertical height of the two holes is $h$. The resultant force of reaction of water flowing out of vessel is-

physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

  1. $h\rho gA$

  2. $\cfrac{1}{2}h\rho g A$

  3. $2h\rho gA$

  4. $\cfrac{1}{4}h\rho g A$

Show answer
Correct Option: C

Hydraulic lifts are used to lift the heavy loads primarily because :

physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

  1. they look good

  2. they amplify the force applied by a large extent

  3. they are sturdy

  4. they have better tensile strength

Show answer
Correct Option: B
Explanation:

Hydraulic lifts works on the pascals law principle 

therfore it amplify the force applied by a large extent  
hence option (B) is correct

A large cylindrical tank has a hole of area $A$ at its bottom and water is poured into the tank through a tube of cross-section area $A$ ejecting water at the speed $v$. Which of the following is true?

physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

  1. Water level in tank keeps on rising

  2. No water can be stored in the tank

  3. Water level will rise to height $v^{2}/2g$ and then stop

  4. The water level will be oscillating

Show answer
Correct Option: C
Explanation:

In the beginning ,the water escaping the vessel at the rate $\sqrt{2gh}$ and filling it at the rate of $vA$


So the rate of change in the water level of the vessel at any instant is 
($v-\sqrt{2gh})A$ 
where  'h' is the instantaneous height 

Since h is quite small in the beginning, the water is escaping the vessel at a slow rate but filling at a much larger rate as such    ($v-\sqrt{2gh})A$  ,is positive  (increase in water level)

soon after the filling of water  occurs, the height increases and the rate of escape become larger 
But when the rate =rate of filling ,then an equilibrium state is reached and there is no net change  in the height of the water 

the equilibrium is reached when ($v-\sqrt{2gh})A$ =0  $\rightarrow$  $h=\dfrac{v^2}{2g}$

State the principle on which a hydraulic press works. 

physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

  1. Archimedes principle

  2. Bernoulli's principle

  3. Pascals Law

  4. None of these

Show answer
Correct Option: C
Explanation:

It works on the principle of Pascals law.
Pascal's law: Pressure on a confined fluid is transmitted undiminished and acts with equal force on equal areas and at 90 degrees to the container wall.
Uses of hydraulic press - Hydraulic presses are commonly used for forging, clinching, moulding, blanking, punching, deep drawing, and metal forming operations.

The principle of fluid pressure that is used in hydraulic brakes or lifts is that.

physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

  1. pressure is the same at all levels in a fluid.

  2. increases of pressure are transmitted equally to all parts of a fluid.

  3. the pressure at a point in a fluid is due to the weight of the fluid above it.

  4. increases of pressure can only be transmitted through fluids.

Show answer
Correct Option: B
Explanation:

Answer is B.
Hydraulics is the use of a liquid under pressure to transfer force or motion, or to increase an applied force. The pressure on a liquid is called HYDRAULIC PRESSURE. And the brakes which are operated by means of hydraulic pressure are called HYDRAULIC BRAKES. These brakes are based on the principle of Pascals law. 
The Pascal law states that 'The pressure exerted anywhere in a mass of confined liquid is transmitted undiminished in all directions throughout the liquid'.

In a hydraulic lift, used at a service station the radius of the large and small piston are in the ratio of 20 : 1. What weight placed on the small piston will be sufficient to lift a car of mass 1500 kg ? 

physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

  1. 3.75 kg

  2. 37.5 kg

  3. 7.5 kg

  4. 75 kg

Show answer
Correct Option: A
Explanation:

The answer is A.

Pressure is the amount of force acting per unit area. That is, P=F/A.
where:
p is the pressure,
F is the normal force,
A is the area of the surface on contact. Let us consider A = $\pi { r }^{ 2 }$.
Therefore, $\dfrac { { F } _{ 1 } }{ { \pi { r } _{ 1 } }^{ 2 } } =\dfrac { { F } _{ 2 } }{ { \pi { r } _{ 2 } }^{ 2 } } $.


In this case, $\dfrac { 1500 }{ { 20 }^{ 2 } } =\dfrac { W }{ { 1 }^{ 2 } } ,\quad W\quad =\quad 3.75\quad kg.$
Hence, weight to be placed on the small piston sufficient to lift a car of mass 1500 kg is 3.75 kg.

Hydraulic press is based upon :-

physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

  1. Archimede's principle

  2. Bernoulli's theorem

  3. Pascal's law

  4. Reynold's number

Show answer
Correct Option: C
Explanation:

Hydraulic press works on Pascal's principle, which states - “pressure throughout a closed system is constant.” 


In hydraulic press, there is a piston at one end of the system having a small cross-sectional area. This piston is driven by a lever. The other end of the system has small-diameter tubing. 
When a fluid is displaced, either of the pistons is pushed inward. Small piston displaces a smaller amount of volume than the large piston for a given distance of movement. The displaced fluid volume is proportional to the ratio of pistons' head areas. Hence, smaller piston moves a large distance to move large piston significantly. The large piston movement distance is equal to small piston movement distance divided by the ratio of the pistons' head areas.