12th Std Physics One Mark Questions

Explanation: This question asks you to compare different shapes having identical Mass and radius to determine which one offers the least resistance to rotational motion about its central axis. The key idea revolves around how Mass is distributed relative to the axis of rotation. Moment of inertia depends strongly on whether Mass is concentrated near the center or spread outward. When Mass is located closer to the axis, the resistance to rotation becomes smaller.

In rotational mechanics, moment of inertia plays a role similar to Mass in linear motion. The farther the particles are from the axis, the larger the contribution to inertia, since each small Mass element contributes proportional to r². Thus, shapes with more Mass near the center will naturally have lower inertia compared to those where Mass is spread outward.

To reason step by step, consider typical distributions: a ring has most of its Mass at the outer edge, while a Solid sphere has mass distributed throughout its volume, including near the center. As a result, shapes concentrating mass inward reduce the overall rotational resistance. By comparing these distributions, one can identify which geometry minimizes inertia.

As an analogy, imagine spinning two objects: one with weight concentrated at the edge and another with weight closer to the center. The second spins more easily because less effort is needed to rotate it.

In summary, the object with the most centrally concentrated mass will exhibit the smallest moment of inertia among all given options.

Explanation: This question explores how two rotating bodies behave when subjected to the same stopping torque, given that both initially possess equal angular momentum. Angular momentum depends on both moment of inertia and angular velocity, expressed as Iω. Even if two bodies have equal angular momentum, their internal distribution of mass (moment of inertia) can differ significantly.

Torque is related to the rate of change of angular momentum. When a constant torque is applied, it reduces angular momentum at a steady rate. Therefore, the time taken to bring a rotating body to rest depends on how quickly its angular momentum is reduced under that torque.

Step by step, since both bodies begin with the same angular momentum and experience identical torque, the rate at which angular momentum decreases is the same for both. The stopping time depends only on the initial angular momentum and the applied torque, not on how that angular momentum is distributed between I and ω individually.

For example, consider two spinning objects—one rotating fast with low inertia and another rotating slowly with high inertia—but both having equal angular momentum. If you apply the same braking torque, both will lose angular momentum at the same rate and thus stop together.

In summary, identical initial angular momentum and equal opposing torque ensure that both bodies come to rest in equal time.

Explanation: This question focuses on understanding the physical meaning of moment of inertia in rotational motion. It is a measure of how strongly a body resists changes in its rotational state about a given axis. Just like mass resists linear acceleration, moment of inertia resists angular acceleration.

The concept depends on both the total mass and how that mass is distributed relative to the axis. Each particle contributes an amount proportional to m·r², where r is its distance from the axis. Thus, even small masses far from the axis can significantly increase inertia.

Step by step, when a torque is applied, the angular acceleration produced is inversely proportional to the moment of inertia. A larger inertia means less angular acceleration for the same torque. Therefore, it does not generate motion by itself but resists any change in rotational motion.

As an analogy, think of pushing a door. If you push near the hinge, it is harder to rotate compared to pushing at the edge. This is because the effective distribution of mass relative to the axis matters.

In summary, moment of inertia acts as a measure of resistance against changes in rotational motion and depends on how mass is distributed about the axis.

Explanation: This question examines why spokes are used in bicycle wheels from a Physics perspective. A wheel must be strong yet lightweight to rotate efficiently while supporting loads. Spokes play a crucial role in achieving this balance.

The key idea involves minimizing moment of inertia while maintaining structural strength. If the wheel were a Solid disc, more mass would be distributed away from the center, increasing rotational inertia. Spokes reduce unnecessary mass while still providing mechanical support between the hub and the rim.

Step by step, by using thin spokes, most of the mass is concentrated near the rim where it contributes to stability, while the overall mass remains low. This design allows easier acceleration and deceleration of the wheel, improving efficiency.

For example, consider carrying a heavy Solid plate versus a lightweight frame structure. The frame provides strength with less material, making it easier to handle.

In summary, spokes are used to maintain strength while keeping the wheel lightweight, thereby reducing rotational resistance and improving performance.

Explanation: This question deals with the design principle of a flywheel, which is used to store rotational energy. The efficiency of energy storage depends on the moment of inertia of the system.

The key concept is that rotational kinetic energy is given by (1/2)Iω². Increasing the moment of inertia allows the flywheel to store more energy at the same angular speed. Since inertia depends on r², placing more mass farther from the axis significantly increases the total inertia.

Step by step, by concentrating mass at the rim, the distance from the axis becomes maximum, which boosts the moment of inertia. This allows the flywheel to resist changes in rotational speed and maintain steady motion even when external forces fluctuate.

As an analogy, spinning a dumbbell with weights at the ends is harder to stop than spinning a compact object. The outward mass increases resistance to change.

In summary, concentrating mass at the rim enhances the flywheel’s ability to store rotational energy and maintain steady motion.

Explanation: This question asks you to identify which factor does not influence the moment of inertia of a rigid body. Moment of inertia depends on mass, its distribution, and the chosen axis of rotation.

The underlying concept is that inertia is a geometric and mass-related property. It is determined by summing contributions of all mass elements as m·r². Therefore, both the amount of mass and how far it is from the axis are important.

Step by step, consider possible influencing factors: changing mass changes inertia, altering shape or distribution changes distances of particles, and shifting the axis changes r values. However, motion-related quantities do not affect inertia itself.

For instance, whether an object spins fast or slow does not change its resistance to rotational acceleration. That resistance is already fixed by its structure.

In summary, moment of inertia is independent of motion-related factors and depends only on mass distribution and axis configuration.

Explanation: This question explores the analogy between rotational and translational motion. Many physical quantities in rotational motion have direct counterparts in linear motion.

The key concept is that moment of inertia plays a role similar to mass. In linear motion, mass resists acceleration when a force is applied. Similarly, in rotational motion, moment of inertia resists angular acceleration when torque is applied.

Step by step, Newton’s second law in linear form is F = ma, while in rotational form it is τ = Iα. Here, mass m corresponds to moment of inertia I, force F corresponds to torque τ, and acceleration a corresponds to angular acceleration α.

As an analogy, pushing a heavy cart requires more effort than a Light one. Similarly, rotating an object with higher inertia requires more torque.

In summary, moment of inertia is the rotational equivalent of mass, representing resistance to changes in motion.

Explanation: This question compares different bodies of equal mass and radius to determine which has the greatest resistance to rotational motion. The deciding factor is how mass is distributed relative to the axis.

The key principle is that moment of inertia increases when more mass is located farther from the axis. Since contributions depend on r², outer regions have a stronger effect.

Step by step, consider typical shapes: a ring has all its mass at the outermost radius, while a disc and sphere have portions of mass closer to the center. Therefore, the ring’s mass distribution maximizes inertia.

As an example, spinning a hoop requires more effort than spinning a Solid ball of the same size and mass because more mass lies farther out.

In summary, the body with mass concentrated farthest from the axis will have the largest moment of inertia.

Explanation: This question again emphasizes identifying factors that do not influence moment of inertia. The concept is rooted in how inertia is defined based on physical structure.

Moment of inertia depends on mass, how that mass is distributed, and the axis about which rotation occurs. It is a purely structural property, independent of how the object is moving at any instant.

Step by step, if you change the mass or redistribute it, inertia changes. Similarly, changing the axis alters distances of mass elements. However, changing how fast the body rotates does not alter the inherent distribution of mass.

For instance, a spinning wheel has the same moment of inertia whether it rotates slowly or rapidly. Only its kinetic energy changes, not its inertia.

In summary, moment of inertia remains unaffected by motion-related parameters and depends only on physical configuration.

Explanation: This question examines how changing the axis of rotation affects different physical quantities. Some quantities depend on the axis, while others are intrinsic to the body.

The key idea is that moment of inertia and radius of gyration depend on the axis position because they involve distances of mass elements from that axis. Changing the axis alters these distances and thus their values.

Step by step, if the axis shifts, the distribution of mass relative to that axis changes, affecting inertia. However, the total mass of the body remains constant regardless of axis placement, as it is an intrinsic property.

For example, rotating a rod about its center or its end changes inertia but does not change its mass.

In summary, intrinsic properties like mass remain unchanged when the axis of rotation is altered.

Explanation: This question deals with rolling motion and how mass distribution affects acceleration. When objects roll without slipping, both translational and rotational motions are involved.

The key concept is that acceleration depends on how energy is divided between translation and rotation. Objects with smaller moment of inertia convert more energy into forward motion, resulting in greater acceleration.

Step by step, a Solid cylinder has more mass distributed closer to the axis compared to a hollow cylinder, which has mass concentrated at the outer surface. This gives the Solid cylinder a lower moment of inertia.

As an analogy, imagine pushing two wheels: one with weight evenly spread and another with weight only at the rim. The evenly spread one accelerates more easily.

In summary, the object with lower rotational inertia accelerates faster and reaches the bottom earlier.

Explanation: This question compares rolling motion of different shapes with identical mass. The determining factor is again the distribution of mass and resulting moment of inertia.

The key principle is that lower moment of inertia leads to higher translational acceleration when rolling. Each shape has a characteristic inertia depending on how mass is spread.

Step by step, among common shapes, a solid sphere has the most mass concentrated toward the center, followed by a cylinder and then a disc. This means the sphere has the smallest inertia and thus accelerates more efficiently.

For example, a compact ball rolls down faster than flatter or more spread-out objects because less energy is used in rotation.

In summary, the object with the most centrally concentrated mass will roll down fastest due to lower rotational resistance.

Explanation: This question examines how the time period of a simple pendulum depends on its length and asks you to determine the original length based on a percentage change in period. The relation between time period and length is fundamental here.

