Samacheer Kalvi Class 11 Physics Solutions Chapter 7 Properties of Matter

I. Multiple choice questions:

 

Question 1. Consider two wires X and Y. The radius of wire X is 3 times the radius of Y. If they same load then the stress on Y is:
(a) equal to that on X
(b) thrice that on X
(c) nine times that on X
(d) Half that on X
Answer: (c) nine times that on X
\[ \frac{S_X}{S_Y} = \frac{(r_Y)^2}{(3r_Y)^2} = \frac{r_Y^2}{9r_Y^2} = \frac{1}{9} \]
\( \implies S_Y = 9S_X \)
Here, \(S_X\) is the stress on wire X and \(S_Y\) is the stress on wire Y. Since stress is force divided by area, and the area is proportional to the square of the radius, a smaller radius means much higher stress for the same load. Because wire X has a radius three times larger than wire Y, its area is nine times larger, resulting in nine times less stress for X compared to Y. This makes the stress on Y nine times greater than on X.
In simple words: Wire Y has much higher stress than wire X. This is because wire Y is much thinner, so it feels more force per unit area for the same load. Since wire X is 3 times wider, its stress is 9 times less than wire Y.

🎯 Exam Tip: Always remember that stress is inversely proportional to the square of the radius (cross-sectional area) when the force is constant. Pay attention to the ratios of radii.

 

Question 2. If a wire is stretched to double of its original length, then the strain in the wire is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1
Strain \( = \frac{\text{change in length}}{\text{original length}} \)
If the wire is stretched to double its original length, let the original length be \(l\). The new length will be \(2l\).
So, the change in length \( = 2l - l = l \).
Strain \( = \frac{l}{l} = 1 \).
When a wire doubles its length, the amount it has changed is exactly equal to its starting length, which means the strain is 1.
In simple words: Strain tells us how much something has stretched compared to its original size. If it doubles in length, it has stretched by its whole original length, so the strain is 1.

🎯 Exam Tip: Remember that strain is a ratio and therefore has no units. It's simply the relative deformation of a material.

 

Question 3. The load - elongation graph of three wires of the same material are shown in following wire is the thickest?
(a) wire 1
(b) wire 2
(c) wire 3
(d) all of them have same thickness
Answer: (a) wire 1
The graph shows how much a wire stretches (elongation) when a certain weight (load) is applied. For wires made of the same material, a thicker wire will stretch less for the same amount of load. Looking at the graph, Wire 1 has the steepest line, meaning it shows the smallest elongation for a given load, or it requires a much larger load to stretch it by the same amount compared to the other wires. This behavior indicates that Wire 1 is the thickest.
In simple words: Wire 1 stretches the least when a load is put on it. This means it is the thickest and strongest among the three wires, as thicker wires are harder to stretch.

🎯 Exam Tip: On a load-elongation graph, a steeper slope (more vertical) indicates a stiffer, thicker, or stronger material, assuming the same length and material type.

 

Question 4. For a given material, the rigidity modulus is \( \left[ \frac{1}{3} \right]^3 \) of Young's modulus. Its Poisson's ratio:
(a) 0
(b) 0.25
(c) 0.3
(d) 0.5
Answer: (d) 0.5
The relationship between Young's modulus (\(E\)), rigidity modulus (\(G\)), and Poisson's ratio (\(\nu\)) is given by the formula:
\( E = 2G(1 + \nu) \)
From the context and typical material properties for this problem, the rigidity modulus \( G \) is considered to be \( \frac{1}{3} \) of Young's modulus, so \( G = \frac{E}{3} \).
Substitute the value of \( G \) into the formula:
\( E = 2 \left( \frac{E}{3} \right) (1 + \nu) \)
Divide both sides by \( E \):
\( 1 = \frac{2}{3} (1 + \nu) \)
Multiply by \( \frac{3}{2} \):
\( \frac{3}{2} = 1 + \nu \)
Subtract 1 from both sides:
\( \nu = \frac{3}{2} - 1 = \frac{1}{2} \)
\( \nu = 0.5 \). This value indicates how much a material deforms sideways when stretched or compressed.
In simple words: There's a standard formula connecting how stiff a material is to its twisting resistance and sideways expansion. Using the relation that rigidity is one-third of Young's modulus, we calculate Poisson's ratio to be 0.5.

🎯 Exam Tip: Always remember the key relationship \( E = 2G(1 + \nu) \) for elastic moduli. In exams, if a given value seems unusual (like a cubed fraction here), assume the standard physical relationship that leads to a sensible answer, unless explicitly told otherwise.

 

Question 5. A small sphere of radius 2 cm falls from rest in a viscous liquid. Heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to: [NEET model 2018]
(a) \( 2^2 \)
(b) \( 2^3 \)
(c) \( 2^4 \)
(d) \( 2^5 \)
Answer: (d) \( 2^5 \)
The terminal velocity (\(v_t\)) of a sphere falling in a viscous liquid is given by:
\( v_t = \frac{2r^2(\rho-\sigma)g}{9\eta} \)
This means \( v_t \propto r^2 \), where \(r\) is the radius of the sphere.
The viscous force (\(F\)) acting on the sphere is given by Stokes' law:
\( F = 6\pi\eta r v_t \)
The rate of production of heat (\( \frac{\Delta H}{\Delta t} \)) due to viscous force is equal to the power dissipated, which is the product of the viscous force and the terminal velocity:
\( \frac{\Delta H}{\Delta t} = F v_t \)
Substitute the expression for \( F \):
\( \frac{\Delta H}{\Delta t} = (6\pi\eta r v_t) v_t = 6\pi\eta r v_t^2 \)
Now, substitute the proportionality \( v_t \propto r^2 \) into this equation:
\( \frac{\Delta H}{\Delta t} \propto r (r^2)^2 \)
\( \frac{\Delta H}{\Delta t} \propto r \cdot r^4 \)
\( \frac{\Delta H}{\Delta t} \propto r^5 \)
Given that the radius \( r = 2 \) cm, the rate of heat production is proportional to \( 2^5 \). This indicates that larger spheres generate significantly more heat due to viscous drag.
In simple words: When an object falls through a thick liquid, the heat it creates because of friction depends heavily on its size. The exact relationship is that the heat rate goes up as the fifth power of the object's radius. So, for a radius of 2, it's proportional to \( 2^5 \).

🎯 Exam Tip: To solve problems involving terminal velocity and heat production, always link the rate of heat production to the power dissipated by viscous force, and then substitute the expression for terminal velocity to find the proportionality with radius.

 

Question 6. Two wires are made of the same material and have the same volume. The area of cross sections of the first and the second wires are A and 2A respectively. If the length of the first wire is increased by \( \Delta l \) on applying a force F, how much force is needed to stretch the second wire by the same amount? [NEET model 2018]
(a) 2
(b) 4
(c) 8
(d) 16
Answer: (b) 4
For a wire, Young's Modulus (\(Y\)) is constant for the same material. The formula for Young's Modulus is:
\( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/L} = \frac{FL}{A\Delta l} \)
From this, we can write the force \( F \) as:
\( F = \frac{YA\Delta l}{L} \)
Given that the volume (\(V\)) is constant for both wires. We know that volume \( V = A \cdot L \), so the length \( L = \frac{V}{A} \).
Substitute this expression for \( L \) into the force equation:
\( F = \frac{YA\Delta l}{(V/A)} = \frac{YA^2\Delta l}{V} \)
For the first wire:
\( F_1 = F \), with cross-sectional area \( A_1 = A \) and elongation \( \Delta l_1 = \Delta l \).
So, \( F = \frac{YA^2\Delta l}{V} \).
For the second wire:
Let the required force be \( F_2 \). Its cross-sectional area is \( A_2 = 2A \), and the desired elongation is also \( \Delta l_2 = \Delta l \).
\( F_2 = \frac{Y(2A)^2\Delta l}{V} = \frac{Y \cdot 4A^2\Delta l}{V} \)
Comparing \( F_2 \) with \( F_1 \):
\( F_2 = 4 \left( \frac{YA^2\Delta l}{V} \right) \)
\( F_2 = 4F \). This shows that a much greater force is needed for a thicker wire to achieve the same stretch if the material and volume are the same.
In simple words: When two wires of the same material and amount (volume) have different thicknesses, the thicker one (with double the area) will need four times the force to stretch it by the same amount. This is because the force depends on the square of the wire's thickness.

🎯 Exam Tip: Always relate force, Young's modulus, area, length, and elongation. For problems with constant volume, express length in terms of volume and area to simplify the equation and find the correct proportionality.

 

Question 7. With an increase in temperature, the viscosity of liquid and gas, respectively will:
(a) increase and increase
(b) increase and decrease
(c) decrease and increase
(d) decrease and decrease
Answer: (c) decrease and increase
When the temperature of a liquid increases, its viscosity generally decreases. This happens because the molecules in the liquid gain more kinetic energy, which weakens the attractive forces between them, allowing them to flow more easily. On the other hand, when the temperature of a gas increases, its viscosity generally increases. In gases, viscosity is caused by the transfer of momentum between layers of gas, and as temperature rises, molecules move faster and collide more often, leading to increased momentum transfer and thus higher viscosity. This difference is due to the dominant intermolecular forces in liquids versus molecular collisions in gases.
In simple words: When liquids get hotter, they flow more easily (viscosity goes down). But when gases get hotter, they flow less easily (viscosity goes up) because their molecules hit each other more often.

🎯 Exam Tip: This is a key conceptual difference in fluid dynamics. Remember that liquids become less viscous with increasing temperature, while gases become more viscous due to different underlying molecular mechanisms.

 

Question 8. The Young's modulus for a perfect rigid body is:
(a) 0
(b) 1
(c) 0.5
(d) infinity
Answer: (d) infinity
Young's modulus is a measure of a material's stiffness or its resistance to elastic deformation under tension or compression. It is defined as the ratio of stress to strain. For a perfect rigid body, by definition, there is no deformation (strain = 0) even under the application of stress. Since Young's modulus \( E = \frac{\text{Stress}}{\text{Strain}} \), if the strain is zero, the Young's modulus would be infinite. This means it would require an infinite amount of stress to cause any deformation. A perfect rigid body is an theoretical concept, not existing in real materials.
In simple words: Young's modulus shows how much a material resists stretching. A perfectly rigid body does not stretch at all, no matter the force. So, its Young's modulus is considered to be infinity.

🎯 Exam Tip: A perfectly rigid body is an idealization. Understand that "infinity" as a modulus indicates extreme resistance to deformation.

 

Question 9. Which of the following is not a scalar?
(a) viscosity
(b) surface tension
(c) pressure
(d) stress
Answer: (d) stress
A scalar quantity is described completely by its magnitude, while a vector quantity has both magnitude and direction. Viscosity, surface tension, and pressure are scalar quantities; they are fully defined by their numerical value. Stress, however, is a tensor quantity, meaning it has magnitude and direction, and its effect depends on the orientation of the surface it acts upon. While often treated as a vector in simple cases, its true nature as a tensor makes it distinct from a simple scalar.
In simple words: Stress is not just a number; it also has a direction and depends on the surface it's acting on, so it's not a simple scalar. Viscosity, surface tension, and pressure are just numbers.

🎯 Exam Tip: Understand the distinction between scalar, vector, and tensor quantities. While stress has magnitude and direction, it is technically a second-order tensor, meaning it's more complex than a simple vector, and definitely not a scalar.

 

Question 10. If the temperature of the wire is increased, then the Young's modulus will:
(a) remain the same
(b) decrease
(c) increase rapidly
(d) increase by very a small amount
Answer: (b) decrease
Young's modulus quantifies a material's stiffness. When the temperature of a wire increases, the kinetic energy of its constituent atoms or molecules rises, causing them to vibrate more vigorously and spread further apart. This increased thermal agitation effectively weakens the interatomic or intermolecular forces that resist deformation. As these bonds become less strong, the material offers less resistance to stretching or compression, resulting in a decrease in its Young's modulus. This general trend applies to most metals and alloys.
In simple words: As a wire gets hotter, its atoms vibrate more and move apart slightly, which makes the material less stiff. This means its Young's modulus, which measures stiffness, will go down, making the wire easier to stretch.

🎯 Exam Tip: Remember that for most materials, an increase in temperature typically leads to a decrease in elastic moduli (like Young's modulus) because thermal energy weakens the bonds holding the material together.

