Excess Pressure in Bubbles

Q.

When two soap bubbles of radius $r_1$ and $r_2$ ($r_2$ > $r_1$) coalesce, the radius of curvature of common surface is

1. 

2. 

3. 

4. 

Q. The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface.Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the densityof mercury, the depth of the lake is

Q. A vessel contains a mixture of 7g of nitogen and 11g of carbon dioxide at temp.290k .If pressure of the mixture is 1atm (=1.01×10^5N/m^{2 }then density is(R=8.31 j/mol k)

Q. A room has floor dimensions of 4.0 m x 5.0 m and height of 3.0 m.(a):Find the weight of the air in the room at 20^° c(b)What is the weight of an equal volume of water?(cWhat is the total downward force on the floor surface due to air pressure of 1.0 atm?

Q.

Surface tension of water is$0.072N/m$. The excess pressure inside a water drop of a diameter $1.2mm$ is:

1. $240N/m^2$

2. $24N/m^2$

3. $0.06N/m^2$

4. $60N/m^2$

Q. An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm2 of air trapped in itis brought to the surface of the lake. The volume if the trapped air will be(atmospheric pressure 70 cm of Hg 13.6 g/cm)(1) 350 cm3. The bell(2) 300 cm(3) 250 cm3(4) 22 cm

Q. If the excess pressure inside a soap bubble of radius $1\text{cm}$ is balanced by an oil
($\rho =0.8{\text{g/cm}}^3)$ column of height $2\text{mm}$, then the surface tension of soap solution will be
[Take $g=10{\text{m/s}}^2$]

1. $0.02\text{N/m}$
2. $0.04\text{N/m}$
3. $0.09\text{N/m}$
4. $0.08\text{N/m}$

Q. A soap bubble, having radius $1\text{mm}$, is blown from a detergent solution having a surface tension of $2.5\times {10}^{-2}\text{N/m}$. The pressure inside the bubble equals at a point $Z_o$ below the free surface of water in a container. Taking $g=10{\text{m/s}}^2$, density of water $={10}^3{\text{kg/m}}^3$, the value of $Z_0$ is
[Take $g=10{\text{m/s}}^2$]

1. $5\text{cm}$
2. $8\text{cm}$
3. $1\text{cm}$
4. $3\text{cm}$

Q. What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10¯¹ N m¯¹. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.

Q. As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36, 000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = $6.0\times {10}^{24}kg$, radius = 6400 km.

Q. Determine the surface tension forces in case a plate of radius 10cm is placed on a water surface (surface tension of water = $7\times {10}^{-2}N/m;\pi =\frac{22}{7})$

1. 8800 dyne

2. 2200 dyne

3. 6600 dyne

4. 4400 dyne

Q. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10¯² N m¯¹ ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 105 Pa).

Q. Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6 kg/m3 and angle of contact is 135^° then what is the ratio of surface tension for water and mercury? (Angle of contact for water and glass is 0^°)

Q. A glass tube, sealed at both ends , is 100cm long .it lies horizontally with the middle 10cm containing mercury.the two ends of the tube contain air at 27degreeC. and at a pressure 76cm of mercury.the air column on one side is maintained at 0degreeC and the other side is maintained at 127degreeC.calcyulate the length of the air column on the cooler side.neglect the changes in the volume of mercury and of the glass.

Q. What will be the diameter (in $\text{mm}$) of a water droplet, the pressure inside which is $0.05{\text{N/cm}}^2$ greater than the outside pressure? (Take surface tension as $0.075\text{N/m}$).

1. $3$
2. $0.3$
3. $0.6$
4. $6$

Q. One of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is $4.22\times {10}^{8}m$. Show that the mass of Jupiter is about one-thousandth that of the sun.

Q.

What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10^–2 N m^–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 10⁵ Pa).

Q. How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5\times {10}^{8}km$.

Q.

Find the excess pressure inside a mercury drop of radius 2.0 mm. The surface tension of mercury = 0.464 N/m

1. 928 N/m²

2. 464 N/m²

3. 464 N/m

4. 928 N/m

Q. What should be the pressure inside a air bubble of radius $0.1\text{mm}$ situated $1\text{m}$ below the water surface?
(Take surface tension of water $T=7.2\times {10}^{-2}\text{N/m}$ and $g=10{\text{m/s}}^2$)

1. $3.5\times {10}^5\text{Pa}$
2. $1.11\times {10}^5\text{Pa}$
3. $2.11\times {10}^5\text{Pa}$
4. $3\times {10}^5\text{Pa}$

Q. Air is pushed into a soap bubble of radius r to double its radius. If the surface tension of the soap solution in S, the work done in the process is
(a) 8 π r² S
(b) 12 π r² S
(c) 16 π r² S
(d) 24 π r² S

Q. Let us assume that our galaxy consists of $2.5\times {10}^{11}$ stars each of one solar mass. How long will a star at a distance of 50, 000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be ${10}^5$ ly.

Q. A bubble of air has radius $0.2\text{mm}$ in water. If the bubble had been formed $20\text{cm}$ below the water surface on a day when the atmospheric pressure was $1.013\times {10}^5\text{Pa}$, then what would have been the total pressure inside the bubble? (surface tension of water = $73\times {10}^{-3}\text{N/m}$)

1. $1.052\times {10}^5\text{Pa}$
2. $1.039\times {10}^5\text{Pa}$
3. $1.022\times {10}^5\text{Pa}$
4. $1.152\times {10}^5\text{Pa}$

Q. The excess pressure inside the first soap bubble is three times that inside the second bubble. Then, the ratio of the volume of the first bubble to that of the second bubble will be

1. $1:27$
2. $3:1$
3. $1:3$
4. $1:9$

Q. air is blown inside a soap bubble. what will be the effect of pressure inside a soap bubble.??

Q. The excess of pressure inside a soap bubble than that of the outer pressure is

1. $\frac{2T}{r}$
2. $\frac{4T}{r}$
3. $\frac{T}{2r}$
4. $\frac{T}{r}$

Q.

On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius with the bottom of the vessel. If r << R and the surface tension of water is T, the value of r just before bubbles detach is (density of water is ${\rho }_w$).

1. $R^2\sqrt{\frac{{\rho }_{w}g}{T}}$

2. $R^2\sqrt{\frac{3{\rho }_{w}g}{T}}$

3. $R^2\sqrt{\frac{{\rho }_{w}g}{6T}}$

4. $R^2\sqrt{\frac{2{\rho }_{w}g}{3T}}$

Q. A water droplet splits into $27$ identical small droplets. The pressure difference between the inner and outer surface of the big droplet will be

1. Same as that of smaller droplet
2. $1/3\text{rd}$ of the pressure difference for smaller droplet.
3. $1/4\text{th}$ of the pressure difference for smaller droplet.
4. None of these

Q. The excess pressure inside a spherical drop of water is $4$ times that of another drop. Then their respective mass ratio is

1. $1:4$
2. $1:64$
3. $1:16$
4. $64:1$

Q. 12. In an isolated container 1 mole of a liquid , 100 ml is present at 1 bar . If pressure is steeply taken to 100 bar find Δ H and Δ U for the process if volume is reduced by 1ml. Options: 1. 2O3(g) → 3O2(g) Δ H = ve, Δ S = +ve 2. Mg(s) + H2(g) → MgH2 Δ H = ve, Δ S = ve 3. Br2() → Br2(g) Δ H = +ve, Δ S = +ve 4. 2Ag() + 3N2(g) → 2AgN3 Δ H = +ve, Δ S = ve