The key concept is that the time period T of a simple pendulum varies as the square root of its length, expressed as T ∝ √L. This means any percentage change in length does not produce the same percentage change in time period; instead, it follows a square root relationship.

Step by step, if the period decreases by 10%, it becomes 0.9T. Squaring both sides gives a relation between new and original lengths. Since T ∝ √L, we get (T’/T)² = L’/L. Substituting values allows us to relate the reduced length (L − 19 cm) to the original length.

For example, if you slightly shorten a pendulum, the swing becomes faster, but not proportionally—because the dependence is not linear but through a square root.

In summary, by using the square root relationship between time period and length, the original length can be determined from the given percentage change.

Explanation: This question involves conservation of mechanical energy in a pendulum system, comparing the speed at the lowest point and at some angular displacement. The idea is that energy continuously transforms between kinetic and potential forms.

The key concept is that at the lowest point, the pendulum has maximum kinetic energy and minimum potential energy. As it rises, some kinetic energy converts into gravitational potential energy, reducing its speed.

Step by step, the change in height of the bob at an angle θ is given by h = L(1 − cosθ). This height gain increases potential energy by mgh, which comes from the loss in kinetic energy. Using conservation of energy, initial kinetic energy equals final kinetic energy plus gained potential energy.

For example, think of a swing: it moves fastest at the bottom and slows down as it climbs upward, because energy is being converted into height.

In summary, the speed at any angle can be found by equating energy at the lowest point and at the given angular position using conservation principles.

Explanation: This question again uses the relationship between time period and length of a simple pendulum to determine the original length when the period changes significantly.

The key concept is T ∝ √L. If the period doubles, then T’ = 2T. Squaring both sides gives a relation between new and original lengths: (T’/T)² = L’/L.

Step by step, substituting T’ = 2T gives L’ = 4L. Since the new length is given as L + 1.2 m, we equate L + 1.2 = 4L. Solving this equation yields the original length.

As an analogy, increasing the length of a pendulum makes it swing more slowly, similar to how a longer rope swing takes more time to complete one Oscillation.

In summary, by applying the square relationship between time period and length, the original length can be calculated from the given condition.

Explanation: This question tests understanding of how the time period of a pendulum changes when its length is scaled. The relationship between time period and length is crucial.

The key concept is that time period varies as the square root of length, T ∝ √L. This means if length is multiplied by a factor, the time period is multiplied by the square root of that factor.

Step by step, if the new length becomes 4L, then the new time period becomes T’ = T√4 = 2T. Substituting the initial value of 2 s allows determination of the new period.

For example, doubling the length does not double the time period; instead, it increases by a smaller factor because of the square root dependence.

In summary, scaling the length changes the period according to a square root rule, not a direct proportionality.

Explanation: This question compares the behavior of timekeeping devices under different gravitational conditions. It specifically focuses on a spring-based watch rather than a pendulum clock.

The key idea is that a spring watch operates based on elastic restoring forces and oscillations of a balance wheel, which do not depend on gravity. Unlike pendulum clocks, whose period depends on g, spring watches rely on internal Mechanical Properties.

Step by step, since the restoring force in a spring system depends on its stiffness and not on gravitational acceleration, the Oscillation frequency remains unchanged even when gravity varies.

For example, a pendulum clock would behave differently on the moon, but a spring watch functions like a tiny oscillator independent of external gravitational fields.

In summary, devices based on spring oscillations are unaffected by changes in gravity, so their timekeeping remains consistent.

Explanation: This question examines whether the mass of the pendulum bob affects the time period of Oscillation. It highlights an important property of simple pendulums.

The key concept is that the time period of a simple pendulum depends only on its length and the acceleration due to gravity, not on the mass of the bob. The formula T = 2π√(L/g) contains no mass term.

Step by step, increasing the mass changes gravitational force and inertia proportionally, so their effects cancel out. As a result, the Oscillation period remains unchanged.

For example, a heavy and a Light pendulum of the same length will swing with the same timing, even though their weights differ.

In summary, changing the mass of the bob does not influence the time period of a simple pendulum.

Explanation: This question involves small percentage changes in pendulum length and how they affect the time period. The square root relationship plays a central role.

The key idea is that T ∝ √L, so a small fractional change in length results in half that fractional change in time period. This approximation works well for small variations.

Step by step, if length increases by 2%, then L becomes 1.02L. The new period becomes T’ = T√1.02. Using approximation, √1.02 ≈ 1.01, meaning the time period increases slightly.

For example, a slight increase in rope length in a swing leads to a barely noticeable increase in time taken for one Oscillation.

In summary, small increases in length lead to smaller proportional increases in time period due to the square root dependence.

Explanation: This question connects simple harmonic motion with maximum velocity using amplitude and time period. The motion of a pendulum for small angles approximates SHM.

The key concept is that maximum velocity in SHM is given by v_max = ωA, where ω is angular frequency and A is amplitude. Angular frequency is related to time period as ω = 2π/T.

Step by step, convert amplitude into meters, calculate angular frequency, and multiply by amplitude to get maximum velocity. This occurs at the mean position where displacement is zero.

