Mechanics Archives

1. What is the length of a simple pendulum, which has a frequency of 0.5

What is the length of a simple pendulum, which has a frequency of 0.5 Hz (take g = 10 m/s²)?

[amp_mcq option1=”1 m” option2=”2 m” option3=”1.5 m” option4=”3 m” correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2024

The correct answer is A) 1 m.

The frequency (f) of a simple pendulum is related to its length (L) and the acceleration due to gravity (g) by the formula: f = 1 / (2π√(L/g)). Rearranging this formula to solve for L gives L = g / (4π²f²). Given f = 0.5 Hz and g = 10 m/s², and using the common approximation π² ≈ 10 when g ≈ 9.8 or 10 m/s², we get L = 10 / (4 * 10 * (0.5)²) = 10 / (40 * 0.25) = 10 / 10 = 1 meter.

The period (T) is the reciprocal of the frequency (T = 1/f = 1/0.5 = 2 seconds). The period formula is T = 2π√(L/g). Squaring both sides gives T² = 4π²(L/g), so L = gT² / (4π²). Using T=2s, g=10m/s², and π²≈10: L = 10 * (2)² / (4 * 10) = 10 * 4 / 40 = 40 / 40 = 1 meter. This confirms the result.

2. A person of mass 50 kg is standing in a lift. If the lift moves in upw

A person of mass 50 kg is standing in a lift. If the lift moves in upward direction with an acceleration of 1 m/s², then the weight of the person will be closest to :

[amp_mcq option1=”490 N” option2=”540 N” option3=”440 N” option4=”50 N” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2023

Download PDF Attempt Online

The correct answer is B.

When a person is in a lift accelerating upwards, the apparent weight of the person is greater than their true weight. The forces acting on the person are the gravitational force (mg) downwards and the normal force (N) exerted by the lift floor upwards. According to Newton’s second law, the net force (N – mg) is equal to mass times acceleration (ma). Thus, the normal force N = mg + ma = m(g + a). This normal force is the apparent weight.

Given mass m = 50 kg and upward acceleration a = 1 m/s². Using the standard value of g ≈ 9.8 m/s²: Apparent weight N = 50 kg * (9.8 m/s² + 1 m/s²) = 50 kg * 10.8 m/s² = 540 N. If we use g ≈ 10 m/s², N = 50 kg * (10 m/s² + 1 m/s²) = 50 kg * 11 m/s² = 550 N. The option closest to 540 N (or 550 N) is 540 N.

3. Which one of the following statements is correct?

Which one of the following statements is correct?

[amp_mcq option1=”Weight of an object may vary from place to place but its mass remains constant.” option2=”Mass of an object may vary from place to place but its weight remains constant.” option3=”Both weight and mass of an object do not vary from place to place.” option4=”Both weight and mass of an object vary from place to place.” correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2021

Download PDF Attempt Online

Mass is an intrinsic property of an object, representing the amount of matter it contains. It remains constant regardless of location. Weight, on the other hand, is the force exerted on an object due to gravity (Weight = Mass x gravitational acceleration). Since gravitational acceleration varies slightly depending on altitude, latitude, and local geological features, and significantly on different celestial bodies, the weight of an object can vary from place to place, while its mass remains constant.

– Mass is the amount of matter; it is constant.
– Weight is the force of gravity (mass x gravity); it varies with gravity.
– Gravitational acceleration varies from place to place.

On the surface of the Earth, the variation in gravitational acceleration is small, so the variation in weight is also relatively small. However, the distinction between mass and weight is fundamental in physics. On the Moon, an object has the same mass as on Earth but weighs significantly less because the Moon’s gravity is weaker.

4. In an atomic gas, the motion of particles (atoms) is governed by the c

In an atomic gas, the motion of particles (atoms) is governed by the collisions. If the gas is ionized, then the motion of created particles may be mainly governed by

[amp_mcq option1=”gravitational force.” option2=”collisions.” option3=”scattering of particles.” option4=”electromagnetic force between the particles.” correct=”option4″]

This question was previously asked in

UPSC CISF-AC-EXE – 2020

Download PDF Attempt Online

In an atomic gas, the atoms are electrically neutral, and their motion is primarily governed by collisions between them. When the gas is ionized, atoms lose or gain electrons, becoming charged particles (ions and free electrons). These charged particles exert strong electrostatic (electromagnetic) forces on each other over relatively long distances compared to the short-range forces involved in neutral collisions. Therefore, the motion of particles in an ionized gas (plasma) is mainly governed by the long-range electromagnetic forces between these charged particles, rather than just collisions.