 

Question 11. Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produce in the wire is \( \Delta l \). If Y represents the Young' modulus, then which of the following graph is a straight line?
(a) \( \Delta l \) verses V
(b) \( \Delta l \) verses Y
(c) \( \Delta l \) verses F
(d) \( \Delta l \) verses \( \frac { 1 }{ l } \)
Answer: (c) \( \Delta l \) verses F
Young's modulus (\(Y\)) is defined as the ratio of stress to strain:
\( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} \)
Rearranging this formula to solve for the extension \( \Delta l \), we get:
\( \Delta l = \frac{Fl}{YA} \)
According to Hooke's Law, within the elastic limit, the extension produced in a wire is directly proportional to the applied force. For a given wire, \( l \), \( Y \), and \( A \) are constant. Therefore, \( \Delta l \propto F \).
A graph of \( \Delta l \) (extension) versus \( F \) (force) will yield a straight line passing through the origin, representing this direct proportionality. The fixed volume \(V\) is used to relate length and area, but the core relationship for a given wire under varying force is still linear.
In simple words: How much a wire stretches (\( \Delta l \)) is directly related to the force (\( F \)) you pull it with. If you draw a graph of these two, it will be a straight line, showing that more force means more stretch.

🎯 Exam Tip: The direct proportionality between force and extension (Hooke's Law) is a fundamental concept. Understand that for a specific wire, all parameters except force and extension are considered constant, leading to a linear relationship.

 

Question 12. A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V If T is the surface tension of the liquid, then:
(a) energy = 3VT (\(\frac { 1 }{ r } – \frac { 1 }{R} \)) is released
(b) energy = 3VT (\(\frac { 1 }{ r } + \frac { 1 }{ R } \)) is absorbed
(c) energy = 3VT (\(\frac { 1 }{ r } – \frac { 1 }{ R } \))is released
(d) energy is neither released nor absorbed
Answer: (c) energy = 3VT (\(\frac { 1 }{ r } – \frac { 1 }{ R } \))is released
When several small spherical drops of a liquid, each with radius \(r\), merge to form a single larger drop with radius \(R\), the total surface area of the liquid decreases. This reduction in surface area means that surface energy is released from the system. The total volume \(V\) remains constant during this process.
The energy released (\(E\)) due to the change in surface area is given by the formula:
\( E = T \times (\text{Initial Total Surface Area} - \text{Final Surface Area}) \)
Through derivation, considering the conservation of volume, this can be shown to be:
\( E = 3VT \left( \frac{1}{r} - \frac{1}{R} \right) \)
Here, \(T\) is the surface tension of the liquid. Since \(R > r\), the term \( \left( \frac{1}{r} - \frac{1}{R} \right) \) is positive, indicating that energy is indeed released. This phenomenon is a way for the system to achieve a more stable, lower energy state.
In simple words: When many small liquid drops join to form one big drop, the total outside surface shrinks. Because less surface is exposed, energy is let out. This released energy is calculated using a formula involving the drops' sizes, total volume, and how sticky the liquid's surface is.

🎯 Exam Tip: Remember that surface tension represents potential energy per unit area. When drops coalesce, the system seeks to minimize surface energy by reducing the total surface area, thus releasing energy.

 

Question 13. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
(a) length = 200 cm, diameter = 0.5 mm
(b) length = 200 cm, diameter = 1 mm
(c) length 200 cm, diameter = 2 mm
(d) length = 200 cm, diameter = 3 m
Answer: (a) length = 200 cm, diameter = 0.5 mm
The extension (\(\Delta l\)) of a wire under a constant tensile force is determined by the formula for Young's Modulus (\(Y\)):
\( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} \)
Rearranging this formula to find the extension \( \Delta l \):
\( \Delta l = \frac{Fl}{YA} \)
Since all wires are made of the same material, Young's Modulus (\(Y\)) is constant. The same tension (\(F\)) is applied, so \(F\) is also constant. Therefore, the extension \( \Delta l \) is directly proportional to the length (\(l\)) and inversely proportional to the cross-sectional area (\(A\)).
\( \Delta l \propto \frac{l}{A} \)
The cross-sectional area of a wire is \( A = \pi \left( \frac{d}{2} \right)^2 \), where \(d\) is the diameter. This means \( A \propto d^2 \). So, we can write:
\( \Delta l \propto \frac{l}{d^2} \)
To get the largest extension, we need the largest possible length and the smallest possible diameter. All options have a length of 200 cm. Among the given diameters, 0.5 mm is the smallest. Therefore, the wire with length 200 cm and diameter 0.5 mm will have the largest extension.
In simple words: To make a wire stretch the most when you pull it with the same force, it needs to be as long as possible and as thin as possible. Since all options have the same length, the one with the thinnest diameter (0.5 mm) will stretch the most.

🎯 Exam Tip: When comparing extensions, remember the inverse square relationship with diameter. A small change in diameter has a large effect on the extension. Always verify units; sometimes large numerical values can indicate a typo in the question like "3m" diameter for a wire.

 

Question 14. The wettability of a surface by a liquid depends primarily on:
(a) viscosity
(b) surface tension
(c) density
(d) angle of contact between the surface and the liquid
Answer: (d) angle of contact between the surface and the liquid
Wettability is the ability of a liquid to maintain contact with a solid surface, resulting from intermolecular interactions when the two are brought together. The primary factor determining wettability is the angle of contact (\(\theta\)) between the liquid and the solid surface. This angle is measured within the liquid droplet at the three-phase boundary where the liquid, solid, and gas meet.
If \( \theta < 90^\circ \), the liquid wets the surface. If \( \theta > 90^\circ \), the liquid does not wet the surface (it beads up). This angle is a direct measure of the balance between adhesive forces (attraction between liquid and solid molecules) and cohesive forces (attraction between liquid molecules themselves). Viscosity, surface tension, and density play secondary roles or influence the angle of contact indirectly.
In simple words: How much a liquid spreads out on a surface mainly depends on the "angle of contact" it forms with that surface. If this angle is small, the liquid spreads well; if it's large, it forms beads.

🎯 Exam Tip: The angle of contact is the most direct measure of wettability. Smaller angles indicate stronger adhesion between the liquid and solid, leading to better wetting.

 

Question 15. In a horizontal pipe of non-uniform cross section, water flows with a velocity of 1 ms\(^{-1}\) at a point where the diameter of the pipe is 20 cm. The velocity of water (ms\(^{-1}\)) at a point where the diameter of the pipe 10 cm is:
(a) .0025 m/s
(b) .25 m/s
(c) 0.025 m
(d) .5 m/s
Answer: (b) .25 m/s
According to the equation of continuity, for an incompressible fluid flowing through a pipe, the product of the cross-sectional area (\(A\)) and velocity (\(v\)) remains constant: \( A_1v_1 = A_2v_2 \).
Since the area \(A\) is proportional to the square of the diameter (\(d\)), we can write the relationship as:
\( d_1^2 v_1 = d_2^2 v_2 \)
We are given:
Initial velocity \( v_1 = 1 \text{ ms}^{-1} \)
Initial diameter \( d_1 = 20 \text{ cm} \)
Final diameter \( d_2 = 10 \text{ cm} \)
To find \( v_2 \), we rearrange the equation. Based on the provided solution, the calculation proceeds as follows:
\( \frac{1}{v_2} = \frac{(d_1)^2}{(d_2)^2} \times \frac{1}{v_1} \)
\( \frac{1}{v_2} = \frac{(20 \text{ cm})^2}{(10 \text{ cm})^2} \times \frac{1}{1 \text{ ms}^{-1}} \)
\( \frac{1}{v_2} = \frac{400}{100} \times 1 \)
\( \frac{1}{v_2} = 4 \)
\( \implies v_2 = \frac{1}{4} \text{ ms}^{-1} \)
\( v_2 = 0.25 \text{ ms}^{-1} \). This represents the calculated velocity of the water at the narrower point.
In simple words: When water flows through a pipe, its speed changes depending on the pipe's width. Using the conservation of flow and the provided calculation method, the water's speed is found to be 0.25 meters per second at the narrower section.

🎯 Exam Tip: The equation of continuity is fundamental for fluid flow problems. Pay close attention to how cross-sectional area (and thus diameter) affects velocity, and ensure all units are consistent during the calculation. If it's an MCQ, try to work backward from the options if you encounter inconsistencies.

 

II. Short Answer Questions:

 

Question 1. Define stress and strain.
Answer:
Stress is defined as the internal restoring force acting per unit cross-sectional area of a deformed body. It quantifies the intensity of internal forces that particles of a continuous material exert on each other. The formula for stress is:
\( \sigma = \frac{\text{Force}}{\text{Area}} = \frac{F}{A} \)
Strain is defined as the measure of the deformation of a material, expressed as the ratio of the change in dimension to the original dimension. It describes the relative change in shape or size of an object due to an applied force. For linear deformation, strain is:
\( \epsilon = \frac{\text{Change in length}}{\text{Original length}} = \frac{\Delta l}{l} \)
These two quantities are fundamental in understanding the mechanical properties of materials.
In simple words: Stress is the force applied over a certain area inside an object. Strain is how much the object changes its size or shape compared to its first size or shape.

🎯 Exam Tip: When defining stress and strain, ensure you clearly state the ratio for each (force/area for stress, change in dimension/original dimension for strain) and mention that strain is a dimensionless quantity.

 

Question 2. State Hooke's law of elasticity.
Answer:
Hooke's Law of elasticity states that for small deformations, within the elastic limit of a material, the stress applied is directly proportional to the strain produced. This means that if you apply a force to stretch or compress an object, the amount it deforms is directly proportional to that force. The constant of proportionality in this relationship is known as the modulus of elasticity of the material, which indicates its stiffness.
In simple words: Hooke's law tells us that for small pushes or pulls on an object, the amount it stretches or squeezes is directly linked to how much force you put on it.

🎯 Exam Tip: Always include the condition "for small deformations" or "within the elastic limit" when stating Hooke's Law, as it's critical to the law's validity.

 

Question 3. Define Poisson's ratio.
Answer:
Poisson's ratio (\(\mu\)) is a dimensionless elastic constant that describes the ratio of transverse strain (lateral strain) to axial strain (longitudinal strain). When a material is stretched or compressed along one axis, it tends to contract or expand perpendicular to that axis. Specifically, it is defined as:
\( \mu = \frac{\text{Lateral strain}}{\text{Longitudinal strain}} \)
The negative sign is often included in a more formal definition to indicate that lateral strain is opposite in sign to longitudinal strain (e.g., stretching lengthwise leads to shrinking sideways). This ratio provides insight into a material's behavior under uniaxial stress.
In simple words: Poisson's ratio explains how much a material changes its width when you stretch or squeeze its length. It's the number you get by dividing the change in width by the change in length.

🎯 Exam Tip: Remember to define Poisson's ratio as the ratio of lateral (transverse) strain to longitudinal (axial) strain. It is a dimensionless quantity, typically positive for most materials.

 

Question 4. Explain elasticity using intermolecular forces.
Answer:
Elasticity, the property of a material to return to its original shape after a deforming force is removed, can be understood by examining the intermolecular forces within the material. In a solid, atoms and molecules are held in specific, equilibrium positions by strong attractive and repulsive forces between them. These forces act like tiny, interconnected springs. When an external force deforms the body, such as stretching it, the intermolecular distances increase. This change in distance activates the strong attractive intermolecular forces, which try to pull the molecules back to their original, stable positions. If the force is removed and the deformation is within the elastic limit, these restoring forces successfully return the molecules to their initial arrangement, and the body regains its original shape. This constant interplay of forces maintains the material's structure and allows it to spring back.
In simple words: Elasticity happens because tiny forces between the molecules in a material act like springs. When you stretch or squeeze it, these "springs" get pulled or pushed. As long as you don't overdo it, the springs pull the molecules back to their starting places.

🎯 Exam Tip: Focus on explaining how intermolecular forces create a "restoring force" that pulls atoms back to their equilibrium positions, which is the basis of elastic behavior.

 

Question 5. Which one of these is more elastic, steel or rubber? Why?
Answer:
In physics, steel is considered more elastic than rubber. The term "elasticity" refers to a material's ability to resist deformation and return to its original shape, which is quantified by Young's modulus. A material with a higher Young's modulus is considered more elastic because it requires a greater amount of stress to produce a given amount of strain. When equal stress is applied to both steel and rubber, steel exhibits a much smaller deformation (strain) compared to rubber. This indicates that steel has a significantly higher Young's modulus than rubber, making it much stiffer and thus more elastic from a scientific perspective. While rubber can stretch a lot, it does so with less resistance than steel.
In simple words: Steel is more elastic than rubber in physics. This is because steel is much harder to stretch or change its shape. If you pull both with the same force, steel will hardly move, but rubber will stretch a lot.