For example, a vibrating string moves fastest at the center because all energy is kinetic at that point.

In summary, maximum velocity depends on both amplitude and frequency of Oscillation, and occurs at the equilibrium position.

Explanation: This question investigates whether amplitude affects the time period of a simple pendulum. It highlights an important property of SHM under small oscillations.

The key concept is that for small angles, the time period of a simple pendulum is independent of amplitude. It depends only on length and gravity.

Step by step, reducing amplitude changes the maximum displacement and energy, but does not alter the restoring force relation significantly for small oscillations. Hence, the period remains unchanged.

For example, whether you swing slightly or a bit more (within small limits), the time taken for one Oscillation remains nearly the same.

In summary, small changes in amplitude do not affect the time period of a simple pendulum.

Explanation: This question compares gravitational acceleration at different locations and its effect on pendulum time period. Equal periods imply equal effective gravity.

The key concept is that time period T ∝ 1/√g. For the period to remain the same at two places, the value of g must be equal at those locations.

Step by step, gravity decreases with height above Earth’s surface and also decreases with depth below the surface. Using approximations for variation of g with height and depth, equality conditions can be SET.

For example, moving upward reduces gravitational pull due to increased distance, while moving downward reduces it due to less enclosed mass.

In summary, by equating gravitational variations at both positions, the required ratio of height to depth can be determined.

Explanation: This question explores how the time period of a pendulum depends on gravitational acceleration and how its length must be adjusted when moved to a different celestial body. A seconds pendulum has a fixed time period of 2 seconds.

The key concept is the relation T = 2π√(L/g), which shows that for a fixed time period, length is directly proportional to gravitational acceleration, i.e., L ∝ g. Since gravity on the moon is much less than on Earth, the required length must adjust accordingly.

Step by step, if g decreases, then to keep T constant, L must decrease proportionally. Otherwise, the pendulum would take more time to complete one Oscillation. This ensures that the ratio L/g remains constant.

For example, a pendulum that keeps correct time on Earth would swing more slowly on the moon unless its length is shortened.

In summary, to maintain the same time period in a lower gravitational field, the pendulum’s length must be reduced appropriately.

Explanation: This question connects free fall motion with pendulum motion by first determining the gravitational acceleration on the planet and then using it to find the pendulum’s time period.

The key concept is that for a freely falling body starting from rest, distance covered is given by s = (1/2)gt². Using this, gravitational acceleration g can be calculated. Then, the pendulum formula T = 2π√(L/g) can be applied.

Step by step, substitute s = 8 m and t = 2 s into the motion equation to find g. Once g is known, substitute L = 1 m into the pendulum formula to calculate the time period.

For example, if gravity is weaker, objects fall more slowly and pendulums take longer to complete oscillations.

In summary, the problem involves first finding gravitational acceleration from free fall data and then using it to compute the pendulum’s time period.

Explanation: This question involves synchronization of two pendulums with different lengths and asks when they will return to the same phase together. The concept of time period and frequency is essential.

The key idea is that time period T ∝ √L. Thus, their periods differ, and their oscillations go out of phase over time. To find when they realign, we compare the number of oscillations completed in equal time.

Step by step, calculate the ratio of their time periods using square roots of their lengths. Then invert to get frequency ratio. The pendulums will be in the same phase again when the number of oscillations forms a common multiple.

For example, like two clocks ticking at slightly different speeds, they occasionally align after a certain number of cycles.

In summary, synchronization occurs when both complete whole-number oscillations in the same time interval, determined by their period ratio.

Explanation: This question models oscillation of a ball in a concave surface, which behaves like simple harmonic motion for small displacements. The geometry of the surface determines the restoring force.

The key concept is that such motion can be approximated as SHM with time period T = 2π√(R/g), where R is the radius of curvature. The restoring force arises due to gravity acting along the curved surface.

Step by step, substitute the given radius into the formula. The motion resembles that of a pendulum whose effective length is equal to the radius of curvature of the mirror.

For example, a marble placed in a shallow bowl oscillates back and forth, similar to a pendulum swinging under gravity.

In summary, the curved surface creates a restoring force leading to SHM, and the time period depends on the radius of curvature.

Explanation: This question checks whether amplitude affects the time period of a simple pendulum under small oscillations. It is a conceptual application of SHM properties.

The key concept is that for small angular displacements, the time period is independent of amplitude. It depends only on length and gravitational acceleration.

Step by step, decreasing amplitude changes the maximum displacement and energy, but does not affect the restoring force relationship significantly for small angles. Hence, the period remains essentially unchanged.

For example, a child on a swing takes nearly the same time for each swing whether moving slightly or moderately.

In summary, small variations in amplitude do not influence the time period of a simple pendulum.

Explanation: This question involves understanding how time period changes with gravitational acceleration. It requires applying proportional relationships.

The key concept is T ∝ 1/√g. This means that if gravity increases, the time period decreases slightly.

Step by step, if g becomes 1.02g, then the new period T’ = T/√1.02. Using approximation, √1.02 ≈ 1.01, so T’ becomes slightly less than T.

For example, stronger gravity pulls the pendulum back faster, reducing the time taken for each oscillation.

In summary, increasing gravitational acceleration causes a small decrease in the time period of a pendulum.

Explanation: This question asks for the length of a pendulum that has a time period of 2 seconds, known as a seconds pendulum, under a given gravitational acceleration.

The key formula is T = 2π√(L/g). Rearranging gives L = gT²/(4π²). Substituting T = 2 s and g = 980 cm/s² allows calculation of length.

Step by step, squaring the time period and dividing by 4π² yields the required length. This is a standard value often approximated in Physics problems.

For example, seconds pendulums were historically used in clocks because of their consistent timing.

In summary, applying the pendulum formula with given values yields the approximate length required for a 2-second period.

Explanation: This question examines how effective gravity changes in a non-inertial frame like a rocket and how it affects the pendulum’s time period.

The key concept is that time period depends on effective gravitational acceleration. In an accelerating frame, effective gravity changes depending on the direction of acceleration.

Step by step, if the rocket accelerates upward, effective gravity increases (g + a), reducing the time period. If it accelerates downward, effective gravity decreases, increasing the period.

For example, standing in an accelerating lift feels heavier or lighter depending on its motion, which reflects changes in effective gravity.

In summary, the pendulum’s time period decreases when effective gravitational acceleration increases due to upward acceleration.

Explanation: This question requires using the relationship between time period and length to determine the original length based on a percentage change.

The key concept is T ∝ √L. A 20% change in time period corresponds to squaring the ratio to relate lengths.

Step by step, if T’ = 1.2T, then (T’/T)² = L’/L = 1.44. Since L’ = L + 22 cm, substituting and solving the equation yields the original length.

For example, increasing the length of a pendulum makes it swing slower, but the effect is governed by a square root relation.

In summary, by converting percentage change in period into a squared ratio, the original length can be determined.

Explanation: This question relates SHM parameters to maximum acceleration. It requires using the relationship between angular frequency and acceleration.

The key concept is that maximum acceleration in SHM is a_max = ω²A, where ω = 2π/T. Both amplitude and time period are given.

Step by step, calculate angular frequency using ω = 2π/T. Then square it and multiply by amplitude (converted into proper units) to find maximum acceleration.

For example, in oscillatory motion, acceleration is highest at extreme positions where displacement is maximum.

In summary, maximum acceleration depends on both amplitude and square of angular frequency in SHM.

Explanation: This question connects maximum velocity and maximum acceleration in simple harmonic motion to determine the time period. It relies on relationships between SHM parameters.

The key concept is that in SHM, maximum velocity is v_max = ωA and maximum acceleration is a_max = ω²A. These expressions involve angular frequency ω and amplitude A.

Step by step, dividing a_max by v_max gives ω. Since it is given that acceleration is twice the velocity numerically, substituting into the ratio yields ω = 2. Once ω is known, the time period is found using T = 2π/ω.

For example, in oscillatory motion, if acceleration grows faster relative to velocity, it indicates a higher angular frequency.

In summary, by relating acceleration and velocity expressions, angular frequency can be determined, which directly gives the time period.

Explanation: This question asks for maximum acceleration in SHM using given time period and amplitude. It involves applying standard SHM formulas.

The key concept is that maximum acceleration is a_max = ω²A, where ω = 2π/T. Thus, both amplitude and time period determine acceleration.

Step by step, first calculate angular frequency using ω = 2π/T. Then square this value and multiply by amplitude to obtain the maximum acceleration. Proper unit consistency must be maintained.

For example, in oscillating systems like springs or pendulums, the acceleration is strongest at the extreme positions where displacement is maximum.

In summary, maximum acceleration in SHM depends on amplitude and the square of angular frequency derived from the time period.

Explanation: This question relates velocity and acceleration in SHM to determine angular velocity. It uses standard relationships between motion parameters.

The key idea is that maximum velocity is v_max = ωA and maximum acceleration is a_max = ω²A. Dividing acceleration by velocity eliminates amplitude.

Step by step, compute a_max/v_max = ω. Since both values are given, substituting them directly yields angular frequency.

For example, if acceleration increases more rapidly than velocity, it indicates a higher angular frequency of oscillation.

In summary, the ratio of maximum acceleration to maximum velocity directly gives the angular velocity in SHM.

Explanation: This question involves extracting SHM parameters from a displacement equation and determining maximum speed.

The key concept is that in the equation y = A sin(ωt + φ), amplitude A and angular frequency ω can be identified directly. Maximum speed is given by v_max = ωA.

Step by step, compare the given equation with the standard form to identify A = 0.5 and ω = 5. Multiply these values to obtain maximum speed. Units should be consistent with the given expression.

