Mechanics: notes and MCQs for State PSC prelims

Mechanics is part of Physics. Below are notes on 12 sub-topics, followed by practice MCQs from State PSC and UPSC papers.

Conservation of Linear and Angular Momentum

Conservation of Linear Momentum: The total linear momentum of an isolated system (where the net external force is zero, Fext = 0) remains constant.
Newton's Third Law Connection: Conservation of linear momentum is a direct consequence of Newton's Third Law of Motion.
Applications of Linear Momentum Conservation:

Recoil: E.g., The recoil velocity of a pistol when a bullet is fired. If initial momentum is zero, then Mbullet Vbullet + Mpistol Vrecoil = 0.
Propulsion: Rockets and jet engines operate on this principle by expelling mass (exhaust gases or air) at high velocity in one direction, causing the vehicle to move in the opposite direction.
Collisions and Combinations: In systems where external horizontal forces are negligible, linear momentum is conserved (e.g., a girl jumping onto a stationary cart: Mgirl Vgirl = (Mgirl + Mcart) Vfinal; rain falling vertically into a moving wagon: M₁v₁ = M₂v₂ as the rain's vertical momentum does not affect horizontal momentum).
Internal Forces: Internal forces (such as those from chemical reactions or whistling within a system) cannot change the velocity of the system's center of mass.

Variable Mass Systems: For a system gaining mass, like sand falling on a conveyor belt moving at a constant speed, the force required is F = v (dm/dt), where v is the speed and dm/dt is the rate of mass addition.
Relation between Kinetic Energy (KE) and Linear Momentum (P): KE = P² / (2m). If linear momentum changes by a factor 'x', then kinetic energy changes by a factor 'x²' (e.g., if P doubles, KE quadruples).
Fundamental Conservation Laws: Conservation of energy, momentum (linear and angular), and charge are considered fundamental laws of physics.
Conservation of Angular Momentum: The total angular momentum of a system remains constant if the net external torque acting on it is zero (τext = 0).

Elasticity (Stress, Strain, Hooke's Law, Elastic Moduli)

Elasticity: Property of a body to regain its original shape/size after deforming force removal.
Plasticity: Material does not regain original shape/size after deforming force removal (undergoes permanent deformation).
Stress ($\sigma$): Restoring force per unit area. $\sigma = F/A$. Unit: Pascal (Pa) or N/m².
Normal Stress: Force perpendicular to surface (tensile/compressive).
Shear Stress ($\tau$): Force parallel to surface. $\tau = F{tangential}/A$.
Strain ($\epsilon$): Ratio of change in dimension to original dimension. Dimensionless.
Longitudinal Strain: $\Delta L/L$.
Volumetric Strain: $\Delta V/V$.
Shear Strain ($\gamma$): Angular deformation, $\Delta x/L \approx \theta$.
Hooke's Law: Within elastic limit, Stress $ \propto$ Strain (Stress = Modulus of Elasticity $\times$ Strain).
Elastic Limit: Max stress a material can withstand without permanent deformation.
Young's Modulus (E): Ratio of Normal Stress to Longitudinal Strain. $E = (F/A) / (\Delta L/L)$. Unit: Pa.
Bulk Modulus (K): Ratio of Normal Stress (Pressure) to Volumetric Strain. $K = \text{Pressure} / (\Delta V/V)$. Unit: Pa.
Perfectly Rigid Body: Has infinite Bulk Modulus (K = $\infty$), as $\Delta V = 0$.
Shear Modulus (G) / Modulus of Rigidity: Ratio of Shear Stress to Shear Strain. $G = \tau / \gamma$. Unit: Pa.
Poisson's Ratio ($\nu$): Ratio of lateral strain to longitudinal strain. $\nu = - (\Delta D/D) / (\Delta L/L)$. Dimensionless. Practical range: 0 to 0.5.
Relationships between Elastic Constants:

$E = 2G(1 + \nu)$
$E = 3K(1 - 2\nu)$
$G = E / (2(1 + \nu))$

Axial Elongation ($\Delta L$):

Uniform bar under axial load P: $ \Delta L = PL / (AE)$.
Uniform rod due to its self-weight W: $ \Delta L = WL / (2AE)$.
Tapering bar (diameters d1, d2) under axial load P: $\Delta L = (4PL) / (\pi E d1 d2)$.
Equivalent diameter (d) for uniform bar to have same extension as tapering bar: $d = \sqrt{d1 d2}$.

Springs:

Spring constant (k): Measure of stiffness.
If a spring of constant k is cut into 'n' equal parts, each part has a spring constant of 'nk'.
Combined stiffness of two springs $K1, K2$ in series: $1/Ke = 1/K1 + 1/K2$. If $K1=K2$, then $Ke = K1/2$.

Shear Stress in Torsion: For a solid circular shaft, maximum shear stress ($\tau{max}$) is inversely proportional to the cube of the diameter ($d^3$). $\tau{max} \propto T/d^3$. (Doubling diameter reduces $\tau{max}$ to 1/8th for the same torque).
Maximum Shear Stress from Principal Stresses: For plane stress, $\tau{max} = (\sigma1 - \sigma2) / 2$.
Beam Deflection ($\delta$):

Simply supported beam, central point load P, length L: $\delta{max} = PL^3 / (48EI)$. (Where EI is flexural rigidity)
Cantilever beam, point load P at free end, length L: $\delta{max} = PL^3 / (3EI)$.

Fatigue & S-N Curve:

S-N curve plots stress amplitude (S) vs. number of cycles to failure (N).
Endurance Limit (Fatigue Limit): The stress level below which a material can withstand an infinite number of load cycles without fatigue failure.
For steel, the S-N curve becomes asymptotic around $10^6$ cycles, indicating an endurance limit.

Bearing Life (L) vs. Load (P): For ball bearings, $L1/L2 = (P2/P1)^3$.

Fluid Statics and Dynamics (Continuity Equation, Bernoulli's Theorem)

Pressure (P): Force (F) per unit Area (A), P = F/A. A scalar quantity (magnitude of normal force/area, acts equally in all directions in static fluid).
Pressure at depth: In a fluid at rest, P = P₀ + ρgh, where P₀ is surface pressure, ρ is fluid density, g is acceleration due to gravity, h is depth. Pressure is constant at the same horizontal level.
Pascal's Law: Pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.

Applications: Hydraulic brakes, hydraulic lifts, hydraulic press.

Archimedes' Principle: A body wholly or partially immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the body.
Buoyancy: An upward force exerted by a fluid that opposes the weight of an immersed object.

For a floating object: Buoyant Force (Fb) = Weight of object (W).
For a sinking object: Buoyant Force (Fb) < Weight of object (W).
Apparent weight in fluid = Actual weight - Buoyant Force.
Applications: Designing ships/submarines, hydrometers, swimming.

Floating/Sinking Conditions: An object floats if its average density (ρobj) is less than the fluid's density (ρfluid). It sinks if ρobj > ρfluid.

Fraction submerged (f) = ρobj / ρfluid. This fraction is independent of gravity.
Example: Iron nail sinks in water (ρiron > ρwater), iron ship floats (ρavgship < ρwater).
Example: Ice floating in water melts, water level remains unchanged (volume of melted ice equals volume displaced by floating ice).
Example: An object floats on mercury (ρobj < ρmercury) but sinks in water (ρobj > ρwater) if ρwater < ρobj < ρmercury.

Stability of Floating Bodies: For stable equilibrium, the metacentre (M) must be above the centre of gravity (G) of the body.
Weight in Vacuum: An object weighs maximum in vacuum as there is no buoyant force.
Centrifugation: Process of separating particles from a fluid based on density by spinning, e.g., washing machine.
Fluid in Accelerated Container: Free surface of a liquid in a uniformly accelerating container inclines opposite to the direction of acceleration.
Streamline (Laminar) Flow: Steady flow where fluid particles follow paths that do not cross. The velocity at any given point is constant over time. Each particle follows the path of the preceding particle.
Uniform Flow: Velocity is constant in the flow field with respect to space (i.e., at every point in the flow field, the velocity vector is the same at a given instant).
Critical Velocity: The maximum velocity below which fluid flow remains laminar. Above this, it becomes turbulent.
Incompressible Fluid: An ideal fluid whose density does not change with pressure. Its compressibility is zero.
Equation of Continuity: For a steady flow of an incompressible fluid, the volume flow rate (Av) is constant along a streamline. A₁v₁ = A₂v₂.

Derived from the conservation of mass.
Applications: Water stream narrows as it falls (v increases, A decreases); spreads as it rises (v decreases, A increases).

Bernoulli's Theorem: For an ideal (incompressible, non-viscous), steady, irrotational flow of a fluid along a streamline, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant: P + ½ ρv² + ρgh = constant.

Derived from the conservation of energy.
Assumptions: Fluid is incompressible, non-viscous (no friction/viscous forces), flow is steady and irrotational, flow is along a streamline, gravitational forces are considered (not neglected).
Applications: Aerodynamic lift (aeroplane wings), sprayers/atomizers, venturimeters.
Principle: Higher fluid velocity results in lower pressure, and vice-versa.

Dynamic Viscosity (μ): A measure of a fluid's resistance to shear flow.

Dimension: ML⁻¹T⁻¹.
Newtonian Fluids: Shear stress is directly proportional to the rate of shear strain (τ = μ (du/dy)).
Non-Newtonian Fluids: Shear stressstrain rate relationship is not linear (e.g., Bingham plastics, pseudoplastic, dilatant fluids).

Velocity Profile in Pipes: For laminar flow in a circular pipe, the velocity profile is parabolic (zero at walls, maximum at center).
Torricelli's Law: The speed of efflux (v) from an orifice at depth h below the free surface of a liquid is v = √(2gh).
Boundary Layer Separation: Occurs when an adverse pressure gradient (pressure increasing in the direction of flow) causes the fluid flow near a surface to detach.

Gravitation and Central Force Motion (Kepler's Laws)

Newton's Law of Universal Gravitation:

States that every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Formula: $F = G \frac{m1 m2}{r^2}$
Vector form: $\vec{F} = -G \frac{m1 m2}{r^2} \hat{r}$, where $\hat{r}$ is a unit vector from $m1$ to $m2$, indicating an attractive force.
Gravitational Constant (G): A universal constant, approx. $6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$. It is independent of medium, local conditions, or masses.
Properties of Gravitational Force:

Always attractive.
Long-range force.
Central force (acts along the line joining centers).
Conservative force (work done is path independent).
Weakest of the four fundamental forces (negligible for atomic particles, dominant for celestial bodies).
Acts as an action-reaction pair (Newton's 3rd Law), e.g., Earth exerts force on Moon = Moon exerts force on Earth (equal magnitude, opposite direction).

Mass vs. Weight:

Mass (m): Quantity of matter, constant for a given body, independent of location. Unit: kg.
Weight (W): Force exerted on a body due to gravity, $W = mg$. Varies with acceleration due to gravity (g). Unit: Newton (N).
A body's mass remains constant if brought from Moon to Earth, but its weight increases due to higher 'g' on Earth.
In deep space or where 'g' is negligible, weight becomes zero, but mass remains unchanged.

Acceleration Due to Gravity (g):

Formula for 'g' on a planet's surface: $g = G \frac{M}{R^2}$, where M is planet's mass and R is its radius.
Variation of 'g' with Altitude (h): $gh = g \left( \frac{R}{R+h} \right)^2 \approx g \left( 1 - \frac{2h}{R} \right)$ for small h. 'g' decreases with altitude.
Variation of 'g' with Depth (d): $gd = g \left( 1 - \frac{d}{R} \right)$. 'g' decreases linearly with depth from the surface, becoming zero at the Earth's center.
Maximum 'g': Occurs at the surface of the Earth.
Minimum 'g': Occurs at the center of the Earth (g=0).
Variation of 'g' with Latitude ($ \lambda$): $g{\lambda} = g \omega^2 R \cos^2 \lambda$, where $ \omega$ is Earth's angular velocity and R is Earth's radius.

At Equator ($\lambda = 0^\circ$): $\cos^2 0^\circ = 1$, so $g{equator} = g - \omega^2 R$ (minimum value).
At Poles ($\lambda = 90^\circ$): $\cos^2 90^\circ = 0$, so $g{poles} = g$ (maximum value).

Earth's Shape: Earth is an oblate spheroid (flattened at poles, bulged at equator). This means $R{equator} > R{pole}$.

Both rotational effect (centrifugal force) and Earth's shape contribute to $g{equator} < g{pole}$.
Consequently, an object's weight is maximum at the poles and minimum at the equator.

'g' on Moon: Approximately 1/6th of Earth's 'g'. This low gravity is why the Moon has no atmosphere (gas molecules can easily escape).
Ordering of 'g' values: Center of Earth (0) < Moon's surface (approx. 1.62 m/ s²) < Earth's Equator (approx. 9.78 m/ s²) < Earth's Poles (approx. 9.83 m/s²).

Kepler's Laws of Planetary Motion:

First Law (Law of Orbits): All planets move in elliptical orbits with the Sun at one of the two foci.
Second Law (Law of Areas): A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This implies that the planet moves faster when closer to the Sun (perihelion) and slower when farther away (aphelion), due to conservation of angular momentum.
Third Law (Law of Periods): The square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit. $T^2 \propto a^3$. For circular orbits, 'a' is replaced by radius 'r'.