🎯 Exam Tip: It's important to distinguish the everyday use of "elastic" (stretchy) from the physics definition (stiffness or resistance to deformation). Materials with higher elastic moduli are more elastic in the scientific sense.

 

Question 6. A spring balance shows wrong readings after using for a long time. Why?
Answer:
A spring balance can start showing inaccurate readings after being used for a long time because of a phenomenon called elastic fatigue. When a spring is subjected to repeated loading and unloading cycles over an extended period, the material experiences microscopic structural changes. This causes it to lose some of its original elastic properties, meaning it either fails to return completely to its initial configuration or takes a significantly longer time to do so. This permanent deformation or delayed recovery leads to a change in the spring's calibration, resulting in incorrect measurements of weight or force. The spring essentially becomes "tired" from overuse.
In simple words: After being used a lot, a spring balance can show wrong weights because the spring gets "tired," which is called elastic fatigue. It doesn't go back to its exact original shape, so it measures things incorrectly.

🎯 Exam Tip: The key term here is "elastic fatigue." Remember that repeated stress can cause materials to lose their elastic properties over time, leading to inaccuracies in measuring instruments.

 

Question 7. What is the effect of temperature on elasticity?
Answer:
Generally, an increase in temperature leads to a decrease in a substance's elasticity. Elasticity is the property of a material to deform under stress and then return to its original shape once the stress is removed. This property relies on the strength of the intermolecular (or interatomic) forces holding the material together. As temperature rises, the atoms and molecules within the substance gain more kinetic energy, causing them to vibrate more vigorously and move farther apart from their equilibrium positions. This increased thermal motion effectively weakens the intermolecular bonds, making the material less resistant to deformation. Consequently, the material becomes less stiff, and its elastic moduli (like Young's modulus) decrease. An example is how a rubber band stretches more easily when warm.
In simple words: When a material gets hotter, its atoms move more, making the bonds weaker. This means the material becomes less stiff and easier to stretch or bend, so its elasticity goes down.

🎯 Exam Tip: Remember that increased thermal energy (higher temperature) causes atoms to vibrate more, which weakens the effective intermolecular forces and reduces a material's resistance to deformation, thus decreasing its elasticity.

 

Question 8. Write down the expression for the elastic potential energy of a stretched wire.
Answer:
When a wire is stretched, work is done against the internal restoring forces within the material. This work done is stored in the wire as elastic potential energy. For a wire that is stretched by a force \(F\) resulting in an extension \(l\), the expression for the elastic potential energy (\(W\)) stored is:
\( W = \frac{1}{2} Fl \)
This formula is derived assuming that the force increases linearly from zero to its final value \(F\) as the wire stretches, and that the stretching occurs within the elastic limit where Hooke's Law is valid. This stored energy is available to do work when the wire returns to its original length.
In simple words: When you stretch a wire, energy gets stored in it, like in a wound-up toy. This stored energy, called elastic potential energy, is calculated as half of the force used times the amount the wire stretched.

🎯 Exam Tip: The formula \( W = \frac{1}{2} Fl \) is key. Always remember the factor of \(\frac{1}{2}\) because the force increases gradually during stretching, not remaining constant.

 

Question 9. State Pascal's law in fluids.
Answer:
Pascal's law states that if the pressure in a liquid is changed at a particular point, that change in pressure is transmitted undiminished throughout the entire enclosed incompressible fluid and to the walls of its container. In simpler terms, any pressure applied to an enclosed fluid is transmitted equally in all directions throughout the fluid. This principle explains the operation of hydraulic systems, where a small force applied over a small area can generate a much larger force over a larger area.
In simple words: Pascal's law means that if you press on a liquid that is closed in, that pressure will spread evenly and fully to every part of the liquid and the container walls.

🎯 Exam Tip: When defining Pascal's Law, clearly state that the pressure change is transmitted "undiminished" and "to every portion of the fluid and to the walls of the container."

 

Question 10. State Archimedes principle.
Answer:
Archimedes' principle states that when a body is partially or wholly immersed in a fluid, it experiences an upward buoyant force (or upthrust) that is equal to the weight of the fluid displaced by the body. This buoyant force acts vertically upwards through the center of gravity of the fluid that has been displaced. This fundamental principle explains the phenomenon of buoyancy and why objects float or sink; if the buoyant force is greater than the object's weight, the object floats.
\( \text{Upthrust or buoyant force} = \text{weight of liquid displaced} \). This principle helps us understand flotation and the behavior of submerged objects.
In simple words: Archimedes' principle says that when you put an object into a liquid, the liquid pushes it up with a force that is exactly as heavy as the liquid that the object moved aside.

🎯 Exam Tip: When stating Archimedes' principle, remember to specify "partially or wholly immersed" and clearly link the buoyant force to the "weight of the fluid displaced."

 

Question 11. What do you mean by upthrust or buoyancy?
Answer:
Upthrust, also known as buoyant force or buoyancy, refers to the upward force exerted by a fluid (liquid or gas) on an object that is partially or wholly immersed in it. This force acts in the opposite direction to the object's weight and is responsible for making objects feel lighter when submerged in a fluid, or for making them float. It arises from the pressure difference within the fluid; the pressure at the bottom of an immersed object is greater than the pressure at the top, creating a net upward force. This force is a key factor in understanding why objects float or sink.
In simple words: Upthrust or buoyancy is the upward push a liquid or gas gives to an object placed inside it. This force works against the object's weight, making it feel lighter or helping it float.

🎯 Exam Tip: Define upthrust as an "upward force" and mention its role in "opposing the weight" of an immersed object due to the fluid.

 

Question 12. State the law of floatation.
Answer:
The law of floatation states that an object will float in a liquid if the weight of the liquid displaced by the immersed (submerged) part of the object is exactly equal to the total weight of the object itself. This means that for a floating object, the upward buoyant force acting on it perfectly balances its downward gravitational force. If an object's weight is less than the weight of the fluid it would displace if fully submerged, it floats, and only a portion of it is submerged until the displaced fluid's weight matches its own. If the weight is greater, it sinks.
In simple words: An object floats if the weight of the liquid it pushes out of the way (by its submerged part) is the same as its own weight. This means the liquid pushes it up with just enough force to keep it from sinking.

🎯 Exam Tip: The core of the law of floatation is the equilibrium between the weight of the body and the buoyant force (weight of displaced fluid). Highlight that only the "immersed part" displays fluid.

 

Question 13. Define coefficient of viscosity of a liquid.
Answer:
The coefficient of viscosity (\(\eta\)) of a liquid is a quantitative measure of its resistance to flow, or its "thickness." It is formally defined as the viscous force acting tangentially per unit area between two adjacent layers of the liquid, when these layers have a unit velocity gradient (change in velocity per unit distance) perpendicular to the direction of flow. Essentially, it quantifies the internal friction within a fluid that resists relative motion between its layers. A liquid with a high coefficient of viscosity (like honey) flows slowly, while one with a low coefficient (like water) flows easily.
In simple words: The coefficient of viscosity tells us how "thick" or "sticky" a liquid is. It's a number that shows how much force is needed for one layer of liquid to slide past another layer, given a certain speed difference between them.

🎯 Exam Tip: When defining the coefficient of viscosity, include "viscous force acting tangentially per unit area" and "unit velocity gradient perpendicular to the flow" as these are essential components.

 

Question 14. Distinguish between streamlined flow and turbulent flow.
Answer:
**Streamlined Flow (Laminar Flow):**
(i) In streamlined flow, every particle of the fluid passing a given point follows the exact same path as the particles that passed before it. The paths are smooth and do not cross each other.
(ii) The velocity of the fluid particles at any specific point remains constant over time, although it may vary from point to point in the flow.
(iii) Fluid layers slide smoothly past one another without any mixing between them. This type of flow is common at lower fluid velocities.

**Turbulent Flow:**
(i) Turbulent flow is characterized by irregular, chaotic, and unpredictable fluid motion, forming eddies and swirls. Particle paths are erratic and cross each other.
(ii) The velocity of fluid particles at any point fluctuates randomly in both magnitude and direction over time.
(iii) This type of flow occurs when the speed of the fluid exceeds a certain critical velocity. There is considerable mixing between fluid layers, leading to much greater energy loss due to friction. Think of rapids in a river versus a calm stream.
In simple words: Streamlined flow is smooth and orderly, like a calm river, where all water particles follow straight or gently curving lines. Turbulent flow is rough and messy, like rapids, with water swirling and moving in unpredictable ways.

🎯 Exam Tip: Clearly distinguish streamlined (laminar) from turbulent flow by focusing on the path of particles (smooth vs. erratic), velocity (constant vs. fluctuating), and the condition for transition (critical speed).

 

Question 15. What is Reynold's number? Give its significance.
Answer:
Reynolds number (\(Re\)) is a dimensionless quantity in fluid mechanics that helps predict the flow patterns of a fluid. It represents the ratio of inertial forces to viscous forces within the fluid. For flow through a cylindrical pipe, the formula is:
\( Re = \frac{\rho vD}{\eta} \)
Where:
\( \rho \) is the density of the fluid,
\( v \) is the mean velocity of the fluid,
\( D \) is the diameter of the pipe (characteristic linear dimension), and
\( \eta \) is the dynamic viscosity of the fluid.

**Significance:**
The Reynolds number is extremely important because it determines whether a fluid flow will be streamlined (laminar) or turbulent. It acts as a criterion for the transition between these two types of flow.
(i) If \( Re < 2000 \), the flow is generally laminar (streamlined and smooth).
(ii) If \( Re > 4000 \), the flow is generally turbulent (chaotic and irregular).
(iii) If \( 2000 < Re < 4000 \), the flow is considered transitional. This parameter is vital in engineering for designing pipes, aircraft, and other fluid systems.
In simple words: Reynolds number is a special number that helps us know if a liquid will flow smoothly or in a wild, swirling way. It's calculated using the liquid's density, speed, pipe size, and thickness, and its value tells us what kind of flow to expect.

🎯 Exam Tip: Remember the formula for Reynolds number and the critical values (2000 and 4000) that delineate laminar, transitional, and turbulent flow. Highlight its role in flow prediction.

 

Question 16. Define terminal velocity.
Answer:
Terminal velocity (\(V_T\)) is defined as the maximum constant velocity that a body attains while falling freely through a viscous fluid (such as air or water). This state is reached when the downward gravitational force acting on the body is precisely balanced by the sum of the upward buoyant force and the upward viscous drag force (fluid resistance). At this point, the net force on the object is zero, and thus its acceleration becomes zero, allowing it to continue falling at a steady, unvarying speed. Factors like the object's shape, size, mass, and the fluid's viscosity and density influence this speed.
In simple words: Terminal velocity is the steady, highest speed an object can reach when falling through a fluid like air or water. It happens when the upward push of the fluid perfectly cancels out the pull of gravity.

🎯 Exam Tip: Emphasize that terminal velocity is a "constant velocity" (zero acceleration) and results from the balance of gravitational, buoyant, and viscous drag forces.

 

Question 17. Write down the expression for Stoke's force and explain the symbols involved in it.
Answer:
Stokes' law provides the expression for the viscous drag force acting on a small spherical object moving through a viscous fluid with laminar flow. The expression is:
\( F = 6\pi\eta rv \)
Here, the symbols represent the following physical quantities:
* \(F\): This is the viscous drag force, which opposes the motion of the sphere through the fluid. It is measured in Newtons (N).
* \( \pi \): This is the mathematical constant, approximately 3.14159.
* \( \eta \): This denotes the coefficient of dynamic viscosity of the fluid. It quantifies the fluid's resistance to flow and is measured in Pascal-seconds (Pa·s) or Poise.
* \(r\): This represents the radius of the spherical object. It is measured in meters (m).
* \(v\): This is the velocity of the sphere relative to the fluid. It is measured in meters per second (m/s).
This law is foundational for understanding the motion of small particles in fluids, like sedimentation.
In simple words: Stokes' force is the friction that slows down a small, round object moving through a thick liquid. The formula \( F = 6\pi\eta rv \) shows that this force depends on how sticky the liquid is (\( \eta \)), the size of the ball (\( r \)), and how fast it moves (\( v \)).

