An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m\(^3\). If the pressure inside the bubble is 2100 N/m\(^2\) greater than the atmospheric pressure, then the surface tension of the liquid in SI units is (use \(g = 10 \, {m/s}^2\)).
Show Hint
- 0.02
- 0.1
- 0.25
- 0.05
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to use the concepts of pressure difference and surface tension. The pressure difference caused by surface tension for an air bubble is given by the formula:
\(P_{\text{inside}} - P_{\text{outside}} = \frac{4T}{r}\)
Here, \(T\) is the surface tension of the liquid, and \(r\) is the radius of the air bubble.
Given that the pressure inside the bubble is 2100 N/m\(^2\) greater than the atmospheric pressure, we can write:
\(P_{\text{inside}} - P_{\text{outside}} = 2100 \, \text{N/m}^2\)
Also, given that the radius \(r = 0.1 \, \text{cm} = 0.001 \, \text{m}\) (converted to meters), we can substitute these values into the equation:
\(2100 = \frac{4T}{0.001}\)
Simplifying for \(T\) gives:
\(T = \frac{2100 \times 0.001}{4} = \frac{2.1}{4} = 0.525 \times 0.1 = 0.05 \, \text{N/m}\)
Therefore, the surface tension of the liquid is \(0.05 \, \text{N/m}\).
The correct answer is 0.05.
Approach Solution -2
Given:
- Radius of the air bubble \( r = 0.1 \, \text{cm} = 0.001 \, \text{m} \)
- Depth below the surface \( h = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Density of the liquid \( \rho = 1000 \, \text{kg/m}^3 \)
- Pressure inside the bubble is 2100 N/m\(^2\) greater than the atmospheric pressure.
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
Step 1: Pressure due to the depth of the liquid
The pressure at a depth \( h \) below the surface is given by the formula: \[ P_{\text{liquid}} = \rho g h. \] Substituting the known values: \[ P_{\text{liquid}} = 1000 \times 10 \times 0.2 = 2000 \, \text{N/m}^2. \]
Step 2: Total pressure inside the bubble
The total pressure inside the bubble is the atmospheric pressure plus the pressure due to the liquid depth, and the additional pressure given in the problem: \[ P_{\text{inside}} = P_{\text{atm}} + P_{\text{liquid}} + 2100 \, \text{N/m}^2. \] Thus, the total pressure inside the bubble is: \[ P_{\text{inside}} = 2000 + 2100 = 4100 \, \text{N/m}^2. \]
Step 3: Using the formula for the pressure difference inside a bubble
The pressure difference between the inside and outside of a spherical bubble is related to the surface tension \( T \) by the formula: \[ \Delta P = \frac{4T}{r}. \] Where \( r \) is the radius of the bubble. Rearranging for the surface tension \( T \): \[ T = \frac{\Delta P \times r}{4}. \] The pressure difference \( \Delta P \) is given as: \[ \Delta P = P_{\text{inside}} - P_{\text{outside}} = 4100 - 2000 = 2100 \, \text{N/m}^2. \] Substituting the known values: \[ T = \frac{2100 \times 0.001}{4} = 0.525 \, \text{N/m}. \]
Final Answer:
The surface tension of the liquid is \( \boxed{0.05} \, \text{N/m} \).
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 ______.