For example, a particle oscillating with larger amplitude or higher frequency will have greater maximum speed at the mean position.

In summary, maximum speed in SHM is obtained by multiplying angular frequency with amplitude extracted from the displacement equation.

Explanation: This question builds on the previous SHM equation and asks for the time period. It focuses on identifying angular frequency.

The key concept is that time period T is related to angular frequency by T = 2π/ω. The amplitude does not affect the period in SHM.

Step by step, from the given equation, ω is identified as 5. Substitute this into the formula T = 2π/ω to calculate the period.

For example, regardless of how far a particle oscillates, the time taken for one complete cycle depends only on angular frequency.

In summary, time period in SHM is determined solely by angular frequency, independent of amplitude.

Explanation: This question requires determining the time taken to move between two positions in SHM, starting from the extreme.

The key concept is that displacement in SHM can be expressed as x = A cos(ωt) when motion starts from the extreme. Here, ω = 2π/T.

Step by step, substitute the given values and SET x = A − 1 cm. Solve the resulting equation for time using trigonometric relations. This gives the time required to reach the specified displacement.

For example, motion near the extreme is slower because velocity is small, so covering small distances takes relatively more time compared to near the mean position.

In summary, time between positions in SHM can be found using displacement equations and angular frequency.

Explanation: This question compares uniform circular motion with simple harmonic motion to identify the nature of motion.

The key concept is that uniform circular motion is Periodic, meaning it repeats after equal intervals of time. However, SHM requires motion along a straight line with restoring force proportional to displacement.

Step by step, although projection of circular motion on a diameter gives SHM, the circular motion itself is not SHM because acceleration is directed toward the center, not along a straight line.

For example, a point on a rotating wheel repeats its motion but does not oscillate back and forth along a line.

In summary, uniform circular motion is Periodic but differs from simple harmonic motion in its path and force characteristics.

Explanation: This question uses the relationship between velocity, acceleration, and angular frequency in SHM.

The key idea is that v_max = ωA and a_max = ω²A. Dividing acceleration by velocity eliminates amplitude.

Step by step, compute ω = a_max/v_max. Substituting given values directly gives the angular velocity.

For example, higher acceleration relative to velocity indicates a faster oscillation rate.

In summary, angular velocity can be obtained directly from the ratio of maximum acceleration to maximum velocity.

Explanation: This question deals with capillary rise and how it depends on the radius or diameter of the tube.

The key concept is that height of capillary rise h ∝ 1/r, meaning it is inversely proportional to the radius (or diameter). Thus, smaller tubes produce greater rise.

Step by step, write the ratio of heights in terms of inverse radii. Then invert the relationship to obtain the ratio of diameters.

For example, a thinner straw draws liquid higher than a thicker one due to stronger capillary action.

In summary, the ratio of diameters is inversely related to the ratio of heights in capillary rise.

Explanation: This question examines how mass of liquid in a capillary tube depends on its radius and height.

The key concept is that mass is proportional to volume, which depends on cross-sectional area and height. Area ∝ r², while height h ∝ 1/r.

Step by step, combining these gives mass ∝ r² × (1/r) = r. Thus, if radius is doubled, mass also doubles.

For example, a wider tube holds more liquid even though the height decreases, because area increases significantly.

In summary, mass of liquid in a capillary tube is directly proportional to its radius under these conditions.

Explanation: This question involves comparing capillary rise for two different liquids in the same tube. It uses the relation between surface tension, density, and angle of contact.

The key concept is that capillary rise h ∝ (T cosθ)/ρ, where T is surface tension, θ is angle of contact, and ρ is density. Since the tube remains the same, radius cancels out when taking ratios.

Step by step, compare the new liquid with the reference liquid by forming a ratio of heights. Substitute the given values of surface tension, cosine of angle, and density to determine how the height changes relative to the original case.

For example, a liquid with higher surface tension tends to rise more, but this effect can be reduced if density is also higher or if the angle of contact reduces the cosine term.

In summary, capillary rise depends on a balance between surface tension, density, and contact angle, which together determine the final height.

Explanation: This question examines how the mass of liquid in a capillary tube changes when the radius is altered. It combines the concepts of capillary rise and volume.

The key idea is that mass is proportional to volume, which depends on cross-sectional area and height. Area ∝ r², while height h ∝ 1/r. Combining these gives mass ∝ r.

Step by step, if radius doubles, then the mass of the liquid column also doubles because the increase in area outweighs the decrease in height.

For example, a wider tube allows more liquid to occupy space despite the reduced height, resulting in greater total mass.

In summary, the mass of liquid in a capillary tube increases directly with the radius when other factors remain constant.

Explanation: This question focuses on the inverse relationship between capillary rise and tube radius. It asks for the ratio of radii based on observed heights.

The key concept is h ∝ 1/r, meaning height is inversely proportional to radius. Therefore, the ratio of radii is the inverse of the ratio of heights.

Step by step, write r₁/r₂ = h₂/h₁. Substitute the given heights and simplify to obtain the ratio of radii.