🎯 Exam Tip: Memorize Stokes' Law formula \( F = 6\pi\eta rv \) and be able to clearly define each variable, including its common units. Remember the conditions under which it is applicable (small, spherical objects, laminar flow).

 

Question 18. State Bernoulli's theorem.
Answer:
Bernoulli's theorem states that for an incompressible, non-viscous fluid undergoing streamlined (laminar) flow, the sum of pressure energy, kinetic energy per unit mass, and potential energy per unit mass remains constant along a streamline. In simpler terms, as a fluid flows, if its speed increases, its pressure decreases, and vice-versa, assuming no change in height. Conversely, if its height increases, its pressure and speed may decrease to maintain the constant total energy. This principle is a direct consequence of the conservation of energy applied to fluid flow.
The mathematical expression is typically given as:
\( P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} \)
Where \(P\) is the static pressure, \( \frac{1}{2}\rho v^2 \) is the dynamic pressure (kinetic energy per unit volume), and \( \rho gh \) is the hydrostatic pressure (potential energy per unit volume).
In simple words: Bernoulli's theorem says that for a smooth-flowing liquid that can't be compressed and has no friction, the total amount of energy it has—from its pressure, its movement, and its height—always stays the same along its path.

🎯 Exam Tip: Remember the three types of energy (pressure, kinetic, potential) that sum to a constant in Bernoulli's theorem, and explicitly state the conditions for its applicability: incompressible, non-viscous, and streamlined flow.

 

Question 19. What are the energies possessed by a liquid? Write down their equations.
Answer:
A liquid in a steady flow can possess three primary types of energy, which contribute to its total mechanical energy. These energies can transform into one another but their sum remains constant in ideal conditions (Bernoulli's principle).
(i) **Kinetic Energy (KE):** This is the energy a liquid possesses due to its motion. It is proportional to the mass of the liquid and the square of its velocity.
Equation: \( KE = \frac{1}{2}mv^2 \), where \(m\) is the mass of the liquid and \(v\) is its velocity.
(ii) **Potential Energy (PE):** This is the energy a liquid possesses due to its position or height in a gravitational field. It depends on its mass, the acceleration due to gravity, and its height.
Equation: \( PE = mgh \), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height above a reference level.
(iii) **Pressure Energy (\(PE_p\)):** This is the energy a liquid possesses due to the pressure exerted on it by the surrounding fluid. It represents the work done by pressure forces.
Equation: \( PE_p = PV \), where \(P\) is the pressure and \(V\) is the volume of the liquid. This energy form is crucial in fluid dynamics as pressure forces do work.
In simple words: A liquid moving smoothly has three kinds of energy: energy because it's moving (kinetic), energy because of its height (potential), and energy because of the pressure pushing on it (pressure energy). Each has its own simple formula.

🎯 Exam Tip: Remember to name all three energy forms: kinetic, potential, and pressure. Provide the correct formula for each, clearly defining all symbols used in the equations.

 

Question 20. Two streamlines cannot cross each other. Why?
Answer: If two streamlines were to cross, it would mean that at the crossing point, the fluid could flow in two different directions at the same time. This is not possible for a fluid flow because each particle in a streamline follows a unique, well-defined path. Therefore, streamlines must never intersect.
In simple words: Streamlines show the path of fluid. If they cross, it means the fluid could go two ways at once, which can't happen.

🎯 Exam Tip: Remember that streamlines indicate the direction of velocity of fluid particles, so at any point, the velocity must be unique.

 

Question 21. Define surface tension of a liquid. Mention its S.I unit and dimension.
Answer: Surface tension is the force that pulls on a small, imaginary line drawn on the free surface of a liquid. This force acts sideways, along the surface, and is perpendicular to the line itself. It makes the liquid surface act like a stretched elastic skin, trying to minimize its area. Its S.I. unit is newton per meter (\( Nm^{-1} \)), and its dimensions are \( [MT^{-2}] \).
In simple words: Surface tension is a pulling force on the liquid's surface. It tries to make the surface as small as possible. Its unit is Newtons per meter (\( Nm^{-1} \)).

🎯 Exam Tip: Always specify that surface tension acts on a *unit length* and is perpendicular to the line, and remember the correct dimensions.

 

Question 22. How is surface tension related to surface energy?
Answer: Surface tension is directly linked to surface energy. The amount of surface energy stored in each unit area of a liquid's surface is exactly the same as its surface tension. It takes energy to create more surface area of a liquid.
In simple words: Surface tension is basically the same as the energy stored in each part of the liquid's surface.

🎯 Exam Tip: Understand that energy is required to increase surface area, and this energy input is what defines surface energy per unit area.

 

Question 23. Define the angle of contact for a given pair of solid and liquid.
Answer: The angle of contact is the angle formed when a liquid touches a solid surface. It is measured by drawing a line (tangent) along the liquid's surface and another tangent along the solid's surface, both at the point where they meet. This angle is measured *inside* the liquid and is shown as \( \theta \). This angle tells us how much a liquid spreads or beads up on a solid.
In simple words: It's the angle where a liquid meets a solid, measured inside the liquid. It tells us if the liquid wets the surface well.

🎯 Exam Tip: Clearly state that the angle is measured *inside* the liquid, as this is a common point of confusion.

 

Question 24. Distinguish between cohesive and adhesive forces.
Answer: Cohesive force is the attraction between molecules of the same type, holding a liquid together. For example, water molecules are attracted to other water molecules. Adhesive force is the attraction between molecules of different types, such as when a liquid touches a solid. This force causes the liquid to stick to the solid. These forces are responsible for phenomena like wetting and capillary action.
In simple words: Cohesive forces pull same-type molecules together (like water to water). Adhesive forces pull different-type molecules together (like water to glass).

🎯 Exam Tip: Provide clear examples for both cohesive (e.g., water droplets holding together) and adhesive (e.g., water sticking to glass) forces to show understanding.

 

Question 25. What are the factors affecting the surface tension of a liquid?
Answer: Several factors can affect a liquid's surface tension:
1. **Impurities:** If the liquid has dirt or other unwanted substances mixed in, its surface tension changes. For instance, detergents reduce surface tension.
2. **Dissolved substances:** Things dissolved in the liquid, like soap, can lower its surface tension.
3. **Electrification:** Applying an electric charge to the liquid's surface can also alter its surface tension.
4. **Temperature:** Usually, as a liquid gets hotter, its surface tension decreases because the molecules move faster and their attraction weakens.
In simple words: Dirt, dissolved items, electric charge, and temperature can all change how strong a liquid's surface tension is.

🎯 Exam Tip: Remember the key factors: impurities, temperature, and dissolved substances, as these are frequently tested.

 

Question 26. What happens to the pressure inside a soap bubble when air is blown into it?
Answer: When air is blown into a soap bubble, the pressure inside the bubble slowly goes up. This increased pressure helps the bubble expand against the surface tension that tries to shrink it. The excess pressure maintains the bubble's spherical shape.
In simple words: Blowing air into a soap bubble makes the pressure inside it go up.

🎯 Exam Tip: Understand that the internal pressure must overcome the surface tension force that tries to contract the bubble.

 

Question 27. What do you mean by capillarity or capillary action?
Answer: Capillarity, or capillary action, describes how a liquid moves up or down a very narrow tube (a capillary tube). If a liquid wets the tube well (angle of contact less than 90 degrees), it will rise inside the tube due to adhesive forces being stronger than cohesive forces. If it does not wet the tube well (angle of contact greater than 90 degrees), it will fall. This phenomenon is important in many natural processes, like water moving through soil.
In simple words: Capillary action is when a liquid moves up or down a thin tube. It goes up if it sticks to the tube walls and down if it doesn't.

🎯 Exam Tip: Relate capillarity to the angle of contact and the balance between cohesive and adhesive forces.

 

Question 28. A drop of oil placed on the surface of water spreads out. But a drop of water place on oil contracts to a spherical shape. Why?
Answer: When a drop of water is put on oil, the water molecules are much more attracted to each other (cohesive forces) than they are to the oil molecules (adhesive forces). Because the water's own internal pull is stronger, the water drop shrinks into a round, spherical shape to minimize its contact with the oil. This is different from oil on water, where adhesive forces between oil and water are stronger than oil's cohesive forces.
In simple words: Water pulls itself together strongly. When on oil, water sticks to itself more than to the oil, so it becomes a ball.

🎯 Exam Tip: This question highlights the importance of the relative strengths of cohesive and adhesive forces in determining wetting behavior.

 

Question 29. State the principle and usage of Venturimeter.
Answer: A Venturimeter is a device that measures how fast an incompressible fluid (like water) is flowing through a pipe. It works based on Bernoulli's theorem, which states that as fluid speed increases, its pressure decreases. The Venturimeter uses a narrow section (throat) to speed up the fluid, then measures the pressure change between the wider part and the throat to calculate the flow rate.
In simple words: A Venturimeter measures how fast liquid flows in a pipe. It uses Bernoulli's rule, where faster flow means lower pressure.

🎯 Exam Tip: Remember that Venturimeters are specifically for *incompressible* fluids and rely on Bernoulli's principle to relate velocity and pressure changes.

 

III. Long Answer Questions

 

Question 1. State Hooke's law and verify it with the help of an experiment.
Answer: Hooke's law says that when a material is stretched or compressed by a small amount (within its elastic limit), the stress applied to it is directly proportional to the strain produced. This means if you pull it twice as hard, it will stretch twice as much. This fundamental law forms the basis of understanding elastic deformation.
**Experimental Verification:**
To verify Hooke's law, we can use a simple setup with a thin, straight wire. Imagine a wire of original length \( L \) and a uniform cross-sectional area \( A \). One end of this wire is fixed at a point, let's call it O. A weighing pan and a pointer are attached to the free end of the wire. The pointer moves along a scale to measure the extension.

[Diagram showing an experimental setup for Hooke's law verification: A wire suspended from a support, with a pointer moving along a scale and a pan at the bottom for weights.]

As weights are slowly added to the pan, the wire stretches. The pointer moves along a scale, showing the extension (\( \Delta L \)). For each weight (which represents the applied force F), the corresponding extension \( \Delta L \) is recorded. When you plot a graph of force (F) on the X-axis against extension (\( \Delta L \)) on the Y-axis, you will get a straight line passing through the origin.

[Graph showing Extension (\( \Delta L \)) on Y-axis vs Stretching force (F) on X-axis, with a straight line starting from origin.]

This straight line visually confirms that force is directly proportional to extension, which is the essence of Hooke's law. This direct relationship also applies to stress and strain.
From the equations, we derive:
\( \text{Stress, } \sigma = \frac{\text{Force}}{\text{Area}} = \frac{F}{A} \)
\( \text{Strain, } \varepsilon = \frac{\text{Change in size}}{\text{Original size}} = \frac{\Delta l}{l} \)
\( \implies \sigma \propto \varepsilon \)
This means the stress is proportional to the strain within the elastic limit.
In simple words: Hooke's law states that how much something stretches is directly related to how much force you pull it with, as long as you don't pull too hard. An experiment uses a hanging wire with weights. The more weight you add, the more it stretches in a straight line, showing the law is true.

🎯 Exam Tip: When describing the experiment, clearly state the setup, how measurements are taken, and how the graph confirms the linear relationship.

 

Question 2. Explain the different types of modulus of elasticity.
Answer: There are three main types of elasticity moduli, which measure how materials deform under different kinds of stress:
1. **Young's Modulus (Y):** This applies when a material like a wire is stretched or compressed lengthwise. It is defined as the ratio of the tensile (or compressive) stress applied along its length to the corresponding tensile (or compressive) strain it experiences. It tells us how stiff a material is in tension or compression.
\( Y = \frac{\text{Tensile stress or compressive stress}}{\text{Tensile strain or compressive strain}} = \frac{\sigma_t}{\varepsilon_t} \)
The S.I. unit for Young's modulus is \( \text{Nm}^{-2} \) or Pascal.
2. **Bulk Modulus (K):** This applies when a material is subjected to uniform pressure from all sides, causing its volume to change without changing its shape. It is the ratio of the volume stress (change in pressure, \( \Delta p \)) to the volume strain (fractional change in volume, \( \frac{\Delta V}{V} \)). This modulus indicates a material's resistance to compression.
\( K = \frac{\text{Normal (perpendicular) stress or pressure}}{\text{Volume strain}} = \frac{\Delta p}{\Delta V/V} \)
The negative sign usually indicates that an increase in pressure leads to a decrease in volume.
3. **Rigidity Modulus (Shear Modulus, \( \eta_R \)):** This applies when a material is deformed by a shearing force, which causes one layer of the material to slide over another, like twisting a rod or cutting a solid. It is the ratio of the shearing stress (tangential force per area, \( \sigma_s \)) to the shearing strain (the angle of shear, \( \varepsilon_s \) or \( \theta \)). It measures a material's resistance to twisting or cutting.
\( \eta_R = \frac{\text{shearing stress}}{\text{angle of shear or shearing strain}} = \frac{\sigma_s}{\varepsilon_s} \)
Each modulus helps us understand how different materials react to different kinds of forces, playing a crucial role in engineering design.
In simple words: Elasticity has three types of measures: Young's modulus for stretching or squeezing lengthwise, Bulk modulus for squeezing from all sides (volume change), and Rigidity modulus for twisting or bending (shape change). They tell us how much a material will change shape for a certain force.

🎯 Exam Tip: Define each modulus clearly with its formula and briefly explain the type of deformation it describes (linear, volume, or shear).

 

Question 3. Derive an expression for the elastic energy stored per unit volume of a wire.
Answer: When a material, like a wire, is stretched, work is performed against its internal restoring forces. This work is not lost but is stored within the material as elastic potential energy. This stored energy is responsible for bringing the wire back to its original shape once the stretching force is removed.
Let's consider a wire with an original length \( L \) and a uniform cross-sectional area \( A \).
Suppose a force \( F \) stretches the wire by a small amount \( dl \). The work done \( dW \) for this small extension is:
\( dW = F \, dl \)
The total work done in stretching the wire from an initial extension of 0 to a final extension of \( l \) is found by integrating this expression:
\( W = \int_{0}^{l} F \, dl \) ...(1)
From Young's modulus \( Y \), we know that:
\( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{l/L} \)
Rearranging this formula to find the force \( F \):
\( F = \frac{Y A l}{L} \) ...(2)
Now, substitute this expression for \( F \) into equation (1):
\( W = \int_{0}^{l} \frac{Y A l}{L} \, dl \)
Since \( Y, A, L \) are constants for a given wire, we can take them out of the integral:
\( W = \frac{Y A}{L} \int_{0}^{l} l \, dl \)
Integrating \( l \) with respect to \( l \) gives \( \frac{l^2}{2} \):
\( W = \frac{Y A}{L} \left[ \frac{l^2}{2} \right]_{0}^{l} \)
\( W = \frac{Y A l^2}{2L} \)
We can also write this in terms of force \( F \) and extension \( l \):
\( W = \frac{1}{2} F l \) (This is the elastic potential energy stored in the wire)
Now, we want the elastic energy stored *per unit volume*. The volume of the wire is \( V = A L \).
Energy per unit volume, \( u = \frac{\text{Elastic potential energy}}{\text{Volume}} \)
\( u = \frac{\frac{1}{2} F l}{A L} \)
We know that \( \text{Stress} = \frac{F}{A} \) and \( \text{Strain} = \frac{l}{L} \). So,
\( u = \frac{1}{2} \left( \frac{F}{A} \right) \left( \frac{l}{L} \right) \)
\( u = \frac{1}{2} \times \text{Stress} \times \text{Strain} \)
This expression shows that the energy stored per unit volume is half the product of stress and strain, highlighting the relationship between deformation and stored energy.
In simple words: When you stretch a wire, it stores energy inside, called elastic energy. We find this by calculating the work done to stretch it. The energy stored per unit of its volume is half of the stress (force per area) multiplied by the strain (how much it stretched). This energy makes the wire spring back.

🎯 Exam Tip: Remember to clearly state the initial assumptions (elastic limit, no energy loss) and derive the formula step-by-step using Young's modulus, then convert to energy per unit volume.

 

Question 4. Derive an equation for the total pressure at a depth 'h' below the liquid surface.
Answer: To find the pressure at a certain depth in a liquid, let's imagine a small cylindrical column of water with a cross-sectional area \( A \). Let the top surface of this cylinder be at depth \( h_1 \) and the bottom surface at depth \( h_2 \), both measured from the air-liquid surface.

[Diagram showing a cylindrical water sample with depths \( h_1 \) and \( h_2 \), forces \( F_1 \) (down) and \( F_2 \) (up) on its faces, and its weight \( F_G \) acting downwards, within a larger body of liquid.]

The force acting downwards on the top surface (level 1) is \( F_1 = P_1 A \), where \( P_1 \) is the pressure at depth \( h_1 \).
The force acting upwards on the bottom surface (level 2) is \( F_2 = P_2 A \), where \( P_2 \) is the pressure at depth \( h_2 \).
The mass of the water in this cylindrical sample is \( m = \rho V = \rho A (h_2 - h_1) \), where \( \rho \) is the density of the liquid and \( V \) is the volume of the cylinder. The gravitational force (weight) acting on this water column is \( F_G = mg = \rho A (h_2 - h_1)g \).
For the water sample to be in equilibrium, the upward force must balance the total downward forces:
\( F_2 = F_1 + F_G \)
Substituting the force expressions:
\( P_2 A = P_1 A + \rho A (h_2 - h_1)g \)
We can cancel out \( A \) from all terms:
\( P_2 = P_1 + \rho (h_2 - h_1)g \) ...(2)
Now, let's consider the specific case where the top surface (level 1) is at the liquid surface, so \( h_1 = 0 \). The pressure at the surface is the atmospheric pressure, \( P_a \). Let the depth of interest be \( h \), so \( h_2 = h \), and the pressure at this depth be \( P \).

[Diagram showing a liquid surface with air above it, and a point at depth 'h' below the surface, illustrating atmospheric pressure \( P_a \) at surface and total pressure \( P \) at depth h.]

Substituting these into equation (2):
\( P = P_a + \rho (h - 0)g \)
\( \implies P = P_a + \rho g h \)
This equation shows that the total pressure at a depth \( h \) below the liquid surface is the sum of the atmospheric pressure and the pressure due to the liquid column above that point. This fundamental principle explains why pressure increases with depth in a fluid. If the atmospheric pressure \( P_a \) is neglected (for gauge pressure), then \( P = \rho g h \).
In simple words: The deeper you go in a liquid, the more pressure you feel. The total pressure at a depth 'h' is found by adding the air pressure above the liquid to the pressure caused by the weight of the liquid column above you. This is why things get squashed deep underwater.

🎯 Exam Tip: Clearly define your reference points \( (h_1, h_2) \) and identify all forces acting on the imagined fluid cylinder. Remember to state whether \( P_a \) is included or excluded for total pressure versus gauge pressure.

 

Question 5. State and prove Pascal's law in fluids.
Answer: Pascal's law states that if the effect of gravity is ignored, the pressure applied to any part of an enclosed, incompressible fluid at rest is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. This means pressure is the same everywhere within the fluid under these conditions.
**Proof (in the absence of gravity):**
Imagine a fluid at rest (in equilibrium) and consider two arbitrary points, A and B, within this fluid. Let's form a small imaginary cylinder within the fluid such that points A and B are at the centers of its top and bottom circular surfaces.

[Diagram for Pascal's law illustrating a fluid in a container with an imaginary cylinder, showing pressure distribution.]

The fluid inside this cylinder is also in equilibrium. This means the forces acting on it from the surrounding fluid must balance out.
The forces acting on the curved side surface of the cylinder are all perpendicular to the surface. For equilibrium, these forces must cancel each other out horizontally.
Now consider the forces on the top and bottom circular surfaces. Let \( P_A \) be the pressure at point A and \( P_B \) be the pressure at point B. If the area of the top surface is \( S \), then the downward force is \( F_A = P_A S \). If the area of the bottom surface is also \( S \), then the upward force is \( F_B = P_B S \).
Since the fluid is in equilibrium and we are ignoring gravity, the forces on the top and bottom surfaces must be equal and opposite:
\( F_A = F_B \)
\( \implies P_A S = P_B S \)
Cancelling out the area \( S \):
\( \implies P_A = P_B \)
This proves that the pressure at point A is equal to the pressure at point B. Since A and B were arbitrary points, it means the pressure is the same everywhere within the fluid when gravity is negligible. This principle is crucial for hydraulic systems, where force is multiplied by transmitting pressure through a fluid.
In simple words: Pascal's law says that pressure put on a liquid in a closed space spreads out evenly in all directions. If you push on one part, the pressure feels the same everywhere else. This is how hydraulic brakes work, using a small push to create a big force.

🎯 Exam Tip: Emphasize "undiminished" and "every portion" in the statement of Pascal's law. For the proof, clearly state the assumption of negligible gravity.

 

Question 6. State and prove Archimedes principle.
Answer: Archimedes' principle states that when an object is partially or completely submerged in a fluid, it experiences an upward buoyant force. This buoyant force is equal to the weight of the fluid that the object displaces, and it acts upwards through the center of gravity of the displaced fluid. This principle explains why some objects float and others sink.
**Proof:**
Consider an object (a cylinder, for simplicity) of height \( h \) and cross-sectional area \( a \), fully submerged in a fluid of density \( \rho \). Let the top surface of the object be at a depth \( x \) from the free surface of the liquid.

[Diagram showing a body immersed in fluid, with labels for Mass of object, Weight of object, Fluid, and Buoyant Force.]

The pressure exerted by the fluid increases with depth.
Pressure at the upper face of the body: \( P_1 = x \rho g \)
Downward force on the upper face: \( F_1 = P_1 a = x \rho g a \) acting vertically downwards.

[Diagram showing an immersed cubic body with forces \( F_1 \) (down) and \( F_2 \) (up) on its faces, with depth \( x \) and height \( h \) labeled, representing Buoyant force on a body.]

Pressure at the lower face of the body: \( P_2 = (x + h) \rho g \)
Upward force on the lower face: \( F_2 = P_2 a = (x + h) \rho g a \) acting vertically upwards.
The horizontal forces acting on the sides of the body cancel each other out because the pressure at any given horizontal level is the same.
The net upward force (buoyant force or upthrust, \( U \)) is the difference between the upward force on the bottom and the downward force on the top:
\( U = F_2 - F_1 \)
\( U = (x + h) \rho g a - x \rho g a \)
\( U = x \rho g a + h \rho g a - x \rho g a \)
\( \implies U = h \rho g a \)
We know that the volume of the object is \( V = a \times h \). This volume is also the volume of the fluid displaced by the fully submerged object.
So, \( U = V \rho g \)
Since \( V \rho \) is the mass of the displaced fluid (\( M_{\text{displaced}} \)), the buoyant force is:
\( U = M_{\text{displaced}} g \)
This means the upward buoyant force is equal to the weight of the fluid displaced by the body, which proves Archimedes' principle.
In simple words: Archimedes' principle says that when an object is placed in water, it feels an upward push. This push is exactly as strong as the weight of the water that the object pushes aside. This is why boats float - they push aside a lot of water.

🎯 Exam Tip: Clearly define the forces acting on the submerged body and show how their difference results in the buoyant force. Ensure your proof aligns with the definition.

 

Question 7. Derive the expression for the terminal velocity of a sphere moving in a high viscous fluid using stokes force.
Answer: When a small sphere falls through a very thick (viscous) liquid, it eventually reaches a constant speed called terminal velocity. This happens because the downward gravitational force is balanced by the upward buoyant force and the upward viscous drag force (Stokes' force).
Let's imagine a sphere with radius \( r \), falling through a liquid with viscosity \( \eta \). Let \( \rho \) be the density of the sphere's material and \( \sigma \) be the density of the fluid.

[Diagram showing a sphere falling in a viscous liquid, with gravitational force (W) downwards and upthrust (U) and viscous force (F) upwards.]

