Mechanical Properties of Fluids Class 12 MCQ

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    mcq on Mechanical Properties of fluids for Class 12 Students

    When a uniform wire of radius r is stretched by a 2kg weight, the increase in its length is 2.00 mm. If the radius of the wire is r/2 and other conditions remain the same, the increase in its length is

    1. 8mm

    2. 4mm

    3. 2mm

    4. 1mm

    Explanation:
    A wire undergoes elongation when a load is applied. If its radius is reduced while keeping the same load and material, the change in extension depends on how cross-sectional area influences deformation.

    The elongation of a wire is inversely proportional to its cross-sectional area and directly proportional to the applied force and original length. Since the area of a circular wire is proportional to r2, reducing the radius changes the area significantly.

    When the radius becomes r/2, the new area becomes (r/2)2 = r2/4. This means the area becomes one-fourth of the original. Because elongation is inversely proportional to area, reducing the area increases elongation proportionally. The load and material remain unchanged, so the only factor affecting the extension is this reduction in area. Thus, the extension increases accordingly compared to the original value.

    Think of stretching a thick versus a thin rubber band with the same force—the thinner one stretches more due to smaller area resisting the force.

    Thus, reducing radius leads to a significant increase in elongation due to the inverse dependence on cross-sectional area.

    Option 1 – 8mm

    An iron wire of length 4m and diameter 2mm is loaded with a weight of 8 Kg. If Young’s modulus Y for Iron is 2×10¹¹ N/m². Then the increase in the length of the wire is

    1. 1mm

    2. 2mm

    3. 0.5mm

    4. 8mm

    Explanation:
    A wire stretches slightly when a load is applied. The amount of stretching depends on its dimensions, the applied force, and the material’s stiffness represented by Young’s modulus.

    The elongation is calculated using ΔL = (F × L) / (A × Y). Here, force is obtained using F = mg, and the cross-sectional area is calculated using A = πr2, where r is half the diameter. All values must be converted into SI units for consistency.

    Substituting the given values into the formula shows that elongation depends directly on force and length, and inversely on area and Young’s modulus. Since iron has a very high Young’s modulus, the denominator becomes large, resulting in only a very small extension. This indicates that iron strongly resists deformation under applied load.

    It is similar to comparing a steel rod and a rubber band—the rod barely stretches due to its high stiffness.

    Hence, the elongation remains very small because of the large value of Young’s modulus and relatively small applied load.

    Option 3 – 0.5mm

    Which law governs the working of spring balance

    1. Robert Hooke’s law

    2. Newton’s Law

    3. Keplor’s Law

    4. Young’s Law

    Explanation:
    A spring balance measures force based on how much a spring stretches when a load is applied. The relationship between applied force and extension determines its working principle.

    According to elasticity principles, within the elastic limit, the extension produced in a spring is directly proportional to the applied force. This proportionality allows consistent measurement of weight or force.

    When a load is attached, the spring elongates until the restoring force balances the applied force. The scale on the spring balance is calibrated using this proportional relationship. As long as the spring is not overstretched, it returns to its original shape when the load is removed, ensuring accurate readings.

    This is similar to stretching a rubber band slightly—the more you pull, the more it extends, but only up to a certain limit.

    Thus, the working of a spring balance depends on a direct proportionality between force and extension within the elastic limit.

    Option 1 – Robert Hooke’s law

    The theoretical value of the Poisson ratio lies between

    1. -1 and 0

    2. -1 and 0.5

    3. 0.5 and 1

    4. None

    Explanation:
    Poisson’s ratio describes how a material behaves laterally when stretched or compressed longitudinally. It relates transverse strain to longitudinal strain.

    For most materials, when stretched, they become thinner in the perpendicular direction, giving a positive Poisson’s ratio. Theoretical limits are derived based on stability conditions and material behavior.