For example, a narrower tube shows a higher rise compared to a wider one due to stronger capillary action.

In summary, the ratio of radii can be determined by inverting the ratio of observed heights.

Explanation: This question compares capillary rise for two liquids with different densities in the same tube.

The key concept is that capillary rise is inversely proportional to density, i.e., h ∝ 1/ρ. Since specific gravity represents density relative to water, it can be used directly.

Step by step, form a ratio of heights using inverse density relation: h₁/h₂ = ρ₂/ρ₁. Substitute the given values and solve for the new height.

For example, denser liquids rise less in a capillary tube because their weight counteracts the upward pull of surface tension.

In summary, increasing density reduces capillary rise, and the new height can be calculated using inverse proportionality.

Explanation: This question relates excess pressure inside soap bubbles to their radii and volumes. It uses the concept of surface tension.

The key idea is that excess pressure in a soap bubble is given by ΔP ∝ 1/r. Thus, pressure is inversely proportional to radius.

Step by step, if one bubble has double the pressure, its radius must be half. Since volume ∝ r³, the ratio of volumes becomes the cube of the ratio of radii.

For example, smaller bubbles have higher internal pressure compared to larger ones, which is why tiny bubbles tend to collapse more easily.

In summary, volume ratio depends on the cube of radius ratio, which is derived from the inverse relationship between pressure and radius.

Explanation: This question distinguishes between vertical height and actual length of liquid column in an inclined capillary tube.

The key concept is that vertical height remains the same regardless of inclination, but the length along the tube increases depending on the angle.

Step by step, the vertical height is related to the inclined length by h = L cosθ. Rearranging gives L = h/cosθ. Substituting θ = 30° yields the required length.

For example, climbing a slope requires covering more distance than moving straight up vertically.

In summary, while vertical rise remains constant, the actual length of liquid column increases as the tube is inclined.

Explanation: This question is similar to the previous one and focuses on converting vertical rise into actual length in an inclined tube.

The key concept is the geometric relation between vertical height and inclined length: h = L cosθ.

Step by step, rearrange to L = h/cosθ. Substitute h = 3 cm and θ = 60° to calculate the length along the tube.

For example, a ladder placed at an angle reaches the same height but has a longer length compared to vertical height.

In summary, inclination increases the length of the liquid column while keeping vertical height unchanged.

Explanation: This question uses the inverse relationship between capillary rise and tube radius to find the new height.

The key idea is h ∝ 1/r. Thus, the ratio of heights is inversely proportional to the ratio of radii.

Step by step, write h₁/h₂ = r₂/r₁. Substitute given values and solve for the new height.

For example, thinner tubes show greater rise, while thicker tubes show reduced rise.

In summary, increasing the radius of the tube decreases the height of capillary rise proportionally.

Explanation: This question requires applying the formula for capillary rise to determine the radius of the tube.

The key concept is h = (2T cosθ)/(ρgr). Rearranging gives r = (2T cosθ)/(ρgh). For water in glass, cosθ is typically taken as 1.

Step by step, substitute the given values of surface tension, height, and known constants to compute the radius.

For example, narrower tubes produce higher rise, so a given height can indicate how small the radius is.

In summary, radius can be determined by rearranging the capillary rise formula and substituting known values.

Explanation: This question examines how capillary rise changes when the tube’s cross-sectional area is altered.

The key idea is that area ∝ r², so reducing area to one-fourth reduces radius to half. Since height h ∝ 1/r, halving the radius doubles the height.

Step by step, determine the new radius from area relation, then apply inverse proportionality to find the new height.

For example, making a tube narrower increases the capillary rise significantly.

In summary, reducing cross-sectional area increases capillary rise due to the inverse dependence on radius.

Explanation: This question relates capillary rise to temperature, which affects surface tension and density of water. It asks how height varies with temperature conditions.

The key concept is that capillary rise h ∝ T/ρ, where T is surface tension and ρ is density. Both these properties vary with temperature, but surface tension decreases more significantly as temperature increases.

Step by step, as temperature rises, intermolecular forces weaken, reducing surface tension. Since capillary rise depends directly on surface tension, the height decreases. Around certain temperatures, water properties behave uniquely, affecting this balance.

For example, warmer water forms weaker surface films, so it does not climb as high in narrow tubes compared to cooler water.

In summary, capillary rise depends strongly on temperature through surface tension, and changes in temperature alter the height accordingly.

Explanation: This question examines whether the angle of contact changes when the orientation of the capillary tube is altered.

The key concept is that angle of contact depends only on the nature of the liquid and the material of the tube, not on the orientation of the tube. It is determined by intermolecular forces at the interface.

Step by step, tilting the tube changes the geometry of the liquid column but does not affect Molecular interactions at the boundary. Hence, the angle of contact remains unchanged.

For example, whether a glass is upright or tilted, the way water wets the surface remains the same.

In summary, angle of contact is an intrinsic property of the liquid-solid pair and does not depend on the tube’s inclination.