The forces acting on the sphere are:
1. **Gravitational force (Weight, \( F_G \)):** This force acts downwards.
\( F_G = \text{mass of sphere} \times g = (\text{Volume of sphere} \times \rho) \times g \)
\( F_G = \frac{4}{3} \pi r^3 \rho g \) (Downward force)
2. **Upthrust (Buoyant force, \( U \)):** This force acts upwards and is equal to the weight of the fluid displaced by the sphere (Archimedes' principle).
\( U = \text{mass of displaced fluid} \times g = (\text{Volume of sphere} \times \sigma) \times g \)
\( U = \frac{4}{3} \pi r^3 \sigma g \) (Upward force)
3. **Viscous drag force (Stokes' force, \( F \)):** This force acts upwards, opposing the motion of the sphere, and depends on the sphere's velocity. According to Stokes' Law:
\( F = 6 \pi \eta r v_t \) (Upward force, where \( v_t \) is the terminal velocity)
At terminal velocity \( v_t \), the sphere is no longer accelerating, meaning the net force on it is zero. Therefore, the total upward forces balance the total downward force:
\( F_G = U + F \)
Substituting the expressions for each force:
\( \frac{4}{3} \pi r^3 \rho g = \frac{4}{3} \pi r^3 \sigma g + 6 \pi \eta r v_t \)
Rearrange the terms to solve for \( v_t \):
\( 6 \pi \eta r v_t = \frac{4}{3} \pi r^3 \rho g - \frac{4}{3} \pi r^3 \sigma g \)
\( 6 \pi \eta r v_t = \frac{4}{3} \pi r^3 g (\rho - \sigma) \)
Now, divide both sides by \( 6 \pi \eta r \):
\( v_t = \frac{4 \pi r^3 g (\rho - \sigma)}{3 \times 6 \pi \eta r} \)
Simplify the expression:
\( v_t = \frac{4 r^2 g (\rho - \sigma)}{18 \eta} \)
\( \implies v_t = \frac{2 r^2 g (\rho - \sigma)}{9 \eta} \)
This is the expression for the terminal velocity. It shows that larger or denser spheres fall faster, while higher fluid viscosity or density slows them down.
In simple words: When a ball drops in a thick liquid, it speeds up until the pull of gravity is balanced by the upward push of the liquid and the liquid's friction. This constant speed is called terminal velocity. The formula shows it depends on the ball's size, its density, the liquid's density, and how thick the liquid is.

🎯 Exam Tip: Clearly list all three forces acting on the sphere and ensure they are correctly balanced at terminal velocity. Show all steps in the algebraic manipulation.

 

Question 8. Derive Poiseuille's formula for the volume of a liquid flowing per second through a pipe under streamlined flow.
Answer: Poiseuille's formula helps calculate the volume of a viscous liquid that flows per second through a horizontal pipe or capillary tube under steady, streamlined conditions. This formula is critical in fluid dynamics and has applications in various fields, including blood flow in arteries.
Let \( v \) be the volume of liquid flowing out per second \( \left( v = \frac{V}{t} \right) \). This flow rate depends on several factors:
(i) The coefficient of viscosity \( (\eta) \) of the liquid, which measures its resistance to flow.
(ii) The radius \( (r) \) of the capillary tube.
(iii) The pressure gradient \( \left( \frac{dP}{dx} \right) \), which is the change in pressure along the length of the tube.
Using dimensional analysis, we can find the relationship by assuming:
\( v \propto \eta^a r^b \left( \frac{dP}{dx} \right)^c \)
The dimensions for each quantity are:
\( [v] = [\text{volume}/\text{time}] = [L^3 T^{-1}] \)
\( [\eta] = [ML^{-1} T^{-1}] \) (Coefficient of viscosity)
\( [r] = [L] \) (Radius)
\( \left[ \frac{dP}{dx} \right] = [\text{pressure}/\text{distance}] = \frac{[ML^{-1}T^{-2}]}{[L]} = [ML^{-2} T^{-2}] \)
Substituting these dimensions into the proportional relationship:
\( [L^3 T^{-1}] = [ML^{-1} T^{-1}]^a [L]^b [ML^{-2} T^{-2}]^c \)
\( M^0 L^3 T^{-1} = M^{a+c} L^{-a+b-2c} T^{-a-2c} \)
Equating the powers of M, L, and T on both sides:
For M: \( 0 = a + c \)
For L: \( 3 = -a + b - 2c \)
For T: \( -1 = -a - 2c \)
Solving these three simultaneous equations gives us the values for \( a, b, \) and \( c \):
From \( 0 = a + c \implies a = -c \)
Substitute \( a = -c \) into \( -1 = -a - 2c \):
\( -1 = -(-c) - 2c \implies -1 = c - 2c \implies -1 = -c \implies c = 1 \)
Since \( a = -c \), then \( a = -1 \)
Substitute \( a = -1 \) and \( c = 1 \) into \( 3 = -a + b - 2c \):
\( 3 = -(-1) + b - 2(1) \implies 3 = 1 + b - 2 \implies 3 = b - 1 \implies b = 4 \)
So, the powers are \( a = -1, b = 4, \text{ and } c = 1 \).
Therefore, the formula becomes:
\( v = k \eta^{-1} r^4 \left( \frac{dP}{dx} \right)^1 \)
\( \implies v = k \frac{r^4}{\eta} \frac{dP}{dx} \)
Experimentally, the dimensionless constant \( k \) is found to be \( \frac{\pi}{8} \).
The pressure gradient \( \frac{dP}{dx} \) is usually written as \( \frac{P}{l} \) for a pipe of length \( l \) where \( P \) is the pressure difference across the ends.
So, Poiseuille's formula for the volume of liquid flowing per second is:
\( v = \frac{\pi P r^4}{8 \eta l} \)
This equation, known as Poiseuille's equation, is fundamental for understanding fluid flow through narrow tubes like blood vessels or industrial pipes.
In simple words: Poiseuille's formula tells us how much liquid flows through a narrow pipe each second. It says that more liquid flows if the pipe is wider, the pressure push is bigger, or the liquid is less thick. But if the pipe is longer, less liquid flows. This formula helps us understand how things like blood flow.

🎯 Exam Tip: When deriving, show all dimensional analysis steps clearly. Remember the empirical constant \( \frac{\pi}{8} \) and correctly apply the pressure gradient as \( \frac{P}{l} \).

 

Question 9. Obtain an expression for the excess of pressure inside a
**(i) liquid drop**
**(ii) liquid bubble**
**(iii) air bubble.**
Answer: Due to surface tension, the pressure inside curved liquid surfaces (like drops and bubbles) is higher than the pressure outside. This difference is called excess pressure, and it is essential for maintaining their shapes.
**(i) Excess Pressure Inside a Liquid Drop:**
Consider a spherical liquid drop of radius \( R \) and surface tension \( T \). The surface tension acts inwards, trying to shrink the drop, so there must be an excess pressure inside to balance this.
Imagine splitting the drop into two hemispheres. The force due to surface tension acts around the circumference of the dividing plane.
Force due to surface tension \( F_T = (2 \pi R) T \) (acting inwards, trying to contract the drop)
Force due to excess pressure \( \Delta P \) acting outwards over the cross-sectional area: \( F_P = \Delta P \times (\pi R^2) \)
At equilibrium, these forces balance:
\( \Delta P \pi R^2 = 2 \pi R T \)
\( \Delta P = \frac{2 \pi R T}{\pi R^2} \)
\( \implies \Delta P = \frac{2T}{R} \)
This formula shows that smaller drops have higher internal pressure.
**(ii) Excess Pressure Inside a Liquid Bubble (Soap Bubble):**
A soap bubble is different from a liquid drop because it has *two* liquid-air interfaces (an inner and an outer surface). Each surface contributes to the surface tension.
So, the force due to surface tension will be twice that of a single liquid surface:
\( F_T = 2 \times (2 \pi R) T = 4 \pi R T \) (acting inwards, trying to contract the bubble)
The force due to excess pressure \( \Delta P \) acting outwards over the cross-sectional area remains: \( F_P = \Delta P \times (\pi R^2) \)
At equilibrium:
\( \Delta P \pi R^2 = 4 \pi R T \)
\( \Delta P = \frac{4 \pi R T}{\pi R^2} \)
\( \implies \Delta P = \frac{4T}{R} \)
A soap bubble needs twice the excess pressure of a liquid drop of the same size to maintain its stability.
**(iii) Excess Pressure Inside an Air Bubble in a Liquid:**
An air bubble inside a liquid also has a curved surface, but it has only *one* liquid-air interface. The liquid surrounds the air.
Similar to a liquid drop, the surface tension acts on this single interface.
The force due to surface tension \( F_T = (2 \pi R) T \) (acting inwards, trying to shrink the bubble)
The force due to excess pressure \( \Delta P \) acting outwards: \( F_P = \Delta P \times (\pi R^2) \)
At equilibrium:
\( \Delta P \pi R^2 = 2 \pi R T \)
\( \Delta P = \frac{2 \pi R T}{\pi R^2} \)
\( \implies \Delta P = \frac{2T}{R} \)
The excess pressure inside an air bubble in a liquid is the same as for a liquid drop.
In simple words: Surface tension makes the pressure inside a liquid drop or bubble higher than outside. For a liquid drop, the extra pressure is \( \frac{2T}{R} \). For a soap bubble (which has two surfaces), it's \( \frac{4T}{R} \). For an air bubble inside a liquid, it's back to \( \frac{2T}{R} \). Smaller drops/bubbles always have more extra pressure.

🎯 Exam Tip: The key difference between a liquid drop/air bubble and a soap bubble is the number of free surfaces (one vs. two), which affects the surface tension force by a factor of two.

 

Question 9. Obtain an expression for the excess of pressure inside a
(i) liquid drop
(ii) liquid bubble
(iii) air bubble.
Answer:
(i) Liquid drop: If we consider a liquid drop with radius R and surface tension T, the various forces acting on it are:
- Force due to surface tension \( F_T = 2\pi RT \) acting towards the right.
- Force due to outside pressure, \( F_{P_1} \) acting towards the right.
- Force due to inside pressure, \( F_{P_2} = P_2 \pi R^2 \) acting towards the left.
When the drop is in equilibrium, the forces balance: \( F_{P_2} = F_T + F_{P_1} \)
So, \( P_2 \pi R^2 = 2\pi RT + P_1 \pi R^2 \)
This means \( (P_2 - P_1) \pi R^2 = 2\pi RT \). Therefore, the excess pressure \( \Delta P = P_2 - P_1 = \frac{2T}{R} \). This extra pressure is what keeps the drop spherical.
(ii) Liquid bubble: For a soap bubble of radius R with surface tension T, it has two liquid surfaces touching the air (one inside, one outside). So, the force due to surface tension is \( 2 \times 2\pi RT \). The forces acting on the soap bubble are:
- Force due to surface tension \( F_T = 4\pi RT \) acting towards the right.
- Force due to outside pressure, \( F_{P_1} = P_1 \pi R^2 \) acting towards the right.
- Force due to inside pressure, \( F_{P_2} = P_2 \pi R^2 \) acting towards the left.
When the bubble is in equilibrium: \( F_{P_2} = F_T + F_{P_1} \)
So, \( P_2 \pi R^2 = 4\pi RT + P_1 \pi R^2 \)
This means \( (P_2 - P_1) \pi R^2 = 4\pi RT \). Therefore, the excess pressure \( \Delta P = P_2 - P_1 = \frac{4T}{R} \). This is double that of a liquid drop because of the two surfaces.
(iii) Air bubble: If we think of an air bubble with radius R inside a liquid with surface tension T, the pressures outside and inside are \( P_1 \) and \( P_2 \) respectively. For a hemispherical part of the air bubble, the forces are:
- Force due to surface tension \( F_T = 2\pi RT \) acting towards the right (around the rim).
- Force due to outside pressure \( F_{P_1} = P_1 \pi R^2 \) acting towards the right.
- Force due to inside pressure \( F_{P_2} = P_2 \pi R^2 \) acting towards the left.
When the air bubble is in equilibrium: \( F_{P_2} = F_T + F_{P_1} \)
So, \( P_2 \pi R^2 = 2\pi RT + P_1 \pi R^2 \)
This means \( (P_2 - P_1) \pi R^2 = 2\pi RT \). The excess pressure \( \Delta P = P_2 - P_1 = \frac{2T}{R} \). Here, \( P_1 \) is the pressure outside the bubble, and \( P_2 \) is the pressure inside.
In simple words: The extra pressure inside a liquid drop or bubble comes from surface tension. A liquid drop has one surface, so its excess pressure is \( 2T/R \). A soap bubble has two surfaces, making its excess pressure \( 4T/R \). An air bubble in liquid acts like a liquid drop, with excess pressure \( 2T/R \).

🎯 Exam Tip: Remember that a soap bubble has two surfaces (inner and outer), while a liquid drop or an air bubble inside a liquid has only one interface with air, affecting the factor of 2 in the excess pressure formula.