In a neutral atomic gas, particle motion is dominated by collisions. In an ionized gas (plasma) containing charged particles, the motion is dominated by long-range electromagnetic forces between these charges.

An ionized gas is also known as plasma, which is often considered the fourth state of matter. The collective behavior of charged particles under the influence of electromagnetic fields is a key characteristic of plasma physics. While collisions still occur in plasma, their influence on overall motion is often less dominant than the electromagnetic forces, especially in hot, tenuous plasmas.

5. Consider the following statement : “When a body is in equilibrium, the

Consider the following statement :
“When a body is in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point.”
Which one of the following laws is described in the above statement ?

[amp_mcq option1=”Law of motion” option2=”Law of moments” option3=”Law of momentum” option4=”Law of magnetism” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

Download PDF Attempt Online

The statement “When a body is in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point” is the definition of rotational equilibrium. This condition is formally known as the Law of Moments, which is part of the requirements for a body to be in complete equilibrium (the other part being translational equilibrium, i.e., zero net force).

The Law of Moments is the principle that governs rotational equilibrium.

Moment (or torque) is the tendency of a force to cause rotation around an axis or pivot point. For a body to be in complete mechanical equilibrium, both the net force and the net moment acting on it must be zero.

6. Which of the following statements about a body under equilibrium is/ar

Which of the following statements about a body under equilibrium is/are correct ?

1. No forces are acting.
2. A number of parallel forces may be acting.
3. The law of moments must apply.

Select the correct answer using the code given below :

[amp_mcq option1=”1, 2 and 3″ option2=”2 and 3 only” option3=”1 and 3 only” option4=”2 only” correct=”option2″]

This question was previously asked in

UPSC CISF-AC-EXE – 2018

Download PDF Attempt Online

A body is under equilibrium if the net force acting on it is zero (translational equilibrium) and the net torque acting on it is zero (rotational equilibrium).
Statement 1: “No forces are acting” is incorrect. Forces can be acting, but they must be balanced, meaning the vector sum of all forces is zero.
Statement 2: “A number of parallel forces may be acting” is correct. For example, if two equal and opposite parallel forces act on a body, they form a couple, which produces a torque. For the body to be in equilibrium, this torque must be balanced by other torques, and the net force must be zero (which these two forces satisfy). More generally, parallel forces can sum to zero net force and zero net torque.
Statement 3: “The law of moments must apply” is correct. The Law of Moments states that for rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point. This is a necessary condition for a body to be in equilibrium.

Equilibrium requires zero net force and zero net torque. The Law of Moments describes the condition for zero net torque.

A body in equilibrium can be either at rest (static equilibrium) or moving with constant velocity (dynamic equilibrium). Both require the net force and net torque to be zero.

7. Newton’s law of motion cannot be applicable to the particles moving at

Newton’s law of motion cannot be applicable to the particles moving at a speed comparable to the speed of

[amp_mcq option1=”light” option2=”sound” option3=”rocket” option4=”bullet train” correct=”option1″]

This question was previously asked in

UPSC CISF-AC-EXE – 2017

Download PDF Attempt Online

Newton’s laws of motion are fundamental principles of classical mechanics. Classical mechanics provides an accurate description of the motion of objects in everyday life. However, these laws break down and are not applicable under certain extreme conditions:
1. When the speed of the object is comparable to the speed of light (approximately 3 x 10^8 m/s). In this regime, motion must be described by Einstein’s theory of special relativity. Relativistic effects like time dilation and length contraction become significant.
2. When the size of the object is very small (atomic or subatomic scales). In this regime, quantum mechanics is required to describe the behavior of particles.

The speed of sound, rockets, and bullet trains are all vastly lower than the speed of light, so classical mechanics (and Newton’s laws) apply accurately to objects moving at these speeds.

– 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).