    Mathematically, Poisson’s ratio is defined as the negative ratio of lateral strain to longitudinal strain. For stable, isotropic materials, its value lies within a specific range to ensure physical consistency. Values outside this range would imply unrealistic deformation behavior, such as expansion in all directions under tension.

    For example, stretching a rubber band makes it thinner, showing a typical positive Poisson’s ratio.

    Thus, Poisson’s ratio must lie within a defined theoretical range for materials to remain physically stable under stress.

    Option 2 – -1 and 0.5

    When a spring is loaded, the strain produced is

    1. shearing

    2. volumetric

    3. longitudinal

    4. none of these

    Explanation:
    When a load is applied to a spring, it undergoes deformation. The type of strain depends on how the dimensions of the spring change due to the applied force.

    Strain is defined as the ratio of change in dimension to the original dimension. In a spring, the primary deformation occurs along its length, meaning it either stretches or compresses along its axis.

    Since the change occurs along the direction of the applied force, the strain is classified based on this directional change. Other types of strain, such as shear or volumetric, involve different kinds of deformation, which are not dominant in a simple spring under load.

    This is similar to pulling a coil spring—the length increases while other dimensions change negligibly.

    Thus, the strain produced in a loaded spring corresponds to deformation along its length.

    Option 1 – shearing

    A wire whose cross-sectional area is 2mm² is stretched by a certain load. If a similar wire of triple the area of cross-section is stretched by the same load, the elongation of the second wire would be.

    1. 0.033mm

    2. 0.03mm

    3. 0.3mm

    4. 0.66mm

    Explanation:
    Two wires made of the same material are subjected to the same load, but they differ in cross-sectional area. The change in elongation depends on how area affects deformation.

    Elongation is inversely proportional to cross-sectional area when other factors such as force, length, and material remain constant. This means a thicker wire (larger area) will stretch less under the same load.

    If the second wire has three times the area of the first, the resistance to stretching increases accordingly. As a result, the elongation decreases in proportion to the increase in area.

    This can be compared to pulling a thick rope versus a thin string—the thicker rope stretches less under the same force.

    Thus, increasing the cross-sectional area reduces elongation due to greater resistance to deformation.

    Option 1 – 0.033mm

    A long spring is stretched by 2 cm and its potential energy is V. If the spring is stretched by 10cm. Its potential energy will be

    1. 25V

    2. V/5

    3. 5V

    4. V/25

    Explanation:
    The energy stored in a spring depends on how much it is stretched or compressed. This energy is known as elastic potential energy.

    The potential energy stored in a spring is proportional to the square of its extension, given by the relation U ∝ x2. This means that even a small increase in extension can lead to a large increase in stored energy.

    When the extension increases from one value to another, the energy changes according to the square of the ratio of extensions. Therefore, increasing the stretch significantly increases the stored energy.

    This is similar to pulling a spring slightly versus pulling it much further—the resistance increases rapidly, storing more energy.

    Thus, the potential energy increases sharply as the extension increases due to its dependence on the square of displacement.

    Option 1 – 25V

    The upper end of the wire is 1m long and 4mm radius clamped. The lower end is twisted by an angle of 30°. The angle of the shear is

    1. 0.12°

    2. 0.0012°

    3. 1.2°

    4. 120°

    Explanation:
    When a wire is twisted, different layers of the material experience angular displacement, leading to shear deformation. The angle of shear measures this distortion.

    The angle of shear is defined as the ratio of the linear displacement at the outer surface to the length of the wire. It depends on the radius, angle of twist, and length of the wire.

    As the wire is twisted, the outermost layer moves the most, creating shear strain. The relationship between angular twist and shear angle helps determine the deformation experienced by the material.

    This is similar to twisting a cylindrical rod where the outer surface undergoes more displacement than the inner core.

    Thus, shear angle depends on geometry and twist applied to the wire.

    Option 1 – 0.12°

    Atomizers work on the principle of

    1. Newton’s Second law

    2. Archimedes principle

    3. Bernoulli’s theorem

    4. Stoke’s theorem

    Explanation:
    Atomizers are devices that convert liquid into fine spray droplets. Their operation depends on how Fluid motion affects pressure.

    When a fast-moving stream of air passes over a liquid surface, it reduces the pressure in that region. This pressure difference causes the liquid to rise and break into fine droplets.

    The principle involves the relationship between Fluid velocity and pressure, where an increase in velocity leads to a decrease in pressure. This allows the liquid to be drawn up and dispersed as a spray.

    A common example is a perfume spray bottle, where squeezing the bulb forces air over the liquid, producing a mist.

    Thus, atomizers work based on pressure differences created by Fluid motion.

    Option 3 – Bernoulli’s theorem

    A pipeline carries oil, at a velocity of 2 m/sec through a 20 cm diameter pipe. At another section the diameter is 15 cm. Then the velocity at this section is

    1. 2.4 m/sec

    2. 3.56 m/sec

    3. 4.2 m/sec

    4. 8.62 m/sec

    Explanation:
    Fluid flow in a pipe must obey conservation principles. When the cross-sectional area changes, the velocity adjusts to maintain continuity.

    According to the equation of continuity, A₁v₁ = A₂v₂, meaning the product of area and velocity remains constant for incompressible flow. Since area is proportional to diameter squared (A ∝ d2), reducing the diameter reduces the area significantly.

    When the pipe narrows, the Fluid must move faster to maintain the same flow rate. Thus, velocity increases as the cross-sectional area decreases.

    This is similar to placing your thumb over a hose—the water flows faster when the opening is reduced.

    Hence, a decrease in diameter leads to an increase in velocity to conserve Mass flow rate.

    Option 2 – 3.56 m/sec

    A water main of 20 cm diameter has a pitot tube fixed into it and the pressure difference indicated by the gauge is 5 cm of water column. Then the rate of flux of water through the main is

    1. 0.031 m³/sec

    2. 0.0031 m³/sec

    3.0.31 m³/sec

    4. 3.1 m³/sec

    Explanation:
    A pitot tube measures the velocity of Fluid flow by comparing pressure differences. The given pressure head helps determine the flow speed, which is then used to calculate discharge.

    The velocity of Fluid is obtained using the relation derived from Bernoulli’s principle, where velocity depends on the height of the liquid column (v ∝ √h). Once velocity is known, the discharge (rate of flow) is calculated using Q = A × v, where A is the cross-sectional area of the pipe.

    First, the pressure difference in terms of height is converted into velocity. Then, using the pipe’s diameter, the area is calculated as A = πr2. Multiplying this area with velocity gives the volumetric flow rate. The process shows how pressure energy is converted into kinetic energy in flowing fluids.

    This is similar to measuring water speed in a pipe by observing how much it pushes upward in a vertical tube.

    Thus, pressure difference helps determine velocity, which in turn gives the flow rate through the pipe.

    Option 1 – 0.031 m³/sec

    According to the theory of relativity, the simultaneity of two events is

    1. Absolute

    2. Relative

    3. Sometimes absolute and sometimes relative

    4. Depending on the importance

    Explanation:
    The concept of simultaneity refers to whether two events occur at the same time. In relativity, this idea depends on the observer’s frame of reference.

    According to Einstein’s theory, time is not absolute but relative. Observers moving at different velocities may disagree on whether two events occur simultaneously. This happens because the speed of Light is constant for all observers, leading to differences in time measurements.

    If two events appear simultaneous in one frame, they may not appear so in another moving frame. This challenges classical Physics, where time was considered universal and the same for all observers.

    An example is two lightning strikes observed from a moving train versus a stationary platform—each observer may perceive different timings.

    Thus, simultaneity is not absolute but depends on the observer’s motion and reference frame.