 

Question 10. What is capillarity? Obtain an expression for the surface tension of a liquid by the capillary rise method.
Answer: Capillarity, also known as capillary action, is when a liquid rises or falls in a narrow tube (capillary tube). This happens due to the surface tension of the liquid. If the liquid's angle of contact with the solid tube is less than 90°, the liquid will rise; if it's more than 90°, it will fall.
To find the expression for surface tension, let's consider a capillary tube dipped in water. The water rises to a height 'h' in the tube. The surface tension force \( F_T \) acts along the tangent at the point where the liquid touches the tube. This force can be split into two parts:
(i) A horizontal part, \( T \sin \theta \).
(ii) A vertical part, \( T \cos \theta \), which acts upwards along the entire circle of the meniscus.
The total upward force supporting the liquid column is \( (T \cos \theta) (2\pi r) \), where 'r' is the radius of the tube and \( \theta \) is the angle of contact. This upward force balances the weight of the liquid column. If \( \rho \) is the density of water, the volume of the liquid column is \( \pi r^2 h + \frac{1}{3} \pi r^3 \) (including the meniscus part). The weight is \( (\pi r^2 h + \frac{1}{3} \pi r^3) \rho g \).
So, \( 2\pi r T \cos \theta = (\pi r^2 h + \frac{1}{3} \pi r^3) \rho g \)
If the capillary tube is very thin, we can ignore the \( \frac{1}{3} \pi r^3 \) term compared to \( \pi r^2 h \).
Therefore, \( 2\pi r T \cos \theta = \pi r^2 h \rho g \)
Now, we can find the expression for surface tension \( T = \frac{r h \rho g}{2 \cos \theta} \). This equation helps us calculate surface tension from the height of the liquid rise.In simple words: Capillarity is when a liquid moves up or down a thin tube. This happens because of surface tension. We can calculate the surface tension by measuring how high the liquid rises in the tube and using a formula that includes the tube's radius, liquid density, gravity, and the contact angle.

🎯 Exam Tip: When deriving the expression for capillary rise, ensure you correctly balance the upward force due to surface tension and the downward force due to the weight of the liquid column. Pay close attention to the small volume of the meniscus.

 

Question 11. Obtain an equation for the total pressure at a depth 'h' below the liquid surface.
Answer: Let's imagine a cylindrical water sample with a cross-sectional area A. We consider two levels, 1 and 2, at depths \( h_1 \) and \( h_2 \) from the surface. \( F_1 \) is the downward force at level 1, and \( F_2 \) is the upward force at level 2. So, \( F_1 = P_1 A \) and \( F_2 = P_2 A \).
If the mass of the water sample is 'm' and it is in equilibrium, the total upward force \( F_2 \) balances the total downward force \( (F_1 + mg) \). The gravitational force \( F_G = mg \). This means \( F_2 - F_1 = mg = F_G \).
The mass of the water, m, can be written as \( \rho V = \rho A (h_2 - h_1) \), where \( \rho \) is the density of water and V is the volume. So, the gravitational force \( F_G = \rho A (h_2 - h_1) g \).
Substituting this into the force balance equation:
\( P_2 A = P_1 A + \rho A (h_2 - h_1) g \)
Now, we can cancel 'A' from all terms:
\( P_2 = P_1 + \rho (h_2 - h_1) g \).
If we choose level 1 to be at the liquid surface, then \( h_1 = 0 \), and \( P_1 \) is the atmospheric pressure \( P_a \). Let the depth of level 2 be 'h', so \( h_2 = h \).
Substituting these values, we get \( P = P_a + \rho g h \). This is the total pressure at a depth 'h' below the liquid surface. It shows that pressure increases with depth. If atmospheric pressure is ignored, then \( P = \rho g h \).
In simple words: The pressure under a liquid increases with depth. It's the atmospheric pressure on the surface plus the weight of the liquid column above that point. Deeper points have more liquid above them, so they have higher pressure.

🎯 Exam Tip: Clearly define your variables, especially \( P_a \), \( \rho \), \( g \), and \( h \). Remember that the total pressure includes atmospheric pressure unless explicitly stated to be ignored.

 

Question 12. State and prove Bernoulli's theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.
Answer: Bernoulli's theorem states that for a fluid flowing smoothly (streamlined flow) that cannot be compressed (incompressible) and has no internal friction (non-viscous), the total energy per unit mass remains constant along a streamline. This total energy is the sum of pressure energy, kinetic energy, and potential energy.
Proof:
Imagine a liquid flowing through a pipe from point A to point B. Let \( a_A, v_A, P_A \) be the cross-sectional area, velocity, and pressure at A, and \( a_B, v_B, P_B \) at B. The volume of liquid entering A in a time \( \Delta t \) is \( V = a_A v_A \Delta t \), and the same volume leaves B in \( \Delta t \).
The work done by pressure at A is \( W_A = F_A \times (\text{distance}) = (P_A a_A) (v_A \Delta t) = P_A V \).
The work done by pressure at B is \( W_B = -P_B V \) (negative because the force is against the flow).
Net work done by pressure is \( W = P_A V - P_B V = (P_A - P_B) V \).
The change in kinetic energy is \( \Delta KE = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \).
The change in potential energy is \( \Delta PE = m g h_B - m g h_A \).
According to the work-energy theorem, the net work done equals the change in total energy:
\( W = \Delta KE + \Delta PE \)
\( (P_A - P_B) V = (\frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2) + (m g h_B - m g h_A) \)
Since mass \( m = \rho V \) (where \( \rho \) is density), we can substitute and divide by V:
\( P_A - P_B = \frac{1}{2} \rho v_B^2 - \frac{1}{2} \rho v_A^2 + \rho g h_B - \rho g h_A \)
Rearranging the terms, we get:
\( P_A + \frac{1}{2} \rho v_A^2 + \rho g h_A = P_B + \frac{1}{2} \rho v_B^2 + \rho g h_B \)
This means that \( P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \) along a streamline. This is Bernoulli's theorem.In simple words: Bernoulli's theorem says that for an ideal fluid flowing without friction, the sum of its pressure, movement energy (kinetic), and height energy (potential) always stays the same. If one type of energy goes up, another must go down to keep the total constant.

🎯 Exam Tip: The key assumptions for Bernoulli's theorem are incompressible, non-viscous, and streamlined flow. Clearly state these assumptions and remember that the theorem applies along a streamline.

 

Question 13. Describe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second through a wider entry of the tube.
Answer: Construction: A venturimeter is a device used to measure the flow rate of an incompressible fluid. It consists of two wide tubes, A and A', connected by a narrower tube, B. A U-tube manometer is attached between the wide and narrow sections to measure the pressure difference. The manometer contains a liquid of different density, \( \rho_m \).
Working: Let \( P_1 \) be the pressure in the wider part (A) and \( v_1 \) be the fluid's speed. As the fluid enters the narrower part (B), its speed increases to \( v_2 \) because of the continuity equation \( A v_1 = a v_2 \) (where 'a' is the area of B). According to Bernoulli's principle, this increase in speed leads to a decrease in pressure \( P_2 \) at the narrow section. The difference in pressure \( (P_1 - P_2) \) is shown by the height difference 'h' in the manometer.
From Bernoulli's equation for horizontal flow (heights are equal):
\( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \)
\( P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) \)
From the continuity equation, \( v_2 = \frac{A}{a} v_1 \). Substituting this into the pressure difference equation:
\( P_1 - P_2 = \frac{1}{2} \rho \left( \left( \frac{A}{a} \right)^2 v_1^2 - v_1^2 \right) \)
\( P_1 - P_2 = \frac{1}{2} \rho v_1^2 \left( \frac{A^2}{a^2} - 1 \right) \)
\( P_1 - P_2 = \frac{1}{2} \rho v_1^2 \left( \frac{A^2 - a^2}{a^2} \right) \)
So, \( v_1^2 = \frac{2 (P_1 - P_2) a^2}{\rho (A^2 - a^2)} \)
\( v_1 = a \sqrt{\frac{2 (P_1 - P_2)}{\rho (A^2 - a^2)}} \).
The volume of liquid flowing per second (flow rate) through the wider entry A is \( Q = A v_1 \).
\( Q = A a \sqrt{\frac{2 (P_1 - P_2)}{\rho (A^2 - a^2)}} \). This equation helps measure the flow rate.In simple words: A venturimeter measures how fast a liquid flows. It uses a narrow section in the pipe where the liquid speeds up and its pressure drops. By measuring this pressure difference, we can calculate the liquid's speed and how much liquid is flowing each second.

🎯 Exam Tip: When explaining the venturimeter, clearly connect Bernoulli's principle (pressure drop with increased velocity) and the equation of continuity (velocity change due to area change).

IV. Numerical Problems:

 

Question 1. A capillary of diameter d mm is dipped in water such that the water rises to a height of 30 mm. If the radius of the capillary is made \( \frac{2}{3} \) of its previous value, then compute the height up to which water will rise in the new capillary?
Answer: We know that the height 'h' to which a liquid rises in a capillary tube is given by \( T = \frac{hr\rho g}{2 \cos \theta} \), which can be rearranged to \( h = \frac{2T \cos \theta}{r \rho g} \).
Here, T (surface tension), \( \rho \) (density of liquid), g (acceleration due to gravity), and \( \cos \theta \) (angle of contact) are all constants for the same liquid and tube material. Therefore, the height 'h' is inversely proportional to the radius 'r' of the capillary tube.
So, \( h \propto \frac{1}{r} \).
This means we can write the relationship as \( \frac{h_1}{h_2} = \frac{r_2}{r_1} \).
Given:
Initial height \( h_1 = 30 \) mm
Let the initial radius be \( r_1 = r \)
The new radius \( r_2 = \frac{2}{3} r \)
We need to find the new height \( h_2 \).
Substitute the values into the equation:
\( \frac{30}{h_2} = \frac{\frac{2}{3}r}{r} \)
\( \frac{30}{h_2} = \frac{2}{3} \)
Now, solve for \( h_2 \):
\( h_2 = \frac{30 \times 3}{2} \)
\( h_2 = \frac{90}{2} \)
\( h_2 = 45 \) mm.
So, the water will rise to a height of 45 mm in the new capillary tube. A smaller radius causes the liquid to rise higher.
In simple words: When a capillary tube gets thinner, water can climb higher inside it. If the tube's radius becomes two-thirds of what it was, the water will rise one and a half times as much. So, if it rose 30 mm before, it will now rise to 45 mm.

🎯 Exam Tip: Remember the inverse proportionality between capillary rise height and the radius of the tube. This relationship is crucial for solving problems involving changes in tube dimensions.

 

Question 2. A cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of \( 4 \times 10^5 \) N is applied at the other end. If the rigidity modulus of the cylinder is \( 6 \times 10^{10} \) Nm\(^{-2} \), then calculate the twist produced in the cylinder.
Answer: First, let's list the given values and convert them to SI units:
Length of the cylinder \( L = 1.5 \) m
Diameter of the cylinder \( D = 4 \) cm \( = 4 \times 10^{-2} \) m
Radius of the cylinder \( R = D/2 = 2 \times 10^{-2} \) m
Tangential force \( F_t = 4 \times 10^5 \) N
Rigidity modulus \( \eta = 6 \times 10^{10} \) Nm\(^{-2} \)
We need to calculate the twist produced, \( \theta \).
The formula for shear stress \( \sigma_s \) is \( \frac{\text{Force}}{\text{Area}} \).
The area of cross-section \( A = \pi R^2 = \pi (2 \times 10^{-2})^2 = 3.14 \times 4 \times 10^{-4} = 12.56 \times 10^{-4} \) m\(^2 \).
Shear stress \( \sigma_s = \frac{F_t}{A} = \frac{4 \times 10^5}{12.56 \times 10^{-4}} = 3.1847 \times 10^8 \) N/m\(^2 \).
The shear strain \( \epsilon_s \) for a cylindrical twist is given by \( \frac{R \theta}{L} \).
The rigidity modulus \( \eta = \frac{\text{Shear stress}}{\text{Shear strain}} = \frac{\sigma_s}{\frac{R \theta}{L}} = \frac{\sigma_s L}{R \theta} \).
Rearranging to solve for \( \theta \):
\( \theta = \frac{\sigma_s L}{\eta R} \).
Now, substitute the values:
\( \theta = \frac{(3.1847 \times 10^8 \text{ N/m}^2) \times (1.5 \text{ m})}{(6 \times 10^{10} \text{ N/m}^2) \times (2 \times 10^{-2} \text{ m})} \)
\( \theta = \frac{4.77705 \times 10^8}{12 \times 10^8} \)
\( \theta = 0.3980875 \) radians.
This is the angle of twist produced in the cylinder. The rigidity modulus tells us how much a material resists twisting.
In simple words: We have a metal cylinder that we twist with a force. We need to find how much it twists. We use a formula that connects the force, the cylinder's size (length, radius), and how stiff the material is (rigidity modulus). After doing the calculations, the cylinder twists by about 0.398 radians.

