An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension \( 0.095 \, \text{J/m}^2 \) and density \( 10^3 \, \text{kg/m}^3 \). The difference between pressure inside the bubble and atmospheric pressure is:
Given \( g = 10 \, \text{m/s}^2 \).
Show Hint
Correct Answer: 1
Approach Solution - 1
To solve the problem of finding the pressure difference in the bubble, we first need to understand the factors influencing the pressure. An air bubble submerged in a liquid experiences two main pressures: the hydrostatic pressure due to the liquid column above it and the pressure due to surface tension.
1. Calculate Hydrostatic Pressure:
The hydrostatic pressure \(P_{\text{hydro}}\) at depth \(h\) is given by:
\(P_{\text{hydro}} = \rho \cdot g \cdot h\)
where:
\(\rho = 1000 \, \text{kg/m}^3\) (density of the liquid),
\(g = 10 \, \text{m/s}^2\) (acceleration due to gravity),
\(h = 0.2 \, \text{m}\) (depth in meters).
Thus:
\(P_{\text{hydro}} = 1000 \cdot 10 \cdot 0.2 = 2000 \, \text{Pa}\).
2. Calculate Pressure Due to Surface Tension:
The pressure inside a bubble due to surface tension \(P_{\text{surface}}\) is given by the formula:
\(P_{\text{surface}} = \frac{2 \cdot T}{r}\)
where:
\(T = 0.095 \, \text{J/m}^2\) (surface tension),
\(r = 0.001 \, \text{m}\) (radius of the bubble).
Thus:
\(P_{\text{surface}} = \frac{2 \cdot 0.095}{0.001} = 190 \, \text{Pa}\).
3. Calculate the Total Pressure Inside the Bubble:
The total pressure inside the bubble \(P_{\text{inside}}\) is:
\(P_{\text{inside}} = P_{\text{atmospheric}} + P_{\text{hydro}} + P_{\text{surface}}\)
Since we are finding the difference between the pressure inside the bubble and atmospheric pressure, we subtract \(P_{\text{atmospheric}}\) from both sides:
The pressure difference is:
\(P_{\text{diff}} = P_{\text{hydro}} + P_{\text{surface}} = 2000 + 190 = 2190 \, \text{Pa}\).
We conclude that the difference between the pressure inside the bubble and atmospheric pressure is 2190 Pa. This precise calculation validates the solution correctly, considering the provided values.
Approach Solution -2
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Surface Tension Questions
- A big drop is formed by coalescing 1000 small droplets of water. What will be the change in surface energy What will be the ratio between the total surface energy of the droplets and the surface energy of the big drop?
- A liquid drop of radius R is divided into 27 identical drops. If the surface tension of the drops is T, then find work done in this process.
- A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be :
- A big drop is formed by coalescing 1000 small identical drops of water. If E1 be the total surface energy of 1000 small drops of water and E2 be the surface energy of single big drop of water, the E1 : E2 is x : 1 where x = ________.
- A thin liquid film formed between a U-shaped wire and a light slider supports a weight of \(1.5 \times 10^{-2}\) \(N\). The length of the slider is \(30\ cm\) and its weight negligible. The surface tension of the liquid film is -
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.