    Option 2 – Relative

    According to Toricelli’s theorem, the velocity of efflux of a liquid through an orifice is

    1. Independent of the height of the liquid above

    2. Independent of acceleration due to gravity

    3. Independent of the density of liquid

    4. Depends on the velocity of liquid

    Explanation:
    Torricelli’s theorem describes the speed at which a liquid flows out of an opening under gravity. It relates the velocity of efflux to the height of the liquid column.

    The velocity is given by v = √(2gh), showing dependence on gravitational acceleration and height of liquid above the orifice. This relation is derived from energy conservation, where potential energy converts into kinetic energy.

    Interestingly, the expression does not include the density of the liquid, meaning the speed of efflux remains the same for different liquids at the same height. This occurs because both pressure and inertia effects scale with density, effectively canceling each other.

    This is similar to different liquids flowing out of identical holes at similar speeds if their heights are the same.

    Thus, velocity depends on height and gravity but not on certain material properties of the liquid.

    Option 3 – Independent of the density of liquid

    The path of the jet of liquid coming out of an orifice is

    1. Parabola

    2. Circle

    3. Hyperbola

    4. Ellipse

    Explanation:
    When liquid flows out of an orifice, it moves under the influence of gravity after leaving the container. Its trajectory is determined by its initial velocity and gravitational acceleration.

    The horizontal component of velocity remains constant, while the vertical component changes due to gravity. This combination results in a curved path similar to projectile motion.

    Mathematically, such motion follows the same equations as a projectile, leading to a specific geometric shape. The path depends on initial velocity and height from which the liquid emerges.

    This can be seen in water flowing from a hole in a tank, forming a curved stream as it falls.

    Thus, the liquid jet follows a predictable curved trajectory governed by projectile motion principles.

    Option 1 – Parabola

    The streamlined flow of a liquid through a pipe becomes turbulent flow when its velocity is

    1. less than the critical velocity

    2. greater than the critical velocity

    3. equal to half the critical velocity

    4. equal to one-third of critical velocity

    Explanation:
    Fluid flow can be smooth (streamlined) or chaotic (turbulent), depending on the velocity and other conditions. The transition between these flows is determined by a critical parameter.

    The Reynolds number is used to predict flow behavior. When Fluid velocity is low, flow remains orderly with layers moving smoothly. As velocity increases, disturbances grow, leading to irregular motion.

    Once the velocity exceeds a certain critical value, the flow becomes turbulent. In this state, eddies and vortices form, and the motion becomes unpredictable.

    An example is water flowing gently in a pipe versus rushing rapidly, where it starts to swirl and mix.

    Thus, exceeding a critical velocity leads to the transition from streamlined to turbulent flow.

    Option 2 – greater than the critical velocity

    The velocity of efflux from the orifice of a vessel with a liquid column of height 0.2 m above is (g-10 m/sec²)

    1. 0.2 m/sec

    2. 2 m/sec

    3. 4m/sec

    4. 0.4 m/sec

    Explanation:
    The velocity of a liquid exiting an orifice depends on the height of the liquid column above it. This relationship is described by Torricelli’s theorem.

    The formula v = √(2gh) is used, where g is acceleration due to gravity and h is the height of the liquid. This equation is derived from the conversion of potential energy into kinetic energy.

    By substituting the given height and gravitational acceleration, the velocity can be determined. The result shows that velocity increases with the square root of height.

    This is similar to water flowing faster from a deeper tank compared to a shallow one.

    Thus, efflux velocity depends on the height of liquid and gravitational acceleration.

    Option 2 – 2 m/sec

    The Vector among the following quantities related to the flow of a liquid

    1. rate of flow

    2. speed of flow

    3. density

    4. current density

    Explanation:
    Physical quantities can be classified as scalar or Vector based on whether they have direction. In Fluid flow, several quantities describe motion and properties of the fluid.

    Scalars like density and speed have magnitude only, while Vectors include both magnitude and direction. Certain flow-related quantities involve directional movement, making them Vector quantities.