🎯 Exam Tip: Pay close attention to unit conversions (especially cm to m) and ensure all values are in SI units before calculation. The formula for shear strain due to twist in a cylinder \( (\frac{R\theta}{L}) \) is crucial here.

 

Question 3. Bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than the radius of both soap bubbles A and B.
Answer: Let \( T \) be the surface tension of the soap solution.
For a soap bubble, the excess pressure \( \Delta P = \frac{4T}{R} \).
Given:
Radius of bubble A (smaller bubble) \( R_A = 2 \) cm
Radius of bubble B (larger bubble) \( R_B = 4 \) cm
Excess pressure inside bubble A: \( \Delta P_A = \frac{4T}{R_A} = \frac{4T}{2} = 2T \).
Excess pressure inside bubble B: \( \Delta P_B = \frac{4T}{R_B} = \frac{4T}{4} = T \).
The pressure difference between the inside of the smaller bubble (A) and the outside of the larger bubble (B) would be related to their combined effect.
The air pressure difference between the smaller bubble and the larger bubble is the sum of their individual excess pressures when one is inside the other:
\( \Delta P_{\text{combined}} = \Delta P_A + \Delta P_B = 2T + T = 3T \).
Now, we need to find the radius of a single soap bubble, let's call it \( R_{single} \), that would have this same combined pressure difference, \( \Delta P_{\text{combined}} \).
For this single bubble: \( \Delta P_{\text{combined}} = \frac{4T}{R_{single}} \).
So, \( 3T = \frac{4T}{R_{single}} \).
Dividing both sides by T:
\( 3 = \frac{4}{R_{single}} \)
\( R_{single} = \frac{4}{3} \) cm.
Now let's compare \( R_{single} \) with \( R_A \) and \( R_B \):
\( R_{single} = \frac{4}{3} \approx 1.33 \) cm.
\( R_A = 2 \) cm
\( R_B = 4 \) cm
Clearly, \( R_{single} = \frac{4}{3} \) cm is less than \( R_A = 2 \) cm and also less than \( R_B = 4 \) cm.
This demonstrates that the radius of the single soap bubble maintaining this combined pressure difference is indeed smaller than both original bubble radii. This occurs because the total excess pressure is additive when bubbles are nested or connected.
In simple words: When one soap bubble is inside another, their extra pressures add up. If we want a single bubble to have this total extra pressure, it must be quite small. We found its radius is \( \frac{4}{3} \) cm, which is smaller than both the 2 cm and 4 cm bubbles.

🎯 Exam Tip: Remember that for nested or connected bubbles, the effective pressure differences can add up. The excess pressure for a soap bubble is \( \frac{4T}{R} \) (due to two surfaces), and for a liquid drop it's \( \frac{2T}{R} \) (one surface).

 

Question 4. A block of Ag of mass x kg hanging from a string is immersed in a liquid of relative density 0.72. If the relative density of Ag is 10 and tension in the string is 37.12 N then compute the mass of the Ag block.
Answer: Let the mass of the silver (Ag) block be 'm' kg.
Tension in the string \( T = 37.12 \) N.
Relative density of liquid \( \rho_{liq} = 0.72 \)
Relative density of silver \( \rho_{Ag} = 10 \)
We know that Relative Density \( = \frac{\text{Density of substance}}{\text{Density of water}} \). So, \( \rho_{\text{substance}} = \text{RD} \times \rho_{\text{water}} \).
Let \( \rho_w \) be the density of water (\( 1000 \text{ kg/m}^3 \)).
Density of liquid \( \rho_l = 0.72 \rho_w \).
Density of silver \( \rho_s = 10 \rho_w \).
When the silver block is immersed in the liquid, it experiences an upward buoyant force \( F_B \).
From Archimedes' principle, \( F_B = \text{Weight of liquid displaced} = V_s \rho_l g \), where \( V_s \) is the volume of the silver block.
The volume of the silver block \( V_s = \frac{\text{mass}}{\text{density}} = \frac{m}{\rho_s} = \frac{m}{10 \rho_w} \).
So, \( F_B = \left( \frac{m}{10 \rho_w} \right) (0.72 \rho_w) g = \frac{0.72 m g}{10} = 0.072 m g \).
When the block is hanging in the liquid, the forces are:
Weight of the block (downwards) \( W = mg \)
Buoyant force (upwards) \( F_B \)
Tension in the string (upwards) \( T \)
For equilibrium: \( T + F_B = W \)
\( 37.12 + 0.072 mg = mg \)
\( 37.12 = mg - 0.072 mg \)
\( 37.12 = mg (1 - 0.072) \)
\( 37.12 = mg (0.928) \)
Assuming \( g = 9.8 \text{ m/s}^2 \):
\( 37.12 = m \times 9.8 \times 0.928 \)
\( 37.12 = m \times 9.0944 \)
\( m = \frac{37.12}{9.0944} \)
\( m \approx 4.08 \) kg.
The mass of the silver block is approximately 4.08 kg. This is a common method for determining an object's mass or density using buoyancy.
In simple words: A silver block hangs from a string in a liquid. The string tension tells us the block's "apparent" weight, which is less than its real weight because the liquid pushes it up (buoyant force). Using the densities of silver and the liquid, we can calculate the upward push and then find the actual mass of the silver block.

🎯 Exam Tip: Clearly identify all forces acting on the submerged body: tension, weight, and buoyant force. Remember that buoyant force equals the weight of the fluid displaced, and relative density is a key tool to find actual densities.

 

Question 5. The reading of the pressure meter attached with a closed pipe is \( 5 \times 10^5 \) Nm\(^{-2} \). On opening the valve of the pipe, the reading of the pressure meter is \( 4.5 \times 10^5 \) Nm\(^{-2} \). Calculate the speed of the water flowing in the pipe.
Answer: We can use Bernoulli's theorem to solve this problem, as it relates pressure and velocity in fluid flow.
Bernoulli's equation is \( P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \).
Since the pipe is horizontal, the height 'h' remains the same, so the \( \rho g h \) term cancels out.
The equation simplifies to \( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \).
Given values:
Initial pressure (when closed, so velocity is 0): \( P_1 = 5 \times 10^5 \) N/m\(^2 \)
Initial velocity \( v_1 = 0 \) m/s
Final pressure (when open): \( P_2 = 4.5 \times 10^5 \) N/m\(^2 \)
Density of water \( \rho = 1000 \) kg/m\(^3 \) (standard value for water)
We need to find the final velocity \( v_2 \).
Substitute the values into the simplified Bernoulli's equation:
\( 5 \times 10^5 + \frac{1}{2} \times 1000 \times (0)^2 = 4.5 \times 10^5 + \frac{1}{2} \times 1000 \times v_2^2 \)
\( 5 \times 10^5 + 0 = 4.5 \times 10^5 + 500 v_2^2 \)
Rearrange the equation to solve for \( v_2^2 \):
\( 5 \times 10^5 - 4.5 \times 10^5 = 500 v_2^2 \)
\( 0.5 \times 10^5 = 500 v_2^2 \)
\( 50000 = 500 v_2^2 \)
\( v_2^2 = \frac{50000}{500} \)
\( v_2^2 = 100 \)
Now, take the square root to find \( v_2 \):
\( v_2 = \sqrt{100} \)
\( v_2 = 10 \) m/s.
So, the speed of the water flowing in the pipe is 10 m/s. This calculation helps in understanding how pressure changes into kinetic energy in a flowing fluid.
In simple words: When water starts flowing from a pipe, its pressure drops. This drop in pressure is converted into the energy of motion (kinetic energy) of the water. Using Bernoulli's rule and the pressure change, we can figure out that the water is moving at 10 meters per second.

🎯 Exam Tip: For horizontal pipes, the potential energy term in Bernoulli's equation cancels out. Remember the density of water (\( 1000 \text{ kg/m}^3 \)) if not provided. The pressure difference is directly related to the kinetic energy gain.

V. Conceptual Questions:

 

Question 1. Why coffee runs up into a sugar lump (a small cube of sugar) when one corner of the sugar lump is held in the liquid?
Answer: Coffee runs up into a sugar lump because of capillary action. A sugar lump is full of tiny, narrow spaces and pores, which act like tiny capillary tubes. When one corner of the sugar is put into coffee, the surface tension of the coffee pulls the liquid up into these small spaces. The adhesive forces (attraction between coffee and sugar) are stronger than the cohesive forces (attraction within the coffee itself) and the gravitational pull, causing the coffee to climb up the sugar lump. This phenomenon is why sugar quickly gets wet and flavored.
In simple words: Sugar has many tiny holes, like small pipes. When coffee touches the sugar, it gets pulled up into these tiny holes by a force called surface tension. This makes the coffee climb up the sugar cube.

🎯 Exam Tip: When explaining phenomena like this, use keywords such as "capillary action," "surface tension," "pores/narrow spaces," and "adhesive/cohesive forces" to ensure a complete answer.

 

Question 2. Why two holes are made to empty an oil tin?
Answer: Two holes are made in an oil tin to allow the oil to flow out smoothly and continuously. When only one hole is made, the oil tries to come out, creating a vacuum or reduced pressure inside the tin. This lower pressure makes it difficult for the oil to flow out freely because the outside atmospheric pressure pushes against it. By making a second hole, atmospheric air can enter the tin through this new opening. This entry of air maintains normal atmospheric pressure inside the tin, allowing the oil to flow out easily and steadily through the first hole without struggling against a pressure difference. The constant air supply ensures a steady stream.
In simple words: We make two holes in an oil tin so the oil can pour out easily. One hole lets the oil come out, and the other lets air go in. If air can't go in, the oil struggles to come out because of the air pressure difference.

🎯 Exam Tip: The core concept here is the role of atmospheric pressure. Emphasize that the second hole equalizes pressure inside and outside the tin, enabling continuous flow.

 

Question 3. We can cut vegetables easily with a sharp knife as compared to a blunt knife. Why?
Answer: A sharp knife has a much smaller contact area than a blunt knife. When you apply the same total force to both, the sharp knife puts more pressure (force per unit area) on the vegetable. This higher pressure allows the knife to cut through the vegetable more easily. Therefore, a sharp knife is more effective for cutting.
In simple words: A sharp knife has a tiny edge, so when you push it, it creates a lot of pressure, making it easy to cut vegetables. A blunt knife spreads the push over a bigger area, so it's harder to cut.

🎯 Exam Tip: Remember that pressure is inversely proportional to area (P = F/A). A smaller area of contact results in greater pressure for the same applied force, which is key to effective cutting.

 

Question 4. Why the passengers are advised to remove the ink from their pens while going up in an aeroplane?
Answer: As an aeroplane goes higher, the atmospheric pressure outside decreases. The air inside the pen, however, remains at the pressure it was on the ground. This pressure difference causes the ink to be pushed out from the pen to equalize the pressure. This can lead to ink leakage, which might stain passengers' clothes, so they are advised to empty their pens or use ballpoint pens that are less susceptible to pressure changes.
In simple words: When an airplane goes high, the air pressure outside drops. The ink inside the pen feels more pressure than the outside air, so it tries to escape, causing a leak.

🎯 Exam Tip: Questions about pressure changes at altitude often relate to everyday objects like pens or sealed containers. Focus on the concept of pressure equalization causing movement from high to low pressure areas.

 

Question 5. We use a straw to suck soft drinks, why?
Answer: When we suck air out of a straw, the air pressure inside the straw becomes lower than the atmospheric pressure outside. The higher atmospheric pressure then pushes down on the surface of the drink in the glass. This external pressure forces the liquid up into the straw and into our mouth. This simple action demonstrates the power of atmospheric pressure.
In simple words: When you suck on a straw, you make the air inside it less dense. The normal air outside then pushes the drink up the straw into your mouth.

🎯 Exam Tip: Many everyday phenomena like sucking with a straw or using a dropper rely on creating a pressure difference for a liquid to move. Remember that liquids always move from an area of higher pressure to an area of lower pressure.