Mechanics Archives

131. Whether an object will float or sink in a liquid, depends on

Whether an object will float or sink in a liquid, depends on

[amp_mcq option1=”mass of the object only” option2=”mass of the object and density of liquid only” option3=”difference in the densities of the object and liquid” option4=”mass and shape of the object only” correct=”option3″]

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UPSC NDA-1 – 2018

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Whether an object floats or sinks in a liquid is determined by comparing the density of the object to the density of the liquid. According to Archimedes’ principle and the concept of buoyancy, an object will sink if its density is greater than the density of the liquid, float if its density is less than the density of the liquid, and be suspended if its density is equal to the density of the liquid. The difference in densities dictates the outcome. Options A, B, and D are insufficient as they do not solely or correctly identify the determining factor. Mass alone doesn’t account for volume, and while shape can influence *how* an object floats (by affecting the volume of liquid displaced before full submersion), the fundamental condition for sinking is based on the object’s average density relative to the liquid’s density.

– Buoyancy force is the upward force exerted by a fluid that opposes the weight of an immersed object.
– An object floats if the buoyant force is equal to or greater than its weight.
– The buoyant force is equal to the weight of the fluid displaced by the object.
– Floating/sinking is determined by comparing the density of the object (or its average density for irregularly shaped objects) to the density of the liquid.

Density is defined as mass per unit volume (ρ = m/V). The weight of an object is W_object = m_object * g = ρ_object * V_object * g. The maximum buoyant force when fully submerged is F_buoyant_max = W_liquid_displaced = ρ_liquid * V_object * g. The object sinks if W_object > F_buoyant_max, which simplifies to ρ_object > ρ_liquid. The object floats if W_object <= F_buoyant_max (partially or fully submerged), which simplifies to ρ_object <= ρ_liquid.

132. Which one of the following statements about gravitational force is NOT

Which one of the following statements about gravitational force is NOT correct?

[amp_mcq option1=”It is experienced by all bodies in the universe” option2=”It is a dominant force between celestial bodies” option3=”It is a negligible force for atoms” option4=”It is same for all pairs of bodies in our universe” correct=”option4″]

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UPSC NDA-1 – 2018

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According to Newton’s Law of Universal Gravitation, the gravitational force (F) between two objects with masses m₁ and m₂ separated by a distance r is given by F = G * (m₁*m₂)/r², where G is the gravitational constant. This formula shows that the gravitational force depends on the masses of the two bodies and the distance between their centers. Therefore, the gravitational force is *not* the same for all pairs of bodies in the universe. It varies depending on their masses and separation.

– Gravitational force is a universal force acting between any two objects with mass.
– The magnitude of the gravitational force depends on the product of the masses and the square of the distance between them.
– Gravity is the weakest of the fundamental forces but is dominant on large scales due to the large masses of celestial bodies.

Statements A, B, and C are correct. Gravity acts on all bodies with mass. It is the dominant force holding together planets, stars, and galaxies. At the atomic and subatomic levels, the masses are extremely small, making the gravitational force between them negligible compared to electromagnetic and nuclear forces.

133. Which one of the following has maximum inertia ?

Which one of the following has maximum inertia ?

[amp_mcq option1=”An atom” option2=”A molecule” option3=”A one-rupee coin” option4=”A cricket ball” correct=”option4″]

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UPSC NDA-1 – 2018

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Inertia is the property of an object that resists changes in its state of motion (either rest or uniform velocity in a straight line). Inertia is directly proportional to the mass of an object. The greater the mass, the greater its inertia. Comparing the given options, a cricket ball has significantly more mass than a one-rupee coin, a molecule, or an atom. Therefore, the cricket ball has the maximum inertia among the given options.

– Inertia is the resistance of an object to changes in its state of motion.
– Inertia is a scalar quantity.
– Inertia is directly proportional to mass.

Mass is the quantitative measure of inertia. For example, it is much harder to start moving a heavy object or stop a heavy object already in motion compared to a light object, because the heavy object has more inertia.

134. What is the net force experienced by a bar magnet placed in a uniform

What is the net force experienced by a bar magnet placed in a uniform magnetic field?