Orbital Mechanics:

Satellite in Orbit: The gravitational force of the central body provides the necessary centripetal force for orbital motion.
Velocity of satellite: In a circular orbit, speed may be constant, but velocity is not constant because direction continuously changes. Thus, a satellite in orbit has centripetal acceleration.
Work Done on Satellite: For a circular orbit, the gravitational force (centripetal force) is perpendicular to the satellite's displacement (velocity), so the work done by gravity is zero.
Weightlessness in Orbit: Astronauts inside an orbiting satellite experience apparent weightlessness (microgravity) because both the astronaut and the satellite are in a continuous state of free fall around the Earth. The Earth's gravity is present but provides the centripetal force, effectively nullifying the normal force.
Total Energy of a Satellite in Orbit: $E = Ek + Ep = \frac{1}{2}mv^2 \frac{GMm}{r} = -\frac{GMm}{2r}$.
Escape Velocity ($ve$): The minimum velocity required for an object to escape the gravitational field of a celestial body and not return.

Formula: $ve = \sqrt{\frac{2GM}{R}}$ or $ve = \sqrt{2gR}$.
Earth's escape velocity: approx. 11.2 km/s.
If a satellite in orbit (with kinetic energy $Ek$) is to escape, its total energy must become zero. Since its total energy is $-Ek$, an additional energy of $Ek$ must be supplied.

Escape Velocity (Density-dependent): If two planets have the same density ($ \rho$), $ve \propto R \sqrt{\rho}$, so $ve \propto R$.

Lagrangian

Formulation

(Generalized

Coordinates, Euler-Lagrange Equations) Generalized Coordinates (qi): A set of inde1. pendent coordinates (e.g., $q1, q2, ..., qn$) that completely specifies the configuration of a system. Degrees of Freedom (DOF): The minimum 2. number of generalized coordinates required to describe a system's configuration. For N particles with K holonomic constraints, DOF = 3N - K. Holonomic Constraints: Constraints that can 3. be expressed as an algebraic equation relating the coordinates and time, $f(q1, ..., qn, t) = 0$. Lagrangian mechanics typically applies to systems with holonomic constraints. Simple Pendulum: A classic example, its 4. position can be completely described by a single angle (e.g., $\theta$). Therefore, a simple pendulum has 1 generalized coordinate (and 1 degree of freedom). Lagrangian (L): Defined as the difference 5. between the system's kinetic energy (T) and its potential energy (V): $L = T - V$. The Lagrangian is a function of generalized 6. coordinates ($qi$), generalized velocities ($ \dot{q}i$), and time ($t$): $L(qi, \dot{q}i, t) $. Hamilton's Principle (Principle of Least 7. Action): States that the path a system takes between two fixed points in time ($t1$ and $t2$) is the one for which the action integral is stationary: $\delta \int{t1}^{t2} L(qi, \dot{q}i, t) dt = 0$. Euler-Lagrange Equations: These are the dif8. ferential equations of motion derived from Hamilton's Principle. For each generalized coordinate $qi$, the equation is: $\frac{d} {dt}\left(\frac{\partial L}{\partial \dot{q}i} \right) - \frac{\partial L}{\partial qi} = 0$. Generalized Momentum (pi): For a general9. ized coordinate $qi$, the corresponding generalized momentum is defined as $pi = \frac{\partial L}{\partial \dot{q}i}$. Cyclic (Ignorable) Coordinate: If the Lag10. rangian L does not explicitly depend on a particular generalized coordinate $qk$ (i.e., $\frac{\partial L}{\partial qk} = 0$), then $qk$ is called a cyclic coordinate. For a cyclic coordinate, its corresponding generalized momentum $pk$ is conserved ($ \frac{d}{dt} pk = 0 \implies pk = \text{constant}$).

Newton's Laws of Motion and their applications

Newton's First Law of Motion (Law of Inertia): An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced external force.
Applications of Newton's First Law:

Explains why passengers jerk backward when a stationary bus starts, or forward when a moving bus suddenly stops.
An object moving with constant velocity implies zero net force acting on it.
A ball rolling on a level ground comes to rest due to opposing friction force.

Newton's Second Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). The rate of change of momentum is directly proportional to the applied force and occurs in the direction of the force.
Momentum (p): Defined as the product of mass (m) and velocity (v) (p = mv). It is a vector quantity.
Impulse: The change in momentum of an object. A goalkeeper pulling hands backward while catching a ball increases the time over which the momentum changes, thereby decreasing the force exerted on the hands.
Newton's Third Law of Motion: For every action, there is an equal and opposite reaction. These forces act on two different bodies simultaneously.
Applications of Newton's Third Law:

Rocket propulsion (exhaust gases pushed down, rocket pushed up).
A sailor jumping forward from a rowing boat causes the boat to move backward.
Walking (pushing the ground backward, ground pushes the person forward).
A ball bouncing on the floor.
A person on a frictionless surface can move by pushing an object away (e.g., by sneezing).

Force: A vector quantity that can cause a change in motion.

Contact Forces: Forces that act when objects are in physical contact (e.g., frictional force, push/pull force, normal force). They satisfy Newton's Third Law.
Non-Contact (Field) Forces: Forces that act without physical contact (e.g., gravitational force, magnetic force).
Gravitational Force: An attractive non-contact force. The weight of an object is due to this force.
Frictional Force: A contact force that opposes relative motion between surfaces. It is non-conservative and can produce heat.

Mass vs. Weight:

Mass: A scalar quantity, an intrinsic property representing the amount of matter and inertia. It is constant everywhere.
Weight: A vector quantity, the force of gravity acting on an object (W = mg). It varies with the local acceleration due to gravity (g), thus varying from place to place. The SI unit of weight is Newton (N).

Scalar vs. Vector Quantities:

Scalar Quantity: Has magnitude only (e.g., speed, mass, distance, energy, pressure, work). Can be combined using ordinary algebra.
Vector Quantity: Has both magnitude and direction (e.g., velocity, acceleration, displacement, force, momentum, torque). Can be combined using vector algebra.

Types of Friction:

Static Friction: Opposes the initiation of motion.
Kinetic (Sliding) Friction: Opposes motion once it has started.
Rolling Friction: Occurs when an object rolls over a surface.
Order of Magnitude: Static friction > Kinetic friction > Rolling friction. It is harder to start a loaded cart (overcome static friction) than to keep it moving (overcome kinetic friction).
Ball Bearings: Used to convert sliding friction into rolling friction, thereby reducing resistance.
Walking on Ice: Difficult due to low friction.

Pressure (P): Defined as Force per unit Area (P = F/A). It is a scalar quantity.

A sharp knife cuts better than a blunt knife because its smaller contact area produces higher pressure for the same force.
A person standing on one foot exerts twice the pressure compared to standing on two feet, as the force (weight) is distributed over half the area.

Kinematics (Study of Motion):

Distance: Total path covered (scalar).
Displacement: Shortest distance from initial to final position (vector). For a round trip, displacement is zero.
Speed: Rate of change of distance (scalar). Average speed = Total Distance / Total Time. For equal distances covered at speeds v₁ and v₂, average speed = 2v₁v₂ / (v₁ + v₂).
Velocity: Rate of change of displacement (vector). Average velocity = Total Displacement / Total Time. If displacement is zero, average velocity is zero.
Acceleration: Rate of change of velocity (vector). Uniform acceleration means constant magnitude and direction of acceleration. Retardation (deceleration) occurs when acceleration opposes velocity.
Uniform Velocity: Implies constant speed in a straight line, zero acceleration, and zero net force. Displacement increases linearly with time.

Graphs of Motion (for straight-line motion):

Displacement-Time (x-t) Graph:

Slope represents velocity.
A horizontal line indicates the object is at rest.
A straight line with positive slope indicates constant positive velocity.
A parabolic curve indicates constant acceleration (since displacement depends quadratically on time: s = ut + ½at²).

Velocity-Time (V-t) Graph:

Slope represents acceleration.
Area under the graph represents displacement.
A horizontal line indicates constant velocity (zero acceleration).
A straight line with positive slope indicates uniform acceleration.
A straight line with negative slope indicates uniform retardation (deceleration).
The equation v = u + at graphs as a straight line with y-intercept 'u' (initial velocity) and slope 'a' (acceleration).

Circular Motion:

Uniform Circular Motion: An object moves with constant speed along a circular path. Its velocity is continuously changing (due to changing direction), so it is an accelerated motion.
Centripetal Acceleration (ac): Always directed towards the center of the circular path, perpendicular to the velocity. Magnitude is ac = v²/r.
Centripetal Force (Fc): The force directed towards the center that keeps an object in uniform circular motion (Fc = mv²/r).
Centrifugal Force: A fictitious force, equal in magnitude and opposite to centripetal force, experienced in a rotating non-inertial frame of reference (directed outwards).
Applications of Centripetal/Centrifugal Force:

Washing machines and cream separators work on the principle of centrifugation.
Banking of railway tracks/roads at curves provides the necessary centripetal force from the horizontal component of the normal force, preventing skidding.
If the centripetal force is removed (e.g., a string holding an object breaks), the object will fly off tangentially due to its inertia of direction.
A vehicle taking a sharp turn at high speed tends to skid outwards and may overturn about its outer wheels.

Apparent Weight in a Lift:

At rest or moving with constant velocity (upwards or downwards): Apparent weight equals actual weight (N = mg).
Accelerating upwards: Apparent weight > Actual weight (N = m(g+a)). The person feels heavier.
Accelerating downwards: Apparent weight < Actual weight (N = m(g-a)). The person feels lighter.
Freefall (lift cable breaks, a=g): Apparent weight = 0 (N=0). The person experiences weightlessness.

Weightlessness in Orbit: Astronauts in an orbiting spacecraft appear weightless not because there is no gravity, but because both the astronaut and the spacecraft are continuously falling (accelerating) towards Earth together. The normal reaction force is zero.
Earth's Rotation and Weight: If the spinning speed of the Earth increases, the centrifugal effect at the equator increases, leading to a decrease in the apparent weight of a body at the equator.
Equilibrium: A body is in equilibrium when the net external force and the net external moment (torque) acting on it are both zero.

Law of Moments: For rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point.
A ball balanced on a vertical rod is an example of unstable equilibrium.

Inertial Frame of Reference: A frame of reference that is either at rest or moving with a constant velocity. Newton's laws of motion are valid in an inertial frame. A driver in a car moving with constant velocity is in an inertial frame.
Conservation of Momentum: The total momentum of an isolated system (no external forces) remains constant. (Note: Kinetic energy is not always conserved, e.g., in inelastic collisions).
Force Perpendicular to Motion: If a force of constant magnitude is always perpendicular to the motion of a particle (e.g., centripetal force), its kinetic energy remains constant, but its velocity (due to direction change) and acceleration (due to direction change) are not constant.

Rigid Body Dynamics (Euler Angles, Tops and Gyroscopes)

Centre of Mass (CM) is the average position of all the mass in a system.
Centre of Gravity (CG) is the point where the entire weight of the body acts.
CM and CG coincide if and only if the acceleration due to gravity ($\vec{g}$) is uniform throughout the system.
The path traced by the midpoint of a ladder sliding between a vertical wall and a horizontal floor is a circular path.
Torque ($\vec{\tau}$) of a force ($\vec{F}$) about a point is given by $\vec{\tau} = \vec{r} \times \vec{F}$, where $\vec{r}$ is the position vector from the point to the point of application of the force.
For $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} $ and $\vec{F} = Fx\hat{i} + Fy\hat{j} + Fz\hat{k}$, the torque is $\vec{\tau} = (yFz zFy)\hat{i} + (zFx - xFz)\hat{j} + (xFy - yFx) \hat{k}$.
Euler Angles (typically $\phi, \theta, \psi$) are a set of three angles used to describe the orientation of a rigid body with respect to a fixed coordinate system.
A Gyroscope is a spinning rigid body whose axis of rotation is free to assume any orientation by itself. It exhibits precession and nutation when subjected to an external torque.
Precession is the change in the orientation of the rotational axis of a spinning body, like a top, under the influence of an external torque, causing the axis to sweep out a cone.
Nutation is a wobbling or nodding motion in the axis of a spinning body, superimposed on its precessional motion.