The speed of light is a fundamental constant in the universe. No object with mass can reach or exceed the speed of light. Particles moving at speeds close to light speed are typically subatomic particles accelerated in particle accelerators.

8. Ball bearings are used in bicycles, cars, etc., because

Ball bearings are used in bicycles, cars, etc., because

[amp_mcq option1=”the actual area of contact between the wheel and axle is increased” option2=”the effective area of contact between the wheel and axle is increased” option3=”the effective area of contact between the wheel and axle is reduced” option4=”None of the above statements is correct” correct=”option3″]

This question was previously asked in

UPSC IAS – 2013

Download PDF Attempt Online

Ball bearings are used in bicycles, cars, etc., because they significantly reduce the effective area of contact between the moving parts (like the wheel and axle), thereby reducing friction.

Ball bearings replace sliding friction with rolling friction. Sliding friction occurs over a larger contact area between the axle and its housing. Ball bearings consist of spherical balls rolling between two races. The contact between the balls and the races is nearly point contact (in theory, or a very small ellipse in practice). This reduces the area experiencing friction and, crucially, replaces high sliding friction with much lower rolling friction.

Rolling friction is generally much less than sliding friction for the same load. By allowing parts to roll over each other via the balls, the overall frictional force opposing motion is greatly reduced, making movement easier and more efficient, and reducing wear and tear.

9. A simple harmonic motion of a particle is represented as, y = 10 cos ω

A simple harmonic motion of a particle is represented as, y = 10 cos ωt 10. The acceleration of the particle at time t = $\frac{\pi}{2\omega}$ will be : (symbols here carry their usual meanings)

[amp_mcq option1=”10 ω” option2=”$-10\omega^2$” option3=”0″ option4=”$\frac{10}{\omega}$” correct=”option3″]

This question was previously asked in

UPSC CAPF – 2023

Download PDF Attempt Online

The correct answer is 0.

The given equation for the simple harmonic motion of a particle is $y = 10 \cos \omega t + 10$. This represents the displacement of the particle from a reference point (in this case, an origin shifted by 10 units). The velocity of the particle is the first derivative of displacement with respect to time: $v = \frac{dy}{dt} = \frac{d}{dt}(10 \cos \omega t + 10) = -10 \omega \sin \omega t$. The acceleration of the particle is the first derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-10 \omega \sin \omega t) = -10 \omega^2 \cos \omega t$. We need to find the acceleration at time $t = \frac{\pi}{2\omega}$. Substituting this value of $t$ into the acceleration equation: $a\left(t=\frac{\pi}{2\omega}\right) = -10 \omega^2 \cos\left(\omega \cdot \frac{\pi}{2\omega}\right) = -10 \omega^2 \cos\left(\frac{\pi}{2}\right)$. Since $\cos\left(\frac{\pi}{2}\right) = 0$, the acceleration at this time is $a = -10 \omega^2 \cdot 0 = 0$.

In simple harmonic motion described by $y = A \cos(\omega t + \phi) + C$, the term $A \cos(\omega t + \phi)$ represents the oscillation about the equilibrium position. The acceleration is proportional to the displacement from the equilibrium position and directed towards it ($a = -\omega^2 (y-C)$). In this case, the equilibrium position is at $y=10$. At $t = \frac{\pi}{2\omega}$, the displacement $y = 10 \cos(\frac{\pi}{2}) + 10 = 10(0) + 10 = 10$. Since the displacement is equal to the equilibrium position, the acceleration is zero, as expected.

10. Which one of the following conservation laws is a consequence of the N

Which one of the following conservation laws is a consequence of the Newton’s third law of motion ?

[amp_mcq option1=”Conservation of energy” option2=”Conservation of momentum” option3=”Conservation of charge” option4=”Conservation of mass” correct=”option2″]

This question was previously asked in

UPSC CAPF – 2023

Download PDF Attempt Online

The correct answer is B. The law of conservation of momentum is a direct consequence of Newton’s third law of motion when applied to a system of particles.

– 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.

Newton’s laws of motion are fundamental to classical mechanics. While conservation of energy, charge, and mass are also fundamental physical laws, the conservation of momentum is the one most directly and explicitly derived from Newton’s third law in the context of interactions between particles.