[amp_mcq option1=”Zero” option2=”Depends upon length of the magnet” option3=”Never zero” option4=”Depends upon temperature” correct=”option1″]

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UPSC NDA-1 – 2018

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A bar magnet has two poles, a North pole and a South pole. When placed in a uniform magnetic field, the field exerts a force on each pole. The force on the North pole is in the direction of the magnetic field, and the force on the South pole is equal in magnitude but in the opposite direction to the magnetic field. Since the field is uniform, the forces on the two poles are equal and opposite, resulting in a net force of zero on the magnet. Although there is no net force, there is usually a torque which tends to align the magnetic dipole moment of the bar magnet with the direction of the uniform magnetic field.

– A bar magnet is a magnetic dipole.
– In a uniform magnetic field, the force on the North pole is equal and opposite to the force on the South pole.
– The net force is the vector sum of the forces on the individual poles.

If the magnetic field were non-uniform, the forces on the two poles would not be equal and opposite, and there would be a non-zero net force on the magnet, in addition to the torque. This is why magnets are attracted to or repelled from other magnets or ferromagnetic materials where the field is non-uniform.

135. An object is moving with uniform acceleration a. Its initial velocity

An object is moving with uniform acceleration a. Its initial velocity is u and after time t its velocity is v. The equation of its motion is v = u + at. The velocity (along y-axis) time (along x-axis) graph shall be a straight line

[amp_mcq option1=”passing through origin” option2=”with x-intercept u” option3=”with y-intercept u” option4=”with slope u” correct=”option3″]

This question was previously asked in

UPSC NDA-1 – 2018

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The given equation of motion is v = u + at. We are asked about the velocity (v) vs. time (t) graph. This equation is in the form of a linear equation, y = mx + c, where v corresponds to y, t corresponds to x, ‘a’ (uniform acceleration) is the slope (m), and ‘u’ (initial velocity) is the y-intercept (c). The y-intercept is the value of y when x=0. In this case, it is the value of velocity (v) when time (t) is zero, which is the initial velocity ‘u’. Therefore, the graph is a straight line with a y-intercept equal to u.

– The equation v = u + at is a linear relationship between velocity (v) and time (t) for uniform acceleration.
– In a linear equation y = mx + c, ‘c’ is the y-intercept and ‘m’ is the slope.
– On a velocity-time graph, the y-axis represents velocity and the x-axis represents time.

The slope of the velocity-time graph represents the acceleration. Since the acceleration ‘a’ is uniform (constant), the graph is a straight line. The graph passes through the origin only if the initial velocity u is zero. The x-intercept would be the time when velocity is zero, which is t = -u/a (if applicable and physically meaningful).

136. If an object moves with constant velocity then which one of the follow

If an object moves with constant velocity then which one of the following statements is NOT correct ?

[amp_mcq option1=”Its motion is along a straight line” option2=”Its speed changes with time” option3=”Its acceleration is zero” option4=”Its displacement increases linearly with time” correct=”option2″]

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UPSC NDA-1 – 2018

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If an object moves with constant velocity, it means that both its speed and its direction of motion are constant. Therefore, its speed does *not* change with time. Statement B says its speed changes with time, which contradicts the definition of constant velocity, making it the incorrect statement.

– Constant velocity implies constant speed and constant direction.
– Acceleration is the rate of change of velocity.
– Displacement is the change in position.

– Motion along a straight line means constant direction. Constant velocity guarantees this.
– Zero acceleration means no change in velocity. Constant velocity means zero acceleration.
– If velocity (v) is constant, displacement (s) is given by s = v*t. For a constant v, s increases linearly with time t.

137. If T is the time period of an oscillating pendulum, which one of the f

If T is the time period of an oscillating pendulum, which one of the following statements is NOT correct ?

[amp_mcq option1=”The motion repeats after time T only once” option2=”T is the least time after which motion repeats itself” option3=”The motion repeats itself after nT, where n is a positive integer” option4=”T remains the same only for small angular displacements” correct=”option1″]

This question was previously asked in

UPSC NDA-1 – 2018

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The time period (T) of an oscillating pendulum is defined as the smallest time interval after which the motion repeats itself. While the motion does repeat after time T, it also repeats after any integer multiple of T (2T, 3T, 4T, etc.). Therefore, the statement that the motion repeats after time T “only once” is incorrect. The oscillation continues to repeat after every interval of T.

– A periodic motion repeats itself after a fixed interval of time called the time period.
– The time period T is the *minimum* time for the motion to repeat.
– The motion repeats after nT, where n is a positive integer.