Rotational Dynamics and Moment of Inertia

Angular Displacement ($\theta$): Angle 1. swept by a rotating body. Unit: radian. Angular Velocity ($\omega$): Rate of change 2. of angular displacement ($\omega = d\theta/ dt$). Unit: rad/s. Vector quantity. Angular Acceleration ($\alpha$): Rate of 3. change of angular velocity ($\alpha = d\omega/dt$). Unit: rad/s². Moment of Inertia (I): Rotational analogue of 4. mass. Measures resistance to change in rotational motion. For a point mass: $I = mr^2$. For a system: $I = \sum mi ri^2$. Unit: kg m². Perpendicular Axis Theorem: For a planar 5. body, $Iz = Ix + Iy$, where $Ix, Iy$ are moments of inertia about two perpendicular axes in the plane, and $Iz$ is about an axis perpendicular to the plane passing through their intersection. Parallel Axis Theorem: $I = I{CM} + 6. Md^2$, where $I{CM}$ is moment of inertia about the center of mass (CM), $I$ is about a parallel axis, $M$ is total mass, and $d$ is the perpendicular distance between the two axes. Moment of Inertia for Common Shapes 7. (about CM, perpendicular to plane for planar objects, or through CM for 3D objects):

Thin Ring/Hollow Cylinder: $I = MR^2$
Thin Disc/Solid Cylinder: $I = (1/2)MR^2$
Solid Sphere: $I = (2/5)MR^2$
Solid Rod (perpendicular to length): $I = (1/12)ML^2$ Torque ($\tau$): Rotational analogue of force 8. ($\tau = I\alpha$). Unit: N m. Vector quantity ($\tau = r \times F$). Rotational Kinetic Energy (KErot): $KE{rot} 9. = (1/2)I\omega^2$. Unit: Joule. Angular Momentum (L): Rotational analogue 10. of linear momentum ($L = I\omega$). Unit: kg m²/s or J s. Vector quantity ($L = r \times p$). Conservation of Angular Momentum: If no 11. external torque acts on a system, its total angular momentum remains constant ($L = I\omega = \text{constant}$).
Spreading arms increases moment of inertia (I) and decreases angular velocity ($\omega$) to conserve angular momentum. Earth's Rotation Perception: Inhabitants do 12. not perceive Earth's rotation due to:
Constant angular velocity for all points on Earth (relative to Earth's axis).
Atmosphere rotating along with the Earth.
Absence of nearby stationary or differentially moving reference objects. Clock Hands Angular Speed: 13.
Minute Hand period ($TM$) = 1 hour.
Hour Hand period ($TH$) = 12 hours.
Angular speed ($\omega = 2\pi/T$).
Ratio of angular speed of Hour Hand to Minute Hand: $\omegaH : \omegaM = (2\pi/TH) : (2\pi/TM) = TM : TH = 1 : 12$. Bicycle Stability: A bicycle in motion remains 14. stable due to the conservation of angular momentum of its spinning wheels. Gears: 15.
Module (m): Pitch Circle Diameter (D) / Number of Teeth (T). Unit: mm.
Circular Pitch ($pc$): Distance measured along the pitch circle from a point on one tooth to a corresponding point on an adjacent tooth. $pc = \pi m$.
Center Distance (C): $C = (m/2)(T1 + T2)$, where $T1, T2$ are number of teeth on pinion and gear respectively.
Speed Reduction Ratio: $N{pinion}/ N{gear} = T{gear}/T{pinion}$. Flywheel: 16.
Coefficient of Speed Fluctuation ($Cs$): $Cs = (\omega{max} \omega{min}) / \omega{mean}$, where $\omega{mean} = (\omega{max} + \omega{min})/2$. Hartnell Governor: A type of spring-loaded 17. governor.
If vertical and horizontal arms of bell crank lever are equal (a=b), then change in spring force ($\Delta Fs$) is $2(F{c2} - F{c1})$.
Change in spring compression ($ \Delta x$) is proportional to change in ball radius ($r2 - r1$). If a=b, $\Delta x = r2 - r1$.
Spring stiffness ($k$) = $\Delta Fs / \Delta x$.

Simple Harmonic Motion and Damped/Forced Oscillations

Simple Harmonic Motion (SHM): A special type of periodic motion where the restoring force (and acceleration) is directly proportional to the displacement from the equilibrium position and acts in the opposite direction ($F \propto -x$, or $a = -\omega^2 x$).
SHM Kinematics:

Displacement: $x(t) = A \sin(\omega t + \phi)$ (or cosine).
Velocity: $v(t) = A\omega \cos(\omega t + \phi) = \pm \omega\sqrt{A^2x^2}$. Maximum at mean position ($x=0$), zero at extreme positions ($x=\pm A$).
Acceleration: $a(t) = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x(t) $. Maximum at extreme positions ($x=\pm A$), zero at mean position ($x=0$).

SHM Energy:

Kinetic Energy (KE): Maximum at the mean position, zero at extreme positions.
Potential Energy (PE): Zero (minimum) at the mean position, maximum at extreme positions.
Total Mechanical Energy (KE + PE): Constant in an ideal (undamped) SHM.
If the oscillating system has frequency $f$, its KE and PE oscillate with frequency $2f$.

Periodicity & Phase:

SHM is a type of periodic motion.
The time period (T) is the least time after which the motion repeats itself. The motion also repeats after $nT$ (where $n$ is a positive integer).
The phase of the oscillating particle is the same at times $t$ and $t+T$, $t+2T$, etc.

Simple Pendulum (approx. SHM for small angles):

Time period $T = 2\pi\sqrt{\frac{L} {g}}$, where L is length and g is acceleration due to gravity.
T increases with increased length (L).
T decreases with increased 'g'.
T is independent of the mass of the bob.
T is independent of the amplitude only for small angular displacements (typically < 10-15 degrees), where the restoring force is proportional to displacement.
For L = 1m and g $\approx 9.8 \text{ m/s}^2$, T $\approx 2$ seconds.
Angular acceleration $\alpha = -(g/L) \sin\theta$. For small $\theta$, $\alpha \approx -(g/L)\theta$.
Impact of effective length: When a person stands up on a swing, their center of mass rises, effectively decreasing the pendulum's length (L), thus decreasing the time period (T) and making the swing go faster.

Spring-Mass System (SHM):

Time period $T = 2\pi\sqrt{\frac{m} {k}}$, where m is mass and k is spring constant.
Natural angular frequency $\omegan = \sqrt{k/m}$.
T (and $\omegan$) is independent of acceleration due to gravity (g). This makes it suitable for measuring mass in space (oscillation method).
Extension of a vertical spring due to mass 'm': $x = mg/k$ (depends on 'g').

Damped Oscillations:

Occur when dissipative forces (e.g., air resistance, friction) reduce the total mechanical energy of an oscillating system, causing the amplitude to decrease over time.
Equation for free vibration with damping: $m\ddot{x} + c\dot{x} + sx = 0$, where $c$ is the damping coefficient.

Forced Oscillations:

Occur when an external periodic force acts on an oscillating system.
Equation for forced vibration with damping: $m\ddot{x} + c\dot{x} + sx = F0 \sin \omega t$.
Critical Damping Coefficient ($cc$): $cc = 2\sqrt{mk} = 2m\omegan$.
Damping Ratio ($\zeta$): $\zeta = c/ cc$.

$\zeta < 1$: Underdamped system (oscillates with decreasing amplitude).
$\zeta = 1$: Critically damped system (returns to equilibrium fastest without oscillation).
$\zeta > 1$: Overdamped system (returns to equilibrium slowly without oscillation).

Resonance: Occurs when the driving frequency ($\omega$) equals the natural frequency ($\omegan$), leading to maximum amplitude.
Magnification Factor: Ratio of the steady-state amplitude of forced vibration to the static deflection due to the forcing amplitude.
Steady-state response ($t \to \infty$): For a stable system under a constant forcing term $F0$, the acceleration and velocity tend to zero, and the displacement $x$ becomes $F0/k$ (where $k$ is stiffness, or $s$ in the equation $m\ddot{x} + c\dot{x} + sx = F0$).

Applications & Concepts:

A body falling through a diametric tunnel across Earth (ignoring air resistance) would execute SHM. With damping (air resistance), it would execute damped SHM.
Clocks: Hour hand moves 30°/hour (0.5°/minute). Minute hand moves 360°/hour (6°/minute).

Special Theory of Relativity (Lorentz Transformations, Relativistic Dynamics)

Postulates of Special Theory of Relativity (Einstein, 1905):

First Postulate (Principle of Relativity): The laws of physics are the same for all observers in all inertial frames of reference.
Second Postulate (Constancy of Speed of Light): The speed of light in a vacuum (c) is the same for all inertial observers, regardless of the motion of the light source or the observer.

Lorentz Transformations: Mathematical equations relating space and time coordinates between two inertial frames moving at a constant relative velocity (v).

For frame S' moving with velocity 'v' along the x-axis relative to frame S:
x' = γ(x - vt)
y' = y
z' = z
t' = γ(t - vx/c²)
Lorentz factor (γ) = 1 / √(1 - v²/c²). (γ > 1 for v > 0).

Relativistic Effects (Derived from Lorentz Transformations):

Length Contraction: The length (L) of an object moving at velocity 'v' relative to an observer appears shorter than its proper length (L₀) measured in its rest frame.
Formula: L = L₀ / γ = L₀√(1 - v²/c²)
Example: A 100m rocket at rest (L₀ = 100m) moving at v = 0.8c will have a length L = 100 √(1 - (0.8)² ) = 100 √(1

as seen by a stationary observer.

Time Dilation: A time interval (Δt) measured in a frame where a clock is moving appears longer than the proper time interval (Δt₀) measured in the clock's rest frame.
Formula: Δt = γΔt₀ = Δt₀ / √(1 - v²/c²)
Relativistic Mass: The mass (m) of an object increases with its velocity (v).
Formula: m = γm₀ = m₀ / √(1 - v²/c²), where m₀ is the rest mass.

Relativistic Dynamics: Describes motion at speeds comparable to the speed of light (v ≈ c).

Limitations of Newtonian Mechanics: Newton's laws of motion are valid only at speeds much less than the speed of light (v << c). They fail at speeds comparable to the speed of light (relativistic speeds).
Relativistic Momentum: p = γm₀v
Mass-Energy Equivalence (Einstein): E = mc²
Where E is total energy, m is relativistic mass, c is speed of light.
Relativistic total energy: E = √(p²c² + m₀²c⁴) = γm₀c²
Relativistic kinetic energy: K = E m₀c² = (γ - 1)m₀c²

Systems of Particles and Collisions

Centre of Mass (CM): The CM of a body can lie outside its physical boundary (e.g., for a ring, hollow sphere).
Linear Momentum: Defined as $p = mv$. It is a vector quantity.
Impulse-Momentum Theorem: Impulse ($J$) equals the change in linear momentum ($ \Delta p$). $J = F{avg} \Delta t = \Delta p$.
Velocity before impact: An object dropped from height $h$ reaches a velocity $v = \sqrt{2gh}$ just before hitting the ground (ignoring air resistance).
Conservation of Linear Momentum: In an isolated system (no external forces), the total linear momentum remains constant before and after a collision. $m1u1 + m2u2 = m1v1 + m2v2$.
Types of Collisions:

Elastic Collision:

Total linear momentum is conserved.
Total kinetic energy is conserved.
The relative speed of approach equals the relative speed of separation ($u1 - u2 = v2 v1$).
Coefficient of restitution, $e=1$.
For a 1D elastic collision between two identical masses, if one is initially at rest, they exchange velocities (the moving mass stops, and the stationary mass moves with the initial velocity of the first).

Inelastic Collision:

Total linear momentum is conserved.
Total kinetic energy is not conserved (some kinetic energy is converted into other forms like heat, sound, or deformation).
Coefficient of restitution, $0 \le e < 1$.

Perfectly Inelastic Collision:

Bodies stick together after collision, moving as a single unit ($v1=v2=v{final}$).
Total linear momentum is conserved.
Kinetic energy loss is maximum.
Coefficient of restitution, $e=0$.

Basis of Momentum Conservation: It is a direct consequence of Newton's Second Law ($F{ext} = dP/dt$) and Newton's Third Law (internal forces cancel out in a system).
Change in Momentum during Bounce: When a ball bounces off a surface, its momentum changes suddenly because its velocity vector reverses direction (even if speed remains constant in a perfectly elastic bounce). Speed, kinetic energy, and potential energy may not change as suddenly or at all, depending on the collision type.
Coefficient of Restitution ($e$): A measure of the "bounciness" of a collision. Defined as the ratio of the relative speed of separation to the relative speed of approach. For a ball bouncing off a fixed surface, $e = \frac{v{rebound}}{v{impact}} = \sqrt{\frac{\text{rebound height}}{\text{drop height}}}$.
Motion in Ionized Gases: In an ionized gas (plasma), particles (ions and electrons) carry electric charges. Their motion is primarily governed by long-range electromagnetic forces between them, rather than solely by short-range collisions.
Coalescence of Drops: When multiple drops coalesce into a single larger drop, the total volume of the liquid remains conserved. If $N$ identical drops of radius $r$ form one large drop of radius $R$, then $N \times (4/3)\pi r^3 = (4/3)\pi R^3$, implying $N r^3 = R^3$.

Work, Energy, and Power

Work (W): Done when a force (F) causes dis1. placement (S). W = F ⋅ S = F S cosθ. Work Unit: Joule (J). 1 Joule = 1 N·m. 2. Positive Work: Force and displacement in 3. same general direction (0° ≤ θ < 90°). E.g., force pulling, gravity on falling object. Negative Work: Force and displacement in 4. opposite general direction (90° < θ ≤ 180°). E.g., friction, air resistance, lifting against gravity, force anti-parallel to displacement. (MCQ 4, 16, 27) Zero Work: Force perpendicular to displace5. ment (θ = 90°), or no displacement (S=0), or no force (F=0). E.g., gravitational force on horizontally moving object. (MCQ 45) Energy: Capacity to do work. Unit: Joule 6. (J). Law of Conservation of Energy: Energy can 7. neither be created nor destroyed; it can only be transformed from one form to another. (MCQ 23, 51) Kinetic Energy (KE): Energy due to motion. 8. KE = (1/2)mv². (MCQ 3) KE and Velocity: If velocity doubles, KE be9. comes 4 times (KE ∝ v²). (MCQ 49) KE and Momentum (p): p = mv. KE = p²/ 10. (2m). Momentum and KE Relation: 11.

If momentum doubles, KE increases by four times. (MCQ 2)
For equal momentum, the lighter body has greater kinetic energy (KE ∝ 1/ m). (MCQ 29)
For equal kinetic energy, the heavier body has greater momentum (p ∝ √m). (MCQ 35) Potential Energy (PE): Stored energy due to 12. position or configuration/shape. (MCQ 17) Gravitational Potential Energy: PE = mgh 13. (mass × acceleration due to gravity × height). Energy Transformation in a Swing: As a 14. swing rises, potential energy increases, and kinetic energy decreases. (MCQ 13) Stored Energy Forms: Nuclear energy, Po15. tential energy, Chemical energy are examples of stored energy. Electrical energy is not typically considered a stored form. (MCQ 14, 50) Energy Conversion in a Falling Object (e.g., 16. Apple): Gravitational Potential Energy → Kinetic Energy → (upon impact and air friction) Heat energy (to air, ground, and object) + Sound energy. (MCQ 15) Power (P): Rate at which work is done or 17. energy is transferred. P = Work/Time = Energy/Time. Unit: Watt (W). 1 Watt = 1 J/s. (MCQ 12, 19) Power Formula (Constant Force/Velocity): 18. P = F ⋅ v (Force × velocity, if force is constant and parallel to velocity). (MCQ 32) Work-Energy Theorem: The net work done 19. on an object equals the change in its kinetic energy (Wnet = ΔKE). Positive net work increases kinetic energy. (MCQ 21, 41, 43) Frictional Force and Energy Change: Fric20. tional force converts kinetic energy into heat energy (thermal energy). (MCQ 1) Conservative Forces: Work done is path-in21. dependent and depends only on initial/final positions. Examples: Gravitational force, Electric force, Spring force, Magnetostatic force. Associated with potential energy. (MCQ 26, 34) Non-Conservative Forces: Work done is 22. path-dependent. Examples: Frictional force, Air drag. These dissipate mechanical energy, usually as heat. (MCQ 26, 34) Simple Machine: Helps a person do the 23. same amount of work with lesser force or greater convenience; it does not reduce the total work done. (MCQ 10, 11, 39) Energy Converters: 24.
Dynamo: Converts mechanical energy into electrical energy. (MCQ 36)
Wind Power: Converts kinetic energy (of wind) into electrical energy. (MCQ 37) Potential Energy from Force: If force F(x) is 25. given, potential energy U(x) = -∫F(x)dx. For F(x) = Ax² - Bx, U(x) = -Ax³/3 + Bx²/2. (MCQ 22) Stopping Distance and Kinetic Energy: If 26. objects have equal kinetic energies and are brought to rest by equal stopping forces, they will have equal stopping distances. (Wstop = Fstop × dstop = KE). (MCQ 38)

4. Motion under Gravity

Practice MCQs: Mechanics

Conservation of Linear and Angular Momentum

A railway wagon (open at the top) of mass M₁ is moving with speed v₁ along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M₂ and speed becomes v₂. Taking the rain to be falling vertically and water stationary inside the wagon, the relation between the two speeds v₁ and v₂ is :
Asked in: UPSC NDA 1 2023
1. v₁ = v₂
2. 1/2 M₁v₁² < 1/2 M₂v₂²
3. M₁v₁ = M₂v₂
4. M₁v₁ < M₂v₂
Show answer
Answer: C. M₁v₁ = M₂v₂
GS&EI ElectEngg 14 2024 1601-2025 GS&EI ElectEngg 14 2024 1601-2025 This question is identical to ID 80852 and ID 80861. In a Schering bridge balance, the unknown capacitance $C_1$ and its loss angle $\delta$ are determined by the following formulas: $C_1 = C_2 \frac{R_4}{R_3}$ $\tan \delta = \omega C_4 R_4$Given values: $C_2 = 100 \mu \text{F}$ $R_3 = 100 \Omega$ $R_4 = 300 \Omega$ $C_4 = 0.5 \mu \text{F}$ $f = 50 \text{ Hz}$Calculation for $C_1$: $C_1 = 100 \mu \text{F} \times \frac{300 \Omega}{100 \Omega} = 100 \mu \text{F} \times 3 = 300 \mu \text{F}$.Calculation for $\delta$: Angular frequency, $\omega = 2\pi f = 2\pi \times 50 = 100\pi \text{ rad/s}$. $\tan \delta = (100\pi \text{ rad/s}) \times (0.5 \times 10^{-6} \text{ F}) \times (300 \Omega)$ $\tan \delta = 100\pi \times 0.04712$.The loss angle $\delta$ is the arctangent of the dissipation factor: $\delta = \arctan(0.04712) \approx 2.698^{\circ}$. Rounding to one decimal place, $\delta \approx 2.7^{\circ}$.Thus, the unknown capacitance is $300 \mu \text{F}$ and its loss angle is $2.7^{\circ}$. This matches option C. The initial horizontal momentum of the wagon is $P_1 = M_1 v_1$. After the rain has collected, the total mass is $M_2$ and the speed is $v_2$. The final horizontal momentum is $P_2 = M_2 v_2$. By conservation of horizontal momentum, $P_1 = P_2$, so $M_1 v_1 = M_2 v_2$.
Sand falls vertically on a conveyor belt at a rate of 0·1 kg/s. In order to keep the belt moving at a uniform speed of 2 m/s, the force required to be applied on the belt is :
Asked in: UPSC NDA 1 2023
1. 0 N
2. 0·2 N
3. 1·0 N
4. 2·0 N
Show answer
Answer: B. 0·2 N
The force required to keep the belt moving at a uniform speed is equal to the rate of change of momentum of the sand added to the belt, which is the product of the rate of mass flow and the belt's velocity.
Let there be an object having some chemicals in it. It starts moving with a uniform velocity v and a chemical reaction starts happening. In this case, which of the following statement/s is/ are correct ? 1. Chemical reactions happening in the system cannot change the velocity v of the center of mass of the object. 2. Chemical reactions happening in the system cannot change the kinetic energy of the particles inside with respect to the center of mass of object. Select the correct answer using the code given below :
Asked in: UPSC NDA 2 2019
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Show answer
Answer: A. 1 only
Statement 1: The velocity of the center of mass of a system can only be changed by external forces acting on the system. Chemical reactions happening within the object involve internal forces between constituent particles. Internal forces cannot change the motion of the center of mass of the system. Therefore, if the object is moving with uniform velocity (implying zero net external force or balanced external forces), a chemical reaction inside will not change the velocity of the center of mass. Statement 2: Chemical reactions often involve changes in the internal energy of the system, which can include changes in the kinetic energy of the particles relative to the center of mass (manifesting as temperature changes). For example, an exothermic reaction releases energy, which increases the kinetic energy of the constituent particles. Therefore, chemical reactions can change the kinetic energy of the particles inside with respect to the center of mass of the object.
Fundamental laws of physics require
Asked in: UPSC NDA 2 2021
1. conservation of energy and non-conservation of charge.
2. conservation of charge and non-conservation of linear momentum.
3. conservation of charge and non-conservation of energy.
4. conservation of energy, momentum and charge.
Show answer
Answer: D. conservation of energy, momentum and charge.
Conservation means that the total amount of the quantity remains constant over time within a closed system, even though it may be transformed or transferred between different forms or parts of the system. These conservation laws are derived from symmetries in nature and are cornerstones of physics.
A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1}$ from a pistol of mass 1 kg. What is the recoil velocity of the pistol?
Asked in: UPSC NDA 2 2022
1. 0·3 m s$^{-1}$
2. 3 m s$^{-1}$
3. −3 m s$^{-1}$
4. −0·3 m s$^{-1}$
Show answer
Answer: C. −3 m s$^{-1}$
- This problem can be solved using the principle of conservation of linear momentum. - The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest. - According to the conservation of momentum, the total momentum of the system after firing must also be zero. - Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol. - Momentum before firing = 0 - Momentum after firing = $m_b v_b + m_p v_p$ - By conservation of momentum: $m_b v_b + m_p v_p = 0$ - Given: $m_b = 10 \text{ g} = 0.01 \text{ kg}$, $v_b = 300 \text{ m/s}$, $m_p = 1 \text{ kg}$. - Substituting the values: $(0.01 \text{ kg})(300 \text{ m/s}) + (1 \text{ kg}) v_p = 0$ - $3 \text{ kg} \cdot \text{m/s} + v_p \text{ kg} = 0$ - $v_p = -3 \text{ m/s}$
If the linear momentum of a moving object changes by two times, then its kinetic energy will change by a factor of
Asked in: UPSC CAPF 2022
1. 2
2. 4
3. 6
4. 8
Show answer
Answer: B. 4
Linear momentum (p) of an object of mass m and velocity v is given by p = mv. Kinetic energy (k) of the same object is given by k = ½ mv². We can express kinetic energy in terms of momentum. From p = mv, we get v = p/m. Substituting this into the kinetic energy equation: k = ½ m(p/m)² = ½ m(p²/m²) = ½ p²/m. So, k is proportional to p² (assuming mass m is constant). If the initial momentum is p₁, the initial kinetic energy is k₁ = ½ p₁²/m. If the linear momentum changes by two times, it means the new momentum p₂ = 2p₁. The new kinetic energy k₂ = ½ p₂²/m = ½ (2p₁)²/m = ½ (4p₁²)/m = 4 (½ p₁²/m) = 4k₁. Thus, the kinetic energy will change by a factor of 4.
Which one of the following conservation laws is a consequence of the Newton's third law of motion ?
Asked in: UPSC CAPF 2023
1. Conservation of energy
2. Conservation of momentum
3. Conservation of charge
4. Conservation of mass
Show answer
Answer: B. Conservation of momentum
- Newton's third law states that for every action, there is an equal and opposite reaction. If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. - When considering a system of two interacting objects, the forces they exert on each other are internal forces. According to the third law, these internal forces cancel out as a pair ($\vec{F}_{\text{AB}} = - \vec{F}_{\text{BA}}$). - Applying Newton's second law ($\vec{F} = m\vec{a} = d\vec{p}/dt$) to the system, the total internal force is zero. Thus, the rate of change of total momentum of the system due to internal forces is zero. - In the absence of external forces, the total momentum of the system remains constant. This is the principle of conservation of momentum.
Rocket works on the principle of the following:
Asked in: UPPSC Upper Mains 2014
1. Avogadro's concept
2. Energy conservation
3. Momentum conservation
4. Bernoulli's theorem
Show answer
Answer: C. Momentum conservation
- Rocket propulsion is an application of Newton's Third Law and the principle of momentum conservation. - The backward momentum of expelled exhaust mass creates forward momentum for the rocket.

Elasticity (Stress, Strain, Hooke's Law, Elastic Moduli)

A sphere of radius 100 cm is cut by a plane at a distance of h from its center, thereby breaking this sphere into two different pieces. The cumulative surface area of these two pieces is 25% more than that of the sphere. What is the value of h?
Asked in: GPSC Prelims CSP 1 2022
1. 50/√2
2. 50√2
3. 75√2
4. None of the above
Show answer
Answer: B. 50√2
The surface area of the original sphere with radius R is $A_{sphere} = 4 \pi R^2$. When the sphere is cut by a plane at a distance h from its center (0
Following are general statements about water drops. (a) Surface tension is responsible for the spherical shape of the water drops. (b) Even in the absence of gravitation, water drops have spherical shape. (c) Formation of water drops occurs, if mass of the water is stretched. Which of the above statement/statements is/are correct?
Asked in: MPSC CS Prelims 2013
1. (a) only
2. (a) and (b) only
3. (c) only
4. (a), (b) and (c)
Show answer
Answer: B. (a) and (b) only
(a) Surface tension is the property of a liquid surface that causes it to behave like a stretched elastic membrane, minimizing its surface area. For a given volume, a sphere has the minimum surface area. Thus, surface tension is indeed responsible for pulling liquid drops into a spherical shape, especially when other forces like gravity are weak. This statement is correct. (b) In the absence of gravity (e.g., in space), surface tension becomes the dominant force determining the shape of a liquid. Without gravity distorting the shape, the liquid will naturally form a perfect sphere to minimize its surface energy (related to surface area). This statement is correct. (c) The formation of water drops from a bulk liquid typically occurs when surface tension forces cause a liquid stream or film to break up or detach from a surface. While 'stretching' a mass of water (e.g., a falling stream) can lead to instabilities that result in drop formation, the statement "Formation of water drops occurs, if mass of the water is stretched" is not a precise or universally accurate description of the primary mechanism. Drops form due to instability driven by surface tension, inertia, and sometimes gravity or air resistance, not just stretching in isolation. This statement is questionable and not as fundamentally correct as (a) and (b).
A spring of spring constant k is cut into two equal halves. The spring constant of each half is now
Asked in: WBPSC CS Prelims 2021
1. k
2. k/2
3. 2k
4. 4k
Show answer
Answer: C. 2k
- For a uniform spring, the product of the spring constant and the length is constant (kL = constant). - When the spring is cut into two equal halves, the length of each half is L/2. - Using the constant product relationship: k' (L/2) = k L - Solving for k': k' = (k L) / (L/2) = 2k.
Bulk Modulus for a perfectly rigid body is
Asked in: ukpsc 2025 2025
1. Infinite
2. Zero
3. Unity
4. Some finite low value
Show answer
Answer: A. Infinite
Bulk modulus is a measure of a substance's resistance to uniform compression.
If the principal stresses in a plane stress problem, are $\sigma_1 = 100$ MPa, $\sigma_2 = 40$ MPa, the magnitude of the maximum shear stress in MPa will be - (5) Question not attempted
Asked in: Assistant Eng. (Mechanical) Competitive Examination, 2023
1. 60 MPa
2. 30 MPa
3. 50 MPa
4. 70 MPa
Show answer
Answer: C. 50 MPa
The maximum shear stress in plane stress is half the difference between the principal stresses.
The Young’s modulus of elasticity of steel is 210 GPa, and the Poisson’s ratio is 0.29. The shear modulus will be : - (5) Question not attempted
Asked in: Assistant Eng. (Mechanical) Competitive Examination, 2023
1. 110.45 GPa
2. 105.33 GPa
3. 81.39 GPa
4. 71.77 GPa
Show answer
Answer: C. 81.39 GPa
The shear modulus, Young's modulus, and Poisson's ratio are related by the formula $G = \frac{E}{2(1 + \nu)}$ for isotropic materials.
Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled, the maximum shear stress developed corresponding to the same torque will be : - (5) Question not attempted
Asked in: Assistant Eng. (Mechanical) Competitive Examination, 2023
1. 240 MPa
2. 60 MPa
3. 30 MPa
4. 90 MPa
Show answer
Answer: C. 30 MPa
The maximum shear stress ($\tau_{max}$) in a solid circular shaft under pure torsion is given by the formula: $\tau_{max} = \frac{16T}{\pi D^3}$, where T is the torque and D is the diameter of the shaft. Given: $\tau_{max1} = 240$ MPa. When the shaft diameter is doubled, the new diameter $D_2 = 2D_1$. The new maximum shear stress $ \tau_{max2} = \frac{16T}{\pi (D_2)^3} = \frac{16T}{\pi (2D_1)^3} = \frac{16T}{\pi 8D_1^3} = \frac{1}{8} \left(\frac{16T}{\pi D_1^3}\right)$. Since $\tau_{max1} = \frac{16T} {\pi D_1^3}$, we have $\tau_{max2} = \frac{1} {8} \tau_{max1}$. $\tau_{max2} = \frac{1}{8} \times 240$ MPa $= 30$ MPa.
A cylindrical rod of length, l and diameter, d, is rigidly fixed at its upper end and hangs vertically. The elongation produced in the rod due to its total self weight, W is: - (5) Question not attempted Where A is area of cross-section of cylindrical rod and E is modulus of elasticity.
Asked in: Assistant Eng. (Mechanical) Competitive Examination, 2023
1. Wl/2AE
2. Wl/AE
3. 2Wl/AE
4. 3Wl/2AE
Show answer
Answer: A. Wl/2AE
The elongation of a rod due to its own self-weight is half of what it would be if the same total weight were applied as a concentrated load at the end, and it is proportional to the length, inversely proportional to the cross-sectional area and Young's modulus.

Fluid Statics and Dynamics (Continuity Equation, Bernoulli's Theorem)

The working principle of a washing machine is
Asked in: BPSC Prelims Exam 67th Re Exam
1. diffusion
2. centrifugation
3. dialysis
4. reverse osmosis
Show answer
Answer: B. centrifugation
The washing machine utilizes the principle of centrifugation during the spin cycle to remove water from clothes by spinning them at high speed, forcing the water out through centrifugal force. Agitation during the wash cycle also involves forces that help separate dirt.
The pressure of a fluid varies with depth h as P = P₀ + pgh, where ρ is the fluid density. This expression is associated with
Asked in: UPSC CDS 1 2018
1. Pascal's law
2. Newton's law
3. Bernoulli's principle
4. Archimedes' principle
Show answer
Answer: A. Pascal's law
- The formula P = P₀ + ρgh quantifies hydrostatic pressure variation with depth. - Pascal's contributions were foundational to understanding fluid pressure at rest and its transmission.
Two identical solid pieces, one of gold and other of silver, when immersed completely in water exhibit equal weights. When weighed in air (given that density of gold is greater than that of silver)
Asked in: UPSC CDS 1 2019
1. the gold piece will weigh more
2. the silver piece will weigh more
3. both silver and gold pieces weigh equal
4. weighing will depend on their masses
Show answer
Answer: B. the silver piece will weigh more
When immersed in water, the apparent weight of an object is its true weight in air minus the buoyant force. The buoyant force is equal to the weight of the water displaced, which is (volume of the object) (density of water) g. The two pieces have equal apparent weight in water: W_air,gold - B_gold = W_air,silver - B_silver. Since both are solid and immersed, B_gold = V_gold density_water g and B_silver = V_silver density_water g. Given that the density of gold is greater than the density of silver, and they displace the same weight of water to have equal apparent weight difference from their air weight, the object with lower density (silver) must have a larger volume to displace the amount of water needed to satisfy the equality in apparent weight. W_air = density Volume g. W_air,gold - V_gold density_water g = W_air,silver - V_silver density_water g V_gold (density_gold - density_water) = V_silver (density_silver - density_water) Since density_gold > density_silver, it follows that (density_gold - density_water) > (density_silver - density_water). For the equation to hold, V_gold must be less than V_silver. Now comparing weight in air: W_air,gold = density_gold V_gold g and W_air,silver = density_silver V_silver g. We know V_gold
What is the effect of pressure of a human body on sand ?
Asked in: UPSC CDS 2 2021
1. Larger while standing than while lying
2. Smaller while standing than while lying
3. Same while standing or lying
4. Larger while standing during the daytime and smaller during the night time while lying
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Answer: A. Larger while standing than while lying
Pressure is inversely proportional to the area of contact for a constant force.
Which one of the following is the conservation law from which the Bernoulli's equation for fluid flow is derived?
Asked in: UPSC Geoscientist 2022
1. Conservation of momentum
2. Conservation of volume
3. Conservation of mass
4. Conservation of energy
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Answer: D. Conservation of energy
- Bernoulli's equation relates pressure, velocity, and elevation of a fluid. - It is a form of the conservation of energy applied to fluid flow. - Assumptions include inviscid, incompressible, and steady flow along a streamline.
The derivation of Bernoulli's equation for fluid flow takes certain assumptions. Which one of the following assumptions is not among them?
Asked in: UPSC Geoscientist 2022
1. Gravitational forces can be neglected
2. Turbulence of fluid flow can be neglected
3. Viscous forces can be neglected
4. Frictional forces can be neglected
Show answer
Answer: A. Gravitational forces can be neglected
Bernoulli's equation is derived from the application of the conservation of energy principle to fluid flow under specific idealizing assumptions. These assumptions typically include that the fluid is incompressible, the flow is steady, inviscid (no viscous forces or internal friction), and often irrotational (negligible turbulence). The derivation considers the work done by pressure forces and gravity on a fluid element as it moves along a streamline, relating changes in pressure, velocity, and height (potential energy due to gravity).
The force of buoyancy on an object floating on water is F, while it is S on an object that sinks in water. The weight of both the objects is W. Which one of the following is always true?
Asked in: UPSC Geoscientist 2023
1. F = W and S = 0
2. Both F and S have upward direction
3. F has upward direction and S has downward direction
4. F = W and S > W
Show answer
Answer: B. Both F and S have upward direction
The force of buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. It is caused by the pressure difference between the top and bottom of the object. This force always acts vertically upwards, regardless of whether the object is floating or sinking.
By dropping a sphere of 30 cm radius into a cylindrical vessel of 80 cm diameter which is partly filled with water, its water level rises by x cm. Then what would be value of x?
Asked in: GPSC Prelims CSP 1 2021
1. 22.5 cm
2. 25.5 cm
3. 28.5 cm
4. data inadequate
Show answer
Answer: A. 22.5 cm
- The volume of water displaced by a submerged object is equal to the volume of the object. - The displaced water in a cylinder forms a cylindrical volume with the cylinder's radius and the height of the water level rise. - Formula for sphere volume: $V = \frac{4}{3}\pi r^3$. - Formula for cylinder volume: $V = \pi r^2 h$.

Gravitation and Central Force Motion (Kepler's Laws)

Arrange the following places in ascending order of local acceleration due to gravity : (i) Polar region of the earth's surface at the sea level (ii) Surface of the moon (iii) Centre of the earth (iv) Equatorial region on the earth's surface at the sea level
Asked in: APSC CCE Prelims Exam 2022
1. (i), (ii), (iii), (iv)
2. (iii), (ii), (iv), (i)
3. (iii), (ii), (i), (iv)
4. (iii), (i), (ii), (iv)
Show answer
Answer: B. (iii), (ii), (iv), (i)
Acceleration due to gravity is zero at the center of the Earth, is significantly lower on the Moon than on Earth, and varies slightly across the Earth's surface, being highest at the poles and lowest at the equator.
The mass of a body is different from its weight because (i) mass is a variable quantity whereas weight is constant (ii) mass varies very little at different places whereas weight varies a lot (iii) mass is constant but weight decreases as the body moves from the Poles to the Equator (iv) mass is a measure of quantity of matter whereas weight is a force Select the correct reason/reasons using the codes given below.
Asked in: APSC CCE Prelims Exam 2023
1. (i) only
2. (ii) and (iv) only
3. (iii) only
4. (iii) and (iv) only
Show answer
Answer: D. (iii) and (iv) only
- Mass is a scalar quantity, a measure of inertia, and is constant regardless of location. - Weight is a vector quantity, the force of gravity acting on mass (W = mg), and varies with the local acceleration due to gravity 'g'. - 'g' varies with altitude, latitude, and local geological structures.
What happens to the gravitational force between two objects if the mass of one object is doubled and the distance between them is also doubled ?
Asked in: UPSC CDS 1 2022
1. The force would remain the same
2. The force would be doubled
3. The force would be halved
4. The force would increase by a factor of 4
Show answer
Answer: C. The force would be halved
Newton's Law of Gravitation is given by F = G (m1 m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
An object weighs 9 N on the surface of the Earth. What would be its weight, when measured on the surface of a planet where the acceleration due to gravity is 9 times that on the Earth ?
Asked in: UPSC CDS 1 2022
1. The weight would remain the same
2. The weight would be equal to 1 N
3. The weight would become 9 times
4. The weight will be reduced to $\frac{1}{9}$ N
Show answer
Answer: C. The weight would become 9 times
Weight is defined as the force of gravity acting on an object, given by the formula W = m g, where W is weight, m is mass, and g is the acceleration due to gravity. The mass of the object remains constant regardless of location. If the acceleration due to gravity (g) on the planet is 9 times that on Earth (g_planet = 9 g_earth), then the weight on the planet will be W_planet = m g_planet = m (9 g_earth) = 9 (m g_earth) = 9 W_earth. Thus, the weight on the planet becomes 9 times the weight on Earth.
If the distance between two objects is increased by two times, the gravitational force between them will
Asked in: UPSC CDS 2 2020
1. remain same
2. increase by two times
3. decrease by two times
4. decrease by four times
Show answer
Answer: D. decrease by four times
According to Newton's Law of Universal Gravitation, the force (F) between two objects with masses m₁ and m₂ separated by a distance (r) is given by F = G(m₁m₂)/r², where G is the gravitational constant. The force is inversely proportional to the square of the distance between the objects. If the distance is increased by two times (r' = 2r), the new force (F') will be F' = G(m₁m₂)/(r')² = G(m₁m₂)/(2r)² = G(m₁m₂)/(4r²) = (1/4) G(m₁m₂)/r² = (1/4) F. Thus, the force decreases to one-fourth of its original value.
If the Moon is brought closer to the Earth such that its distance from the Earth becomes half of the original distance, then the gravitational force of attraction between the Earth and the Moon would :
Asked in: UPSC CDS 2 2023
1. reduce to half of its original value.
2. increase to two times of its original value.
3. remain the same as the original value.
4. increase to four times of its original value.
Show answer
Answer: D. increase to four times of its original value.
According to Newton's Law of Universal Gravitation, the gravitational force (F) between two objects with masses M₁ and M₂ is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between their centers: F = G (M₁ M₂) / r². In this case, M₁ is the mass of the Earth (M_earth) and M₂ is the mass of the Moon (M_moon). Let the original distance be r. The original gravitational force is F = G (M_earth M_moon) / r². If the distance is halved, the new distance is r' = r/2. The new gravitational force F' will be: F' = G (M_earth M_moon) / (r')² F' = G (M_earth M_moon) / (r/2)² F' = G (M_earth M_moon) / (r²/4) F' = 4 [G (M_earth M_moon) / r²] So, F' = 4 F. The gravitational force increases to four times its original value.
The gravitational force ($\vec{F}$) on mass $M$ due to another mass $m$ at a distance $x$ is given by (vector $\vec{x}$ is from mass $M$ to mass $m$ and unit vector $\hat{x}$ is the corresponding unit vector)
Asked in: UPSC Geoscientist 2023
1. $\vec{F} = G\frac{Mm}{x^3}\hat{x}$
2. $\vec{F} = -G\frac{Mm}{x^3}\hat{x}$
3. $\vec{F} = -G\frac{Mm}{x^2}\hat{x}$
4. $\vec{F} = G\frac{Mm}{x^2}\hat{x}$
Show answer
Answer: C. $\vec{F} = -G\frac{Mm}{x^2}\hat{x}$
Newton's law of universal gravitation states that the force of attraction between two masses M and m separated by a distance x is given by $F = G\frac{Mm}{x^2}$. This force is attractive, meaning the force on M is towards m, and the force on m is towards M. The question defines the vector $ \vec{x}$ (and its unit vector $\hat{x}$) as pointing from mass M to mass m. The gravitational force on mass M due to mass m is an attractive force directed from M towards m. Therefore, the force vector on M should be in the same direction as $\hat{x}$. Based strictly on the provided definition and standard physics, the force on M should be $ \vec{F} = G\frac{Mm}{x^2}\hat{x}$ (Option D). However, option C, $\vec{F} = -G\frac{Mm} {x^2}\hat{x}$, implies the force on M is in the direction opposite to $\hat{x}$. Since $\hat{x}$ points from M to m, $-\hat{x}$ points from m to M. Thus, C suggests the force on M is directed from m to M. This would be consistent with an attractive force on M if $\hat{x}$ was defined as the unit vector pointing from m to M. Given that option C is widely cited as the correct answer for this specific question from previous exams, it indicates a likely inconsistency or misstatement in the question's definition of $\hat{x}$ or the intended target mass of the force vector $\vec{F}$. Assuming option C is indeed the intended correct answer, it is most probable that the definition of the unit vector $ \hat{x}$ was meant to be from mass m to mass M. In that (likely intended) case, the force on mass M (towards m) would be in the direction opposite to $ \hat{x}$, hence $\vec{F} = -G\frac{Mm} {x^2}\hat{x}$.
The weight of a body varies inversely as the square of its distance from the centre of planet P. If the radius of P is 4000 kms, how much would a 200 kg object weigh 1000 kms above the surface of the planet P?
Asked in: GPSC Prelims CSP 1 2023
1. 256 Kg
2. 128 Kg
3. 3200 Kg
4. None of the above
Show answer
Answer: B. 128 Kg
The key concept is inverse square variation ($W \propto 1/d^2$) and correctly calculating the distance from the center of the planet, which includes the planet's radius.

Lagrangian Formulation (Generalized Coordinates, Euler-Lagrange Equations)

The number of generalized co-ordinates required to describe the position of a simple pendulum is
Asked in: TNPSC CS Prelims 2015
1. 1
2. 2
3. 3
4. 4
Show answer
Answer: A. 1
Generalized coordinates are a set of independent coordinates that completely specify the configuration of a system. For a simple pendulum, the motion is constrained to an arc of a circle, and its position along this arc is uniquely determined by one angle.
If Pn(x) is Legendre's polynomial, then the value of P2n(0) is given by -
Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024
1. $(-1)^n \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots (2n)}$
2. $\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots (2n)}$
3. $(-1)^n \frac{1 \cdot 3 \cdot 5 \dots (2n+1)}{2 \cdot 4 \cdot 6 \dots (2n)}$
4. $\frac{1 \cdot 3 \cdot 5 \dots (2n+1)}{2 \cdot 4 \cdot 6 \dots (2n)}$
Show answer
Answer: A. $(-1)^n \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots (2n)}$
The value of Legendre polynomials at $x=0$ follows a specific pattern based on whether the index $n$ is even or odd. For even indices ($2n$), the formula involves the double factorial of odd numbers and is multiplied by $(-1)^n$.
The solution of the differential equation $y = 4xp - 16y^3p^2$, ($p = \frac{dy}{dx}$) is -
Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024
1. $y^3 = Cx - C^2$
2. $y = Cx - C^2$
3. $y^2 = Cx - C^2$
4. $y^4 = Cx - C^2$
Show answer
Answer: A. $y^3 = Cx - C^2$
The equation is a non-linear differential equation. Solutions often take the form $y^k = Cx - C^2$ for specific values of $k$. Verifying a proposed solution involves substituting it and its derivative into the original differential equation to ensure the equality holds.
Which one of the following statement is correct? Every scalar matrix is an identity matrix. Every diagonal matrix is a scalar matrix. Matrix multiplication is always commutative. Every scalar matrix is a diagonal matrix.
Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024
1. (1)
2. (2)
3. (3)
4. (4)
Show answer
Answer: D. (4)
Understanding the definitions of different types of matrices (identity, diagonal, scalar) and the properties of matrix operations (specifically, commutativity of multiplication) is crucial. A scalar matrix is a special case of a diagonal matrix.
The particular integral of differential equation $\frac{d^2x}{dt^2} + \frac{g}{b}(x- a)=0$, where a, b and g are, different positive constants is -
Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024
1. a
2. $\frac{ga}{b}$
3. 0
4. $\frac{b}{g}$
Show answer
Answer: A. a
The particular integral of a linear differential equation with constant coefficients is the part of the solution that satisfies the non-homogeneous term. For a constant non-homogeneous term $K$, if the differential operator $f(D)$ evaluated at $D=0$ is non-zero, the particular integral is $K/f(0)$. Alternatively, one can assume a constant solution and substitute it into the equation.
The series $\sum_{n=0}^{\infty} \frac{x^n} {n^2+1}$, $(x>0)$ is divergent, if -
Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024
1. $x>1$
2. $x=1$
3. $x \leq 1$
4. $x<1$
Show answer
Answer: A. $x>1$
The Ratio Test is a powerful tool for determining the convergence or divergence of series, especially those involving powers and factorials. For a series $ \sum a_n$, if $\lim_{n\to\infty} |\frac{a_{n+1}} {a_n}| = L$, then the series converges if $L1$.
The inverse Laplace transform of $\frac{1} {s^2+4s+13}$ is -
Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024
1. $e^{-2t} \cos 3t - \frac{2}{3}e^{-2t} \sin 3t$
2. $e^{-2t} \cos 3t - \frac{1}{3}e^{-2t} \sin 3t$
3. $e^{-2t} \cos 3t + \frac{2}{3}e^{-2t} \sin 3t$
4. $e^{-2t} \cos 3t + \frac{1}{3}e^{-2t} \sin 3t$
Show answer
Answer: C. $e^{-2t} \cos 3t + \frac{2}{3}e^{-2t} \sin 3t$
The inverse Laplace transform of functions involving $s^2+as+b$ can be found by completing the square in the denominator and using the first shifting theorem ($L\{e^{-at}f(t)\} = F(s+a)$) in conjunction with standard transform pairs for $ \cos(\omega t)$ and $\sin(\omega t)$.
The equation $x^4 + 5x^3 - x^2 + 7x + 9 = 0$ has roots -
Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024
1. at most two positive roots
2. at most three positive roots
3. at most three negative roots
4. at most one negative root
Show answer
Answer: A. at most two positive roots
Descartes' Rule of Signs is used to determine the maximum number of positive and negative real roots of a polynomial. For positive roots, count sign changes in the coefficients of $P(x)$. For negative roots, count sign changes in the coefficients of $P(- x)$. The actual number of roots is either equal to the count or less than it by an even integer.

Newton's Laws of Motion and their applications

A cuboid has six sides of different colours. The red side is opposite to black. The blue side is adjacent to white. The brown side is adjacent to blue. The red side is face down. Which one of the following would be opposite to brown?
Asked in: APSC CCE Prelims Exam 2014
1. Red
2. Black
3. White
4. Blue
Show answer
Answer: C. White
- A cuboid has three pairs of opposite faces. - Adjacent faces share an edge; opposite faces do not share any edge. - If two colours are adjacent, they cannot be opposite.
Running at a speed of 60 km per hour, a train passes through a 1.5 km long tunnel in two minutes. What is the length of the train?
Asked in: APSC CCE Prelims Exam 2015
1. 100 meters
2. 250 meters
3. 500 meters
4. 1000 meters
Show answer
Answer: C. 500 meters
When a train passes a stationary object of significant length (like a tunnel or bridge), the total distance covered by the train is the length of the object plus the length of the train itself.
A washing machine works on the principle of
Asked in: APSC CCE Prelims Exam 2020
1. gravitation
2. acceleration
3. centrifugation
4. linear motion
Show answer
Answer: C. centrifugation
Centrifugation is a process that involves the use of centrifugal force (or, more accurately, inertia in a rotating frame) to separate particles from a solution according to their size, shape, density, medium viscosity, and rotor speed. In a washing machine, water is separated from the clothes.
A man can row 7 1/2 km per hour in still water. If in a river running at 1 1/2 km an hour, it takes him 50 minutes to row to a place and back, how far off is the place?
Asked in: BPSC Prelims Exam 63rd
1. 3 km
2. 4 km
3. 2 km
4. 5 km
Show answer
Answer: A. 3 km
- Speed of the man in still water = 7 1/2 km/hr = 15/2 km/hr. - Speed of the stream = 1 1/2 km/hr = 3/2 km/hr. - Speed downstream = Speed in still water + Speed of stream = 15/2 + 3/2 = 18/2 = 9 km/hr. - Speed upstream = Speed in still water - Speed of stream = 15/2 - 3/2 = 12/2 = 6 km/hr. - Total time taken for the round trip = 50 minutes = 50/60 hours = 5/6 hours. - Let the distance to the place be 'd' km. - Time taken to row downstream = d / Speed downstream = d/9 hours. - Time taken to row upstream = d / Speed upstream = d/6 hours. - Total time = Time downstream + Time upstream => d/9 + d/6 = 5/6. - Find a common denominator for 9 and 6, which is 18. Multiply the equation by 18: 18(d/9) + 18(d/6) = 18(5/6). - This simplifies to 2d + 3d = 3 5. - 5d = 15. - d = 15/5 = 3 km.
A train of length 150 meters, moving towards north direction at a speed of 144 km/hr, can cross a 250 meters long bridge in
Asked in: BPSC Prelims Exam 64th
1. 20 seconds
2. 100 seconds
3. 45 seconds
4. 10 seconds
Show answer
Answer: D. 10 seconds
- When a train crosses a bridge, the total distance covered is the sum of the train's length and the bridge's length. - Speed must be converted to consistent units (e.g., m/s) if distance is in meters and time is required in seconds. - The relationship is Time = Distance / Speed.
If x² - y² = 7 and x - y = 1, then the length of a diagonal of a rectangle with length and width respectively x cm and y cm will be
Asked in: BPSC Prelims Exam 65th
1. 5 cm
2. 6 cm
3. 7 cm
4. 8 cm
Show answer
Answer: A. 5 cm
We have a system of two linear equations: $x - y = 1$ and $x + y = 7$. Adding the two equations gives $(x - y) + (x + y) = 1 + 7$, which simplifies to $2x = 8$, so $x = 4$. Substituting $x = 4$ into $x - y = 1$ gives $4 - y = 1$, so $y = 3$. The length and width are 4 cm and 3 cm.
Which one of the following is a scalar quantity?
Asked in: BPSC Prelims Exam 65th
1. Force
2. Pressure
3. Velocity
4. Acceleration
Show answer
Answer: B. Pressure
Pressure is a scalar quantity, while force, velocity, and acceleration are vector quantities.
A train 110 m long is travelling at a speed of 58 kmph. The time in which it will pass a passerby walking at 4 kmph in the same direction is
Asked in: BPSC Prelims Exam 67th Cancelled
1. 6 seconds
2. $7\frac{1}{2}$ seconds
3. $7\frac{1}{3}$ seconds
4. 8 seconds
Show answer
Answer: C. $7\frac{1}{3}$ seconds
The train is 110 m long. Speed of the train ($v_t$) = 58 kmph. Speed of the passerby ($v_p$) = 4 kmph. They are moving in the same direction. When objects move in the same direction, their relative speed is the difference between their speeds. Relative speed ($v_{rel}$) = $v_t - v_p = 58 \text{ kmph} - 4 \text{ kmph} = 54 \text{ kmph}$. To pass the passerby, the train needs to cover a distance equal to its own length relative to the passerby. Distance to be covered = Length of the train = 110 m. We need to convert the relative speed from kmph to m/s. 1 kmph = 1000 meters / 3600 seconds = 5/18 m/s. $v_{rel} = 54 \text{ kmph} = 54 \times \frac{5}{18} \text{ m/s} = 3 \times 5 \text{ m/s} = 15 \text{ m/s}$. Time taken = Distance / Relative speed. Time = 110 m / 15 m/s. Time = $\frac{110}{15}$ seconds. Simplify the fraction: $\frac{110}{15} = \frac{22 \times 5}{3 \times 5} = \frac{22}{3}$ seconds. Convert the improper fraction to a mixed number: $\frac{22}{3} = 7$ with a remainder of 1, so $7 \frac{1}{3}$ seconds.

Rigid Body Dynamics (Euler Angles, Tops and Gyroscopes)

The centre of gravity of a system of rigid bodies coincides with their centre of mass if and only if
Asked in: UPSC Geoscientist 2022
1. their centre of mass is at their geometrical centre
2. the acceleration due to gravity is same throughout the system of bodies
3. the rigid bodies have same uniform mass dens‐ ities
4. the rigid bodies are very large
Show answer
Answer: B. the acceleration due to gravity is same throughout the system of bodies
- CG and CM coincide when gravitational acceleration (g) is constant across the system. - CM depends on mass distribution. - CG depends on both mass distribution and the distribution of the gravitational field.
A person stands at the middle point of a wooden ladder which starts slipping between a vertical wall and the floor of a room, while continuing to remain in a vertical plane. The path traced by a person standing at the middle point of the slipping ladder is:
Asked in: UPPCS Prelims 2004
1. A straight line
2. An elliptical path
3. A circular path
4. A parabolic path
Show answer
Answer: C. A circular path
This is a problem in coordinate geometry and loci. By representing the position of the ladder's ends and applying the Pythagorean theorem, the coordinates of the midpoint can be related, revealing the shape of the path it traces.
Find the torque of a force $\vec{F} = 7\hat{i} + 3\hat{j} - 5\hat{k}$ about the origin. The force acts on a particle whose position vector is $\vec{r} = \hat{i} - \hat{j} + \hat{k}$.
Asked in: Co operative Supervisor & Environment Supervisor (Group C) Exam 2023
1. $2\hat{i} + 12\hat{j} + 10\hat{k}$
2. $10\hat{i} + 2\hat{j} + 12\hat{k}$
3. $12\hat{i} + 10\hat{j} + 2\hat{k}$
4. $6\hat{i} + 5\hat{j} + 9\hat{k}$
Show answer
Answer: B. $10\hat{i} + 2\hat{j} + 12\hat{k}$
The torque ($\vec{\tau}$) of a force ($\vec{F}$) about a point (taken as the origin in this case) is calculated as the cross product of the position vector ($\vec{r}$) from the point to the point of application of the force and the force vector itself: $ \vec{\tau} = \vec{r} \times \vec{F}$. The calculation is performed using the determinant method for vector cross products.

Rotational Dynamics and Moment of Inertia

A planar, irregular shaped object has a mass of 2 kg. Its moment of inertia along the two orthogonal axes in the plane of the object are 3 kg m² and 4 kg m², respectively. The moment of inertia of the object along an axis perpendicular to the plane and passing through a point 0·5 m away from its centre of mass is represented by I. Which one among the following is the correct value of I ?
Asked in: UPSC Geoscientist 2024
1. 7·5 kg m²
2. 7 kg m²
3. 5 kg m²
4. 1 kg m²
Show answer
Answer: A. 7·5 kg m²
By the perpendicular axis theorem, the moment of inertia about the center of mass (assuming the given orthogonal axes intersect at CM) and perpendicular to the plane is I_CM = Ix + Iy = 3 kg m² + 4 kg m² = 7 kg m². By the parallel axis theorem, the moment of inertia I about a parallel axis perpendicular to the plane and at a distance d = 0.5 m from CM is I = I_CM + m d².
Inhabitants are unaware of the speed of rotation of the planet Earth because the angular velocity is constant for each place on the Earth's surface the atmosphere rotates with the Earth there are no nearby objects, either stationary or moving at a rate different from that of the Earth Which of the above is/are the correct explanation(s)?
Asked in: UPSC NDA 1 2022
1. 1 only
2. 1 and 2 only
3. 2 and 3 only
4. 1, 2 and 3
Show answer
Answer: D. 1, 2 and 3
Our perception of motion is relative. Since everything around us (atmosphere, land, buildings) is rotating with us at the same speed, and the rotation is constant, we don't feel the motion or observe significant relative movement.
A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their center of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true ?
Asked in: UPSC NDA 2 2019
1. The ring has higher kinetic energy
2. The disc has higher kinetic energy
3. The ring and the disc have the same kinetic energy
4. Kinetic energies of both the bodies are zero since they are not in linear motion
Show answer
Answer: A. The ring has higher kinetic energy
The moment of inertia of a thin ring about an axis through its center and perpendicular to its plane is I_ring = M R². The moment of inertia of a thin disc about an axis through its center and perpendicular to its plane is I_disc = (1/2) M R². Since I_ring > I_disc (MR² is greater than (1/2)MR²), and ω is the same for both, the rotational kinetic energy of the ring is higher than that of the disc (KE_ring = (1/2)I_ringω² > KE_disc = (1/2)I_discω²).
A solid disc and a solid sphere have the same mass and same radius. Which one has the higher moment of inertia about its centre of mass ?
Asked in: UPSC NDA 2 2019
1. The disc
2. The sphere
3. Both have the same moment of inertia
4. The information provided is not sufficient to answer the question
Show answer
Answer: A. The disc
- Moment of inertia depends on the mass distribution relative to the axis of rotation. - Formulas for the moment of inertia of common shapes are standard results derived from integration. - For objects of the same mass and radius, the object with more mass distributed further from the axis of rotation will have a higher moment of inertia. In the disc, all mass is at a distance up to R from the axis in a plane, whereas in the sphere, mass is distributed in a volume, including closer to the center.
Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁ and r₂ respectively. Their speeds are such that each car makes a complete circle in the same time 't'. The ratio of angular speed of the first to that of the second car is :
Asked in: UPSC CAPF 2015
1. m₁ : m₂
2. 1 : 1
3. r₁ : r₂
4. 1 : 2
Show answer
Answer: B. 1 : 1
Angular speed (ω) is defined as the rate of change of angular displacement. For an object moving in a circle, the angular speed is given by ω = 2π / T, where T is the time period (time taken to complete one revolution). The problem states that both cars make a complete circle in the same time 't'. Therefore, the time period T is the same for both cars. For the first car, ω₁ = 2π / t. For the second car, ω₂ = 2π / t. The ratio of their angular speeds is ω₁ : ω₂ = (2π / t) : (2π / t) = 1 : 1. The masses (m₁ and m₂) and radii (r₁ and r₂) of the cars are irrelevant for determining the ratio of angular speeds when the time period is given to be the same.
Identify the vector quantity from the following:
Asked in: WBPSC CS Prelims 2020
1. Heat
2. Angular momentum
3. Time
4. Work
Show answer
Answer: B. Angular momentum
A vector quantity is a physical quantity that has both magnitude and direction. A scalar quantity has only magnitude.
The velocity ratio of the hour hand and minute hand of a clock is
Asked in: WBPSC CS Prelims 2021
1. 1 : 12
2. 12 : 1
3. 24 : 1
4. 1 : 24
Show answer
Answer: A. 1 : 12
The velocity ratio of the hands of a clock refers to the ratio of their angular velocities. The minute hand completes one full revolution (360 degrees) in 60 minutes. Its angular velocity is $360^\circ / 60 \text{ minutes} = 6^\circ/\text{minute}$. The hour hand completes one full revolution (360 degrees) in 12 hours, which is $12 \times 60 = 720$ minutes. Its angular velocity is $360^\circ / 720 \text{ minutes} = 0.5^\circ/\text{minute}$. The ratio of the angular velocity of the hour hand to the angular velocity of the minute hand is $0.5^\circ/ \text{minute} : 6^\circ/\text{minute} = 0.5 : 6$. Multiplying both sides by 2 to get whole numbers, the ratio is $1 : 12$. The linear velocity of the tip of the hands would depend on their lengths, but the question usually implies angular velocity or proportional linear velocity for tips of equal length, which is the same ratio.
A bicycle in motion does not fall because one of the following is conserved.
Asked in: WBCS Main Paper 4 2022
1. Linear momentum
2. Angular momentum
3. Kinetic energy
4. All of the above
Show answer
Answer: B. Angular momentum
When a bicycle is in motion, its wheels are spinning, creating angular momentum. The conservation of this angular momentum provides a gyroscopic effect that helps stabilize the bicycle and resist tipping. This effect is stronger at higher speeds, making it easier to balance a moving bicycle than a stationary one.

Simple Harmonic Motion and Damped/Forced Oscillations

The following diagram shows a pendulum at different positions. Which one of the following statement is true ? [Image of a pendulum at positions P, Q, R, S, T]
Asked in: UPSC CDS 2 2023
1. The pendulum has minimum potential energy at positions P and T.
2. The pendulum has minimum potential energy at positions Q and S.
3. The pendulum has minimum potential energy at position R.
4. The pendulum has same potential energy at all positions.
Show answer
Answer: C. The pendulum has minimum potential energy at position R.
Potential energy is given by PE = mgh, where m is mass, g is acceleration due to gravity, and h is height. As the pendulum swings, its height above the equilibrium position (lowest point) changes. Potential energy is maximum at the extreme positions (P and T), where the pendulum is momentarily at its highest point and stops before changing direction. Potential energy is minimum at the mean position (R), where the pendulum is at its lowest point and moving with maximum speed.
Which one of the following statements is true regarding the time period (T) of oscillation of a simple pendulum?
Asked in: UPSC Geoscientist 2020
1. T increases with increase in length of the pen‐ dulum
2. T depends on the mass of the pendulum bob
3. T decreases with increase in length of the pen‐ dulum
4. T is independent of the length of the pendulum
Show answer
Answer: A. T increases with increase in length of the pen‐ dulum
The time period of a simple pendulum is proportional to the square root of its length.
The angular acceleration of a simple pendulum at an angle α from the vertical is proportional to
Asked in: UPSC Geoscientist 2023
1. tan α
2. sin² α
3. sin α
4. sin 2α
Show answer
Answer: C. sin α
- Restoring torque τ = -mgl sin α. - Torque τ = Iβ. - Moment of inertia of a simple pendulum I = ml². - Angular acceleration β = τ/I = -(g/l) sin α. - Magnitude of angular acceleration is proportional to sin α.
A wall clock shows that the time is 3.00 AM. What would be the time after the hour hand rotates by 135° from that position?
Asked in: HAS Prelims 2023
1. 6.00 AM
2. 6.30 AM
3. 7.30 AM
4. 8.00 AM
Show answer
Answer: C. 7.30 AM
At 3:00 AM, the hour hand is pointing directly at the '3' mark on the clock. A full circle on a clock face is 360°. There are 12 hours marked on the clock. The angle between consecutive hour marks is 360° / 12 = 30°. The hour hand moves 30° every hour. We are told the hour hand rotates by 135°. To find out how many hours this rotation represents, we divide the angle by the degrees moved per hour: 135° / 30°/hour = 4.5 hours. So, the time is 3:00 0.5 hours 60 minutes/hour = 4 hours and 30 minutes. Starting from 3:00 AM and adding 4 hours and 30 minutes gives 7:30 AM.
A clock shows the time as 12:20. What is the angle the hour hand makes with the minute hand?
Asked in: JPSC CS Prelims 2011
1. 100°
2. 110°
3. 120°
4. 130°
Show answer
Answer: B. 110°
The hour hand moves 0.5° per minute, and the minute hand moves 6° per minute. The angle between the hands is calculated based on their respective positions relative to the 12 mark (0 degrees).
The pendulum of a clock consists of a heavy point mass (bob) tied to a massless string. It is oscillating faster and so the clock is going fast. What option below should be chosen in order to reduce the time of oscillation ?
Asked in: KPSC KAS Prelims 2015
1. The mass of the bob should be increased
2. The length of the pendulum should be in‐ creased
3. The length of the pendulum should be de‐ creased
4. The mass of the bob should be decreased
Show answer
Answer: B. The length of the pendulum should be in‐ creased
The time period (T) of a simple pendulum is given by the formula T = 2π sqrt(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the period of a simple pendulum. A clock goes fast if its pendulum oscillates too quickly, meaning its period is too short. To correct a fast clock and make it run slower (increase its period), the period needs to be increased.
A spring can be used to determine the mass m of an object in two ways : (i) by measuring the extension in the spring due to the object; and (ii) by measuring the oscillation period for the given mass. Which of these methods can be used in a space-station orbiting Earth ?
Asked in: UPSC NDA 1 2015
1. Both
2. Only the extension method
3. Only the oscillation method
4. Neither
Show answer
Answer: C. Only the oscillation method
Measuring extension relies on weight (force due to gravity), which is effectively zero in orbit. Measuring oscillation period relies on inertial mass and spring constant, which are unaffected by gravity.
Which one of the following statements is correct regarding the provided displacement versus time curve for a particle executing simple harmonic motion?
Asked in: UPSC NDA 1 2017
1. Phase of the oscillating particle is same at t=1s and t=3s
2. Phase of the oscillating particle is same at t=2s and t=8s
3. Phase of the oscillating particle is same at t=3s and t=17s
4. Phase of the oscillating particle is same at t=4s and t=10s
Show answer
Answer: B. Phase of the oscillating particle is same at t=2s and t=8s
The phase of simple harmonic motion repeats itself every period T. Therefore, points in time separated by an integer multiple of T have the same phase.

Special Theory of Relativity (Lorentz Transformations, Relativistic Dynamics)

A rocket 100 m long at rest starts to move with uniform motion of 0.8 C. What is its length as seen by an observer at rest?
Asked in: TNPSC CS Prelims 2014
1. 60 cm
2. 80 cm
3. 100 cm
4. 0
Show answer
Answer: A. 60 cm
- This problem involves length contraction, a phenomenon predicted by special relativity. The formula for length contraction is L = L₀ sqrt(1 - v²/ c²), where L₀ is the proper length (length at rest), v is the velocity, and c is the speed of light. - Given L₀ The contracted length L = 100 m 0.6 = 60 m. - However, the options are given in centimeters. If we assume the initial length was 100 cm instead of 100 m, then the contracted length would be L = 100 cm 0.6 = 60 cm. This matches option A. - If the initial length was 100 m, the answer is 60 m, which is 6000 cm. None of the options match 6000 cm. - Given the options, it is most probable that the question either intended the initial length to be 100 cm or there is a unit inconsistency in the options, with 60 cm being the result if the initial length was 100 cm. We proceed with the assumption that leads to a valid option.
Newton's law of motion cannot be applicable to the particles moving at a speed comparable to the speed of
Asked in: UPSC CISF AC EXE 2017
1. light
2. sound
3. rocket
4. bullet train
Show answer
Answer: A. light
- Newton's laws are part of classical mechanics. - Classical mechanics is an approximation that works well for macroscopic objects at relatively low speeds. - It fails when speeds approach the speed of light (requiring relativity) or at very small scales (requiring quantum mechanics).

Systems of Particles and Collisions

For which one of the following does the centre of mass lie outside the body?
Asked in: UPSC CDS 1 2017
1. A fountain pen
2. A cricket ball
3. A ring
4. A book
Show answer
Answer: C. A ring
The center of mass of an object is a point where the weighted average of the positions of all parts of the object is located. While for uniform solid objects with simple geometry (like a sphere or cube), the center of mass is often at the geometric center and inside the object, for objects with holes or non- uniform density, the center of mass can lie outside the physical boundaries of the object. A ring is a hollow circular structure. Its mass is distributed along the circumference, but the geometric center (and thus the center of mass for a uniform ring) is in the empty space at the center of the circle.
Which one of the following is the quantity of transfer of linear momentum to the floor, when a dumbbell of mass 500 g falls from a height of 5 m and stops after hitting the floor (take $g=10 \text{ m/s}^2$)?
Asked in: UPSC Geoscientist 2023
1. 0·5 kg-m/s
2. 5·0 kg-m/s
3. 10·0 kg-m/s
4. 1·0 kg-m/s
Show answer
Answer: B. 5·0 kg-m/s
The transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction. First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 \text{ m/s}^2$, and $h=5 \text{ m}$. $v^2 = 0^2 + 2 \times 10 \times 5 = 100 \text{ m}^2/\text{s}^2$. $v = \sqrt{100} = 10 \text{ m/s}$. The direction is downwards. The mass of the dumbbell is $m = 500 \text{ g} = 0.5 \text{ kg}$. The momentum of the dumbbell just before impact is $p_{initial} = m \times v = 0.5 \text{ kg} \times 10 \text{ m/s} = 5 \text{ kg-m/s}$ (downwards). After hitting the floor, the dumbbell stops, so its final velocity is 0. The momentum of the dumbbell after impact is $p_{final} = m \times 0 = 0$. The change in momentum of the dumbbell is $\Delta p_{dumbbell} = p_{final} - p_{initial} = 0 - (5 \text{ kg-m/s downwards}) = -5 \text{ kg-m/s downwards} = 5 \text{ kg-m/s upwards}$. By the impulse-momentum theorem and Newton's third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell. Momentum transfer to floor = $-\Delta p_{dumbbell} = -(5 \text{ kg-m/s upwards}) = 5 \text{ kg-m/s downwards}$. The question asks for the quantity of transfer, which is the magnitude. The magnitude is 5 kg-m/s.
During inelastic collision between two bodies, which of the following quantities always remains conserved?
Asked in: MPSC CS Prelims 2019
1. Total kinetic energy
2. Total mechanical energy
3. Total linear momentum
4. Speed of each body
Show answer
Answer: C. Total linear momentum
- Total kinetic energy is conserved only in elastic collisions, not inelastic ones. - Total mechanical energy is also not conserved in inelastic collisions due to the conversion of kinetic energy into other forms. - The speed of individual bodies changes during a collision.
A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart?
Asked in: UPSC NDA 1 2022
1. 2.15 m/s
2. 1.89 m/s
3. 1.51 m/s
4. 2.51 m/s
Show answer
Answer: B. 1.89 m/s
- The system consists of the boy and the stationary cart. - Before the jump, the boy has momentum (mass × velocity) and the cart has zero momentum. - After the boy jumps onto the cart, they move together as a single system with a common velocity. - According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned). - Initial momentum = (mass of boy × velocity of boy) + (mass of cart × velocity of cart) = (52 kg × 2 m/s) + (3 kg × 0 m/s) = 104 kg m/s. - Let the final velocity of the combined system (boy + cart) be 'v'. The combined mass is 52 kg + 3 kg = 55 kg. - Final momentum = (combined mass × final velocity) = 55 kg × v. - By conservation of momentum: 104 kg m/ s = 55 kg × v. - v = 104 / 55 ≈ 1.89 m/s.
A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45 m and rises to a height of 0.20 m. If it was in touch with the floor for 0.1 s, the net force it applied on the floor while bouncing is : (take the gravitational acceleration g = 10 m s⁻²)
Asked in: UPSC NDA 1 2024
1. 1.0 N
2. 6.0 N
3. 3.0 N
4. 5.0 N
Show answer
Answer: B. 6.0 N
First, calculate the velocity of the ball just before impact (v₁) and just after impact (v₂). Using energy conservation (v² = 2gh), v₁ = sqrt(2 10 0.45) = 3 m/s (downward) and v₂ = sqrt(2 10 0.20) = 2 m/s (upward). Let's take upward as positive. v₁ = -3 m/s, v₂ = +2 m/s. The change in momentum of the ball is kg m/s (upward). The average net force on the ball during contact is F_net_on_ball = Δp / Δt = 0.5 Ns / 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N - mg. So, 5 N = N - (0.1 kg 10 m/s²) = N - 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton's third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.
A metallic bob X of mass m is released from position A. It collides elastically with another identical bob Y placed at rest at position B on a horizontal frictionless table. The angle AOB is 30°. How high does the bob X rise immediately after the collision?
Asked in: UPSC NDA 1 2024
1. To the same height as that of position A on the other side in the same trajectory
2. To half the height as that of position A on the other side along the same trajectory
3. The same height at position A
4. It stops at position B
Show answer
Answer: D. It stops at position B
- The collision is described as elastic. In an elastic collision, both momentum and kinetic energy are conserved. - The colliding bobs are identical, meaning they have the same mass (m). - Bob Y is at rest at position B before the collision. Bob X has a certain velocity (V) just before the collision at B due to its swing from A. - For a head-on elastic collision between two objects of equal mass where one object is initially at rest, the moving object comes to rest, and the stationary object moves off with the initial velocity of the moving object.
Conservation of momentum in a collision between particles can be understood on the basis of:
Asked in: UPSC NDA 2 2015
1. Newton's first law of motion
2. Newton's second law of motion only
3. Both Newton's second law of motion and Newton's third law of motion
4. Conservation of energy
Show answer
Answer: C. Both Newton's second law of motion and Newton's third law of motion
Conservation of momentum for a system relies on the principle of action-reaction (Newton's third law) and the relationship between force and change in momentum (Newton's second law).
When a ball bounces off the ground, which of the following changes suddenly ? (Assume no loss of energy to the floor)
Asked in: UPSC NDA 2 2019
1. Its speed
2. Its momentum
3. Its kinetic energy
4. Its potential energy
Show answer
Answer: B. Its momentum
Velocity is a vector quantity, having both magnitude (speed) and direction. Momentum is mass multiplied by velocity (p = mv), so it is also a vector quantity. When a ball bounces, its downward velocity just before impact changes to an upward velocity just after impact. Even if the speed remains the same (in a perfectly elastic collision with no energy loss), the change in direction causes a significant change in the velocity vector, and thus a sudden change in the momentum vector.

Work, Energy, and Power

Which of the following energy changes involves frictional force?
Asked in: BPSC Prelims Exam 68th
1. Potential energy to sound energy
2. Chemical energy to heat energy
3. Kinetic energy to heat energy
4. More than one of the above
Show answer
Answer: C. Kinetic energy to heat energy
Frictional force is dissipative, converting mechanical energy, particularly kinetic energy associated with relative motion, into heat.
If the linear momentum of a moving object gets doubled due to application of a force, then its kinetic energy will
Asked in: UPSC CDS 2 2020
1. remain same
2. increase by four times
3. increase by two times
4. increase by eight times
Show answer
Answer: B. increase by four times
The linear momentum (p) of an object is given by the product of its mass (m) and velocity (v), i.e., p = mv. The kinetic energy (KE) is given by KE = (1/2)mv². We can express KE in terms of momentum: KE = (1/2)m(p/m)² = p²/(2m). If the momentum is doubled (p' = 2p), the new kinetic energy (KE') will be KE' = (p')²/(2m) = (2p)²/(2m) = 4p²/ (2m) = 4 (p²/(2m)) = 4 KE.
If an object of mass 10 kg is moving with a uniform speed of 10 m/s, then the linear momentum and the kinetic energy of the object, respectively, are
Asked in: UPSC CDS 2 2022
1. 100 N.s and 500 J
2. 100 N.s and 1000 J
3. 200 N.s and 500 J
4. 200 N.s and 1000 J
Show answer
Answer: A. 100 N.s and 500 J
- Linear momentum (p) is given by the product of mass (m) and velocity (v). The unit of linear momentum is kg.m/s, which is equivalent to N.s (Newton-second). Given m = 10 kg and v = 10 m/s, p = m v = 10 kg 10 m/s = 100 kg.m/s = 100 N.s.- Kinetic energy (KE) is given by the formula 0.5 m v². The unit of kinetic energy is Joules (J). Given m = 10 kg and v = 10 m/s, KE = 0.5 10 kg (10 m/s)² = 0.5 10 kg 100 m²/s² = 5 100 J = 500 J.Therefore, the linear momentum is 100 N.s and the kinetic energy is 500 J.
In which of the following situations will an applied force do negative work on a body?
Asked in: UPSC Geoscientist 2023
1. The applied force and displacement of the body are at 135° to each other
2. The applied force and displacement of the body are parallel to each other
3. The applied force and displacement of the body are perpendicular to each other
4. The applied force and displacement of the body are at 45° to each other
Show answer
Answer: A. The applied force and displacement of the body are at 135° to each other
Work done by a constant force $\vec{F}$ on a body that undergoes a displacement $\vec{d}$ is given by the dot product: $W = \vec{F} \cdot \vec{d} = | \vec{F}| |\vec{d}| \cos \theta$, where $\theta$ is the angle between the force and displacement vectors. Work is negative when $\cos \theta$ is negative. This occurs when the angle $\theta$ is between 90° and 270° (exclusive of 90° and 270° where work is zero). Let's examine the options: A) Angle is 135°. $\cos 135° = -1/\sqrt{2}$. Since $\cos \theta$ is negative, the work done is negative. B) Angle is 0° (same direction) or 180° (opposite direction). If $ \theta=0°$, $\cos 0° = 1$, work is positive. If $ \theta=180°$, $\cos 180° = -1$, work is negative. So, parallel force and displacement can result in negative work (if opposite directions), but doesn't always. C) Angle is 90°. $\cos 90° = 0$. Work is zero. D) Angle is 45°. $\cos 45° = 1/\sqrt{2}$. Since $\cos \theta$ is positive, work is positive. Option A is the only situation listed that guarantees the work done by the applied force is negative.
A frictionless pulley has a rope of negligible mass passing over it. Blocks of mass 15 kg and 5 kg are attached at the two ends of the rope and held stationary. When the masses are released to move freely, their speed in m s-1, after the rope has slipped by 0.25 m over the pulley, is closest to : (take g = 10.0 m s-2)
Asked in: UPSC Geoscientist 2024
1. 5.0
2. 2.2
3. 1.6
4. 1.2
Show answer
Answer: C. 1.6
- This problem can be solved using the conservation of mechanical energy, as the pulley is frictionless and the rope has negligible mass (ideal Atwood machine). - Let the initial height of the 15 kg mass be $y_1$ and the 5 kg mass be $y_2$. When the rope slips by 0.25 m, the 15 kg mass moves down mass moves up by 0.25 m (new height $y_2 + 0.25$). - The initial potential energy (PE) can be set relative to the initial positions. $\text{PE}_{initial} = m_1 g y_1 + m_2 g y_2$. - The final potential energy is $\text{PE}_{final} = m_1 g (y_1 - 0.25) m_2 g y_2 + 0.25 m_2 g$. - The change in potential energy is $\Delta \text{PE} = \text{PE}_{final} - 0.25 g (m_2 - m_1)$. - Given $m_1 = 15$ kg, $m_2 = 5$ kg, and $g = 10.0$ m/s², $\Delta \text{PE} = -25$ J. The system loses 25 J of potential energy. - The initial kinetic energy (KE) is zero as the masses are held stationary. $\text{KE}_{initial} = 0$. - Let $v$ be the speed of the masses after moving 0.25 m. Both masses move with the same speed. The final kinetic energy is $\text{KE}_{final} = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 = \frac{1}{2} (m_1 + m_2) v^2$. - $\text{KE}_{final} = \frac{1}{2} (15 + 5) v^2 = \frac{1}{2} (20) v^2 = 10 v^2$. - By the conservation of mechanical energy, $\Delta \text{KE} + \Delta \text{PE} = 0$. - $(10 v^2 - 0) + (-25) = 0$. - $10 v^2 = 25$. - $v^2 m/s. - Comparing with the options, 1.6 m/s is the closest value.
A ball of mass 100 g falls freely from a height of h = 20 m and hits the ground at a speed of 1·4 √gh. Take g = 10 m s⁻². Which one among the following is the correct value of the work done on the ball by air friction ?
Asked in: UPSC Geoscientist 2024
1. 0·4 J
2. 0·5 J
3. 0·3 J
4. 1 J
Show answer
Answer: A. 0·4 J
Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m/s²)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = ½ m v_f², where v_f = 1.4√gh. With g=10
A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm², then what is the volume of the spherical ball? [π = 3.14]
Asked in: GPSC Prelims CSP 1 2021
1. 107.04 cm³
2. 109.04 cm³
3. 113.04 cm³
4. None of the above
Show answer
Answer: C. 113.04 cm³
The curved surface area of a hemisphere is $2\pi r^2$. The volume of a sphere is $\frac{4}{3}\pi r^3$.
The volume of globe varies as the cube of its radius. Three solid globes of diameters 1$ \frac{1}{2}$, 2 and 2$\frac{1}{2}$ metres are melted and formed into a new solid globe. What is the diameter of the new globe?
Asked in: GPSC Prelims CSP 1 2022
1. 2.5 meters
2. 3 meters
3. 3.5 meters
4. None of the above
Show answer
Answer: B. 3 meters
The volume of a sphere is proportional to the cube of its diameter (V ∝ d³). When objects are melted and combined, their volumes are additive. The total volume of the melted globes equals the volume of the new globe.

Conservation of Linear and Angular Momentum

Elasticity (Stress, Strain, Hooke's Law, Elastic Moduli)

Fluid Statics and Dynamics (Continuity Equation, Bernoulli's Theorem)

Gravitation and Central Force Motion (Kepler's Laws)

Lagrangian

Formulation

(Generalized

Newton's Laws of Motion and their applications

Rigid Body Dynamics (Euler Angles, Tops and Gyroscopes)

Rotational Dynamics and Moment of Inertia

Simple Harmonic Motion and Damped/Forced Oscillations

Special Theory of Relativity (Lorentz Transformations, Relativistic Dynamics)

Systems of Particles and Collisions

Work, Energy, and Power

4. Motion under Gravity

Practice MCQs: Mechanics

Conservation of Linear and Angular Momentum

Elasticity (Stress, Strain, Hooke's Law, Elastic Moduli)

Fluid Statics and Dynamics (Continuity Equation, Bernoulli's Theorem)

Gravitation and Central Force Motion (Kepler's Laws)

Lagrangian Formulation (Generalized Coordinates, Euler-Lagrange Equations)

Newton's Laws of Motion and their applications

Rigid Body Dynamics (Euler Angles, Tops and Gyroscopes)

Rotational Dynamics and Moment of Inertia

Simple Harmonic Motion and Damped/Forced Oscillations

Special Theory of Relativity (Lorentz Transformations, Relativistic Dynamics)

Systems of Particles and Collisions

Work, Energy, and Power

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