    For example, velocity has both magnitude and direction, and related derived quantities may also carry directional properties. Identifying Vector quantities is important for analyzing flow behavior in different directions.

    This is similar to distinguishing between speed (how fast) and velocity (how fast and in which direction).

    Thus, quantities involving directional flow characteristics are treated as Vectors.

    Option 4 – current density

    The ratio of areas of cross sections of a horizontal pipe through which water flows is 2 : 1. The ratio of velocities of water at those points is

    1. 1 : 2

    2. 2 : 1

    3. 4 : 1

    4. 1 : 4

    Explanation:
    Fluid flowing through a pipe must obey conservation of Mass. When the area changes, velocity adjusts to maintain a constant flow rate.

    According to the continuity equation, A₁v₁ = A₂v₂. This means velocity is inversely proportional to cross-sectional area. If one section has a larger area, the velocity there will be lower.

    Given the ratio of areas, the corresponding velocities must adjust inversely to maintain the same discharge. This relationship ensures that the same volume of fluid passes through each section per unit time.

    This is similar to water flowing faster through a narrow section of a pipe compared to a wider section.

    Thus, velocity changes inversely with cross-sectional area in fluid flow.

    Option 1 – 1 : 2

    The equation of continuity is the counterpart of

    1. law of conservation of momentum

    2. law of conservation of Mass

    3. law of conservation of energy

    4. law of conservation of angular momentum

    Explanation:
    The equation of continuity describes how fluid flows through a system without accumulation or loss. It ensures that the flow remains consistent across different sections.

    Mathematically, it is expressed as A₁v₁ = A₂v₂, meaning the Mass flow rate remains constant. This principle is derived from a fundamental conservation law in Physics.

    It implies that the amount of fluid entering a section must equal the amount leaving it, assuming no storage or leakage. This is essential for analyzing steady fluid flow.

    A simple example is water flowing through a pipe with varying diameter, where flow rate remains constant.

    Thus, the continuity equation is based on a fundamental conservation principle governing fluid motion.

    Option 2 – law of conservation of Mass

    The divergence of the velocity of an expanding gas is

    1. Negative

    2. Zero

    3. Positive

    4. Does not exist

    Explanation:
    Divergence of velocity indicates how fluid is expanding or compressing at a point. It measures whether fluid is spreading out or converging.

    In an expanding gas, fluid particles move away from each other, increasing the volume. This outward motion implies that more fluid is leaving a region than entering it.

    Mathematically, divergence becomes positive when there is a NET outward flow. This reflects expansion and decrease in density in the region.

    An example is air expanding when heated, causing it to spread outward.

    Thus, divergence helps identify whether a fluid is expanding or compressing at a given point.

    Option 3 – Positive

    The rate of flow (Mass per second) through a pipe of cross-section 0.02 m² with water flowing at a velocity of 5 m/sec is

    1. 100 kg/sec

    2. 1000 kg/sec

    3. 0.1 kg/sec

    4. 1 kg/sec

    Explanation:
    The Mass flow rate represents how much Mass of fluid passes through a cross-section per unit time. It depends on the fluid’s density, velocity, and the pipe’s cross-sectional area.

    Mass flow rate is given by ṁ = ρ × A × v, where ρ is density, A is cross-sectional area, and v is velocity. For water, density is typically taken as 1000 kg/m³. This formula combines the idea of volume flow rate (A × v) with density to convert it into Mass per second.

    First, calculate the volume flow rate using the given area and velocity. Then multiply this value by the density of water to obtain the mass flow rate. This shows how both geometry and fluid properties influence the rate of mass transfer.

    This is similar to calculating how much water passes through a pipe every second based on how fast it moves and how wide the pipe is.

    Thus, mass flow rate depends on area, velocity, and density, linking fluid motion with the amount of substance flowing.

    Option 3 – 0.1 kg/sec

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