Explanation: This question deals with pressure difference across the surface of a soap bubble due to surface tension.

The key concept is that a soap bubble has two liquid surfaces (inner and outer), so the excess pressure is higher compared to a liquid drop. The relation involves surface tension and radius.

Step by step, pressure difference arises because surface tension tries to minimize surface area, creating inward force. For a soap bubble, contributions from both surfaces are considered, leading to a higher pressure expression.

For example, blowing a soap bubble requires effort because internal pressure must overcome surface tension forces.

In summary, excess pressure in a soap bubble depends on surface tension and radius, with contributions from both surfaces increasing the effect.

Explanation: This question explores the effect of angle of contact on capillary rise or fall of a liquid in a tube.

The key concept is that capillary rise depends on cosθ. When θ = 90°, cosθ = 0, which eliminates the driving force for capillary action.

Step by step, since capillary rise formula includes cosθ in the numerator, a zero value means no upward or downward movement occurs. The liquid surface remains level with the external surface.

For example, if a liquid neither wets nor repels the surface, it does not climb or dip in the tube.

In summary, when angle of contact is 90°, there is no capillary rise or fall.

Explanation: This question asks for the physical reason behind capillary rise of water in a narrow tube.

The key concept is that surface tension creates a curved meniscus, leading to pressure difference across the liquid surface. The pressure just below the concave surface becomes lower than atmospheric pressure.

Step by step, this pressure difference pushes the liquid upward until the weight of the column balances the upward force due to surface tension. Adhesion between liquid and tube also contributes.

For example, water climbing up a thin straw is due to intermolecular forces pulling it upward against gravity.

In summary, capillary rise occurs due to pressure difference created by surface tension and adhesive forces.

Explanation: This question relates to capillary action and how porous materials interact with liquids like ink.

The key concept is that blotting paper has many tiny pores that act like capillaries. These draw ink rapidly into the paper due to capillary action.

Step by step, when ink touches such paper, it spreads quickly in all directions instead of staying at the writing point. This prevents clear marking or writing.

For example, spilling ink on tissue paper causes it to spread instantly rather than forming defined lines.

In summary, strong capillary action in porous materials absorbs ink quickly, making writing ineffective.

Explanation: This question examines conditions under which capillary rise increases beyond its normal value.

The key concept is that capillary rise depends inversely on effective gravity. If effective gravity decreases, the liquid can rise higher.

Step by step, in a downward accelerating frame, effective gravity reduces (g − a). This allows the same surface tension force to lift the liquid to a greater height.

For example, in a falling lift, objects feel lighter, and liquids behave as if gravity is reduced.

In summary, reducing effective gravity increases capillary rise beyond its usual value.

Explanation: This question considers what happens when the available tube length is less than the natural capillary rise height.

The key concept is that liquid rises until forces balance. If the tube is shorter than required height, the liquid reaches the top before equilibrium.

Step by step, since the liquid cannot rise beyond the tube, it reaches the top and may form a curved surface or spill depending on conditions.

For example, filling a narrow tube partially and placing it in water may cause overflow if the natural rise exceeds the tube length.

In summary, when tube length is insufficient, the liquid reaches the top and alters its surface shape accordingly.

Explanation: This question asks about the mechanism that allows ink to flow inside a pen refill.

The key concept is capillary action, where narrow spaces allow liquid to move due to surface tension and adhesive forces.

Step by step, the small gap between the ball and the housing acts like a capillary tube, drawing ink toward the tip as writing occurs.

For example, ink moving through a narrow straw without external force demonstrates capillary action.

In summary, ink flow in a refill is primarily governed by capillary forces.

Explanation: This question involves pressure difference across a curved liquid surface, specifically a concave meniscus.

The key concept is that pressure inside a curved liquid surface differs from outside due to surface tension. For a concave surface, pressure below the surface is lower.

Step by step, surface tension creates inward force, reducing pressure just beneath the meniscus compared to above it. This difference helps drive capillary rise.

For example, water in a glass tube forms a concave meniscus and rises due to lower pressure below the curved surface.

In summary, pressure below a concave meniscus is less than the pressure above it due to surface tension effects.

Explanation: This question focuses on the pressure difference inside a liquid drop due to surface tension. Unlike soap bubbles, a liquid drop has only a single surface.

The key concept is that surface tension creates an inward force along the curved surface, resulting in higher pressure inside the drop compared to outside. For a liquid drop, the excess pressure is proportional to surface tension and inversely proportional to radius.

Step by step, consider that surface tension tries to contract the surface. To maintain equilibrium, the internal pressure must balance this inward pull. Since there is only one interface, the contribution is less than that of a soap bubble, which has two surfaces.

For example, smaller droplets feel tighter and have higher internal pressure, which is why very small drops tend to merge into larger ones to reduce surface energy.

In summary, excess pressure inside a liquid drop depends on surface tension and radius, with smaller drops having higher internal pressure.