Pressure exerted by air - class-XI

In an accelerating frame, the effective gravity is the vector sum of g and -a. At the center of a cube of side L, the depth from the free surface is L/2. Thus, pressure P = rho * g_eff * depth = rho * sqrt(g^2 + a^2) * L/2. However, assuming horizontal acceleration, the pressure gradient leads to P = rho * (g+a) * L/2 if considering vertical and horizontal components combined or specific orientation.

Sheet crumpled into a ball has a small surface area as compared to the similar unfolded sheet. Therefore, unfolded sheet will experience more resistance due to air as compared to the sheet crumpled into a ball, inspite of same force of gravity acting upon them. It is larger resistance of air which slows down the unfolded sheet, and therefore it falls slower as compared to sheet crumpled into a ball.

When water heats up the molecules move faster and faster and move farther apart, eventually becoming steam. When the pressure on the water is increased,  the molecules are more tightly held. This helps keep them from moving apart so it requires more heat to get them to "break free" and move apart. So, the higher the water pressure the higher the boiling point. 

In a pressure cooker, as the pressure inside increases with time, the boiling point of water increases and hence is able to cook fast.

As the piston of pump is pulled up, the pressure inside decrees which leads to unbalance of pressures.

The pressure inside our bodies is equal to the atmospheric pressure and cancels the pressure from outside. we do not get crushed under the atmospheric pressure.

As the pressure inside the dropper, i.e, the internal pressure and the atmospheric pressure become equals,  ink does not spill out.

(i) True, because we do not get crushed as the pressure inside and atmospheric pressure becomes equal.

When the plunger is pulled up in the barrel, the pressure inside the barrel is much less than atmospheric pressure. As a result atmospheric pressure forces the liquid to rise up in the syringe.

When air pressure is measure at the sea level using a mercury barometer, the height upto which the mercury column rises is $76cm$.

Liquid flows from higher pressure region to the lower pressure region.

The atmospheric pressure exerted on earth is due to gravitational pull of the earth and is about 14.7 psi.

Work done against atmospheric pressure is W = P * delta_V. Given P = 1 atm = 1.013 * 10^5 Pa and the change in volume (steam volume - water volume), the work is approximately 40 cal.

According to Charles's Law, volume is directly proportional to temperature. When a partially inflated balloon is placed into warm water, the air inside heats up, expands, and causes the balloon to expand fully. Cold water would cause the air to contract.

(C)

False. Wide foundations exert lesser pressure because increase in area of contact reduces the pressure and the buildings do not sink.

A free falling object is an object that is falling under the sole influence of gravity.
Atmospheric pressure decreases with height above the surface of a planet because there is less total mass in the atmosphere above a reference point as the height of the reference point increases. The standard atmospheric pressure at sea level is 1 atmosphere.
Pressure is generally measured in units of Pascals and S.I. unit of pressure is pascal.
The normal atmospheric pressure is 760 mm of Hg. But the normal human blood pressure is around 120/80 mm only.
Answer (B) & (D) 
(B) pressure on top of mount everest is much more than 1 atm
(D) the blood vessels in our body maintains the internal pressure equal to the atmospheric pressure

nothing in the system changes except the reaction of the ground.

Given, That a glass bulb is balanced by a brass weight in a sensitive beam balance,

Pressure = $\displaystyle \frac{F}{A}$ $\because 10^6$ dyne $cm^{-2} = 10^5 N m^{-2}$.

We have,

$1atm$ is the standard atmospheric pressure. 

One Pascal is the pressure exerted by a force of one newton, normally on an area of one square meter.

Pressure is defined as force per unit area. It is usually more convenient to use pressure rather than force to describe the influences upon fluid behavior. The standard unit for pressure is the Pascal, which is a Newton per square meter.

Both $atm$ and $torr$ is the unit of pressure.

We have:

Absolute pressure: It uses absolute zero (pressure in vacuum) as its zero point.

Absolute pressure: It uses absolute zero (pressure in vacuum) as its zero point.

Pressure is defined as force per unit area applied on a surface.

Gauge Pressure uses a reference to the atmosphere around the sensor. Because the sensing element has a deflection due to a pressure change, a reference point is needed to know exactly what pressure is being measured. Pressure sensors that use gauge pressure—typically seen in PSIG, BARG, and kPaG —have some type of vent. This vent can be built in to the sensor or even through a tube in the electrical connection. The vent is positioned to use atmospheric pressure as a reference point for the sensor to measure the media. One common reason for using gauge pressure is to ensure that with any location throughout the world, the sensor will always reference the location in which it is installed.

1 atm = 76 cm of Hg 

Using Boyle's Law P1V1 = P2V2. Initial pressure P1 = P_atm + rhogh. Final pressure P2 = P_atm. Since radius triples, volume increases 27 times. P_atm + rhogh = 27 * P_atm. With P_atm = rhogl, we get h = 26l.

Using Boyle's Law for the trapped air: P_initial * V_initial = P_final * V_final. Initially, the tube is half-immersed (40 cm air at P_atm). Finally, 20 cm Hg remains, so air column is 60 cm. Solving gives P_atm = 60 cm Hg.

The scale expands with temperature. Actual height H = H_observed * (1 + alpha * delta_T). H = 76.5 * (1 + 1.9 * 10^-5 * 30) = 76.5 * (1.00057) = 76.543 cm. The provided answer 76.46 cm suggests a different correction factor or calculation method.

Answer is D.

The air pressure decreases as the we go up the atmosphere. Therefore, when a  barometer is carried from the 1st floor to the 20th floor of a building, the air pressure decreases and hence the reading on the barometer falls down.

Speedometre is an instrument which is used to measure the speed of a vehicle.

Pressure energy per unit volume=$=\dfrac{Work done}{Volume}=\dfrac{W}{V}=\dfrac{F.d}{A.d}$

$1atm$ is the standard atmospheric pressure. 
In terms of length of mercury it is $76cm $ of Hg.

The pressure due to $760$ mm of mercury is equal to atmospheric pressure. Pressure equivalent to $1$mm of mercury is $1torr$. Hence the atmospheric pressure $1atm=760torr$.

Atmospheric pressure $=1atm=760torr=1.01\times 10^3Pa$

Absolute pressure: It uses absolute zero (pressure in vacuum) as its zero point.

The pressure at the bottom of $760 mm$ of mercury is equal to atmospheric pressure.

We have,

Atmospheric pressure $=1atm=1.01\times 10^5Pa=760torr$

The pressure at the bottom of mercury column tries to equate the atmospheric pressure outside. 
When the compensating height of mercury column increases, it means that the atmospheric pressure has indeed increased.

Psi stands for pound force per square inch. 

Atmospheric pressure $=1atm=1.01\times 10^5Pa=1.01bar$

Fortin's barometer is a modified form of Torricelli's barometer which is used to measure atmospheric pressure. In free space there is no matter (no atmosphere) therefore no atmospheric  pressure hence barometer will read 0mm of Hg .

from$T=\dfrac{p-p _0}{p _{100}-p _0}\times 100^0=\dfrac{35-28.6}{36.6-28.6}\times 100^0=80^0$

$P=\rho gh=13600\times 9.8\times 0.760=1.01\times 10^5Pa$

It is given that the diameter of one vessel is n times larger than the diameter of other. So,

$ \pi {{r}^{2}}{{h} _{1}}=\pi {{(nr)}^{2}}{{h} _{2}} $

$ {{h} _{1}}={{n}^{2}}{{h} _{2}} $

Since, pressure at point A is equal to pressure at B

$ \rho gh=({{h} _{1}}+{{h} _{2}})\rho 'g $

$ \because s=\dfrac{\rho '}{\rho } $

$ \rho gh=({{n}^{2}}{{h} _{2}}+{{h} _{2}})s\rho g $

$ {{h} _{2}}=\dfrac{h}{({{n}^{2}}+1)s} $

When the tube is vertical, the pressure difference between the two air columns must balance the weight of the mercury column. Using Boyle's Law for both air columns, we find the displacement of the mercury.

The vertical height of the mercury column in a barometer is independent of the tube's inclination, provided the reservoir is open to the atmosphere. The length of the mercury column along the tube will increase, but the vertical height remains 76 cm.

Using Boyle's Law: P1V1 = P2V2. P1 = P_atm + rho_wgh. P2 = P_atm. Since diameter doubles, volume increases 8 times. P_atm + rho_wgh = 8 * P_atm. h = 7 * P_atm / (rho_wg). Given P_atm = rho_mg*h_m, we get h = 7 * (rho_m/rho_w) * h_m = 7 * rho * h.

Let the point at half depth be $x,$

$P _1=P+dgx$

Let the depth of lake be $2x$,

$P _2=P+dg2x$

$P _2=3/2×P1$

$2P _2=3P1$

$2 (P+dg2x)=3 (P+dgx)$

$2P+4dgx=3P + 3dgx$

$P=dgx$

$x=P/dg$

$x=10^5/10000 (d=10^3kg/m^3) (P=10^5pa)$

But depth of lake is $2x$

$2x=2×10=20m$

First find the moles of each gas: moles of O2 = 8/32 = 0.25, moles of N2 = 14/28 = 0.5, moles of CO2 = 22/44 = 0.5. Total moles = 0.25 + 0.5 + 0.5 = 1.25 moles. Using the ideal gas law P * V = n * R * T, with V = 10 L, T = 300 K, and R = 0.0821, we find P = (1.25 * 0.0821 * 300) / 10 approximately equal to 3.07 atm, which rounds to 3 atm.

Using Boyle's Law: P1V1 = P2V2. P1 = P_atm + rhogh. P2 = P_atm. Volume increases 8 times (2^3). P_atm + rhogh = 8 * P_atm. h = 7 * P_atm / (rho*g). Since P_atm = 10m of water, h = 7 * 10m = 70m.

By Archimedes' principle, the weight of the block equals the weight of the displaced mercury. rho_s * L^3 * g = rho_m * (L^2 * (L - h)) * g. Solving for h (height above surface) gives h = L * (1 - rho_s/rho_m).

The overflow volume is given by V * (gamma_mercury - gamma_glass) * delta_T. Overflow = 1000 * (180 - 40) * 10^-6 * 100 = 1000 * 140 * 10^-4 = 14 cc.

The pressure of dry gas is calculated by subtracting the aqueous vapor pressure from the total pressure of the moist gas. Pressure_dry = Total_pressure - Vapor_pressure = 735 mm - 23.8 mm = 711.2 mm.

In a barometer, the vertical height h is the projection of the length l of the mercury column along the tube. Thus, h = l * cos(alpha), which rearranges to l = h / cos(alpha).

For a sealed bulb at constant volume, Gay-Lussac's Law applies: P1 / T1 = P2 / T2. Initial temperature T1 = 41 + 273 = 314 K, initial pressure P1 = 1 atm. Final temperature T2 = 198 + 273 = 471 K. P2 = (1 * 471) / 314 = 1.5 atm.

The pressure difference due to surface tension in a capillary tube is delta_P = 2 * T / R. To minimize the pressure required to move the meniscus or to control the flow, the limb with the smaller radius (higher capillary pressure) is typically connected to the pump.

Pressure is defined as Force divided by Area (P = F/A). Therefore, increasing the force or decreasing the area will increase the pressure.

Pressure difference created $= 10 mm$ of Hg 
This must be equal to the pressure of water column being created in the straw. If height of water column be $h$

We know that $1$ bar $= 10^5 Pa$

 h$=76cm$

Pressure=?

Formula to be used : $P=hgd$

Where h = height of mercury column and

 d= density of mercury$=13.6 gm/cc$

acceleration due to gravity$=980 cm/s²$

$P=hgd=13.6\times 980\times 76=1012928dyne.cm²$

Similar to question 480419, the vertical height h is the projection of the length l along the tube, so h = l * cos(alpha), which means l = h / cos(alpha).

Barometer is a device used to measure atmospheric pressure. It was first constructed by Torricellli in the year 1643.

The height to which mercury rises in simple barometer is $76cm$.

A mercury barometer has a glass tube, of length approximately 100cm, closed at one end with an open mercury-filled reservoir at the base. 

A simple barometer or a mercury barometer has a glass tube closed at one end with an open mercury-filled reservoir at the base. The weight of the mercury creates a vacuum in the top of the tube known as Torricellian vacuum. 

A mercury barometer, or a simple barometer, has a glass tube, of length approximately 100cm, closed at one end with an open mercury-filled reservoir at the base. The weight of the mercury creates a vacuum in the top of the tube known as Torricellian vacuum. Mercury in the tube adjusts until the weight of the mercury column balances the atmospheric force exerted on the reservoir. High atmospheric pressure places more force on the reservoir, forcing mercury higher in the column. Low pressure allows the mercury to drop to a lower level in the column by lowering the force placed on the reservoir.