For a simple pendulum undergoing Simple Harmonic Motion (SHM) at small angles, the time period is given by T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. For larger angles, the period is slightly longer and depends on the amplitude, making the statement D (T remains the same only for small angular displacements) correct for the conditions under which the simple pendulum formula is typically derived.

138. Which one of the following statements is true for the relation $F = \f

Which one of the following statements is true for the relation $F = \frac{Gm_1m_2}{r^2}$ ? (All symbols have their usual meanings)

[amp_mcq option1=”The quantity G depends on the local value of g, acceleration due to gravity” option2=”The quantity G is greatest at the surface of the Earth” option3=”The quantity G is used only when earth is one of the two masses” option4=”The quantity G is a universal constant” correct=”option4″]

This question was previously asked in

UPSC NDA-1 – 2017

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The question asks which statement is true for the gravitational force relation $F = \frac{Gm_1m_2}{r^2}$.

The equation $F = \frac{Gm_1m_2}{r^2}$ is Newton’s Law of Universal Gravitation. It describes the force of attraction between any two point masses $m_1$ and $m_2$ separated by a distance $r$. $G$ is the constant of proportionality known as the Universal Gravitational Constant.
Let’s examine the options:
A) The quantity G depends on the local value of g, acceleration due to gravity: Incorrect. ‘g’ varies from place to place (e.g., poles vs. equator, altitude), while G is constant everywhere in the universe. ‘g’ on Earth’s surface is given by $g = \frac{GM_{Earth}}{R_{Earth}^2}$, showing that ‘g’ depends on G, not the other way around.
B) The quantity G is greatest at the surface of the Earth: Incorrect. G is a constant and does not vary with location. It has the same value on the surface of the Earth, in space, on the Moon, etc.
C) The quantity G is used only when earth is one of the two masses: Incorrect. G is used in the formula to calculate the gravitational force between *any* two masses in the universe, whether they are planets, stars, or small objects.
D) The quantity G is a universal constant: Correct. The value of G has been measured experimentally and is found to be constant throughout the universe, regardless of the nature of the masses, their distance, or the surrounding medium. Its approximate value is $6.674 \times 10^{-11} \, N m^2/kg^2$.

The term “universal constant” implies that the value is the same everywhere and at all times. This universality of G is a fundamental principle of Newtonian gravity. Measuring G accurately is challenging due to the weakness of the gravitational force between laboratory-sized objects.

139. Which one of the following statements is correct regarding the provide

Which one of the following statements is correct regarding the provided displacement versus time curve for a particle executing simple harmonic motion?

[amp_mcq option1=”Phase of the oscillating particle is same at t=1s and t=3s” option2=”Phase of the oscillating particle is same at t=2s and t=8s” option3=”Phase of the oscillating particle is same at t=3s and t=17s” option4=”Phase of the oscillating particle is same at t=4s and t=10s” correct=”option2″]

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For a particle executing simple harmonic motion (SHM), the phase of oscillation at time $t$ is given by $\phi(t) = \omega t + \phi_0$, where $\omega = 2\pi/T$ is the angular frequency and $T$ is the period. The phase is the same at two different times $t_1$ and $t_2$ if their phase difference is an integer multiple of $2\pi$, i.e., $\phi(t_2) – \phi(t_1) = 2n\pi$, which simplifies to $\omega(t_2 – t_1) = 2n\pi$, or $(t_2 – t_1) = nT$, where $n$ is an integer. This means the phase is the same at times separated by an integer multiple of the period T. In option B, the time difference is $8s – 2s = 6s$. If the period T of the motion is a divisor of 6s (e.g., T=6s, T=3s, T=2s, T=1s), then 6s is an integer multiple of T, and the phase would be the same at these times. Assuming the question is valid and option B is the intended correct answer, it implies that 6s is an integer multiple of the period T, while the time differences in options A (2s) and C (14s) are not integer multiples of T. (Note: Option D also has a time difference of 6s, which could imply it is also correct if the period divides 6s. Without the actual graph, definitively determining the period and confirming only B is correct is not possible. However, based on the provided answer choices and typical SHM properties, a period that makes 6s a multiple is implied).

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.

The displacement, velocity, and acceleration of a particle in SHM are all periodic functions with the same period T. The phase determines the instantaneous state of the oscillation (displacement, velocity, and direction of motion).

140. The speed of a car travelling on a straight road is listed below at su

The speed of a car travelling on a straight road is listed below at successive intervals of 1 s :

Time (s) | 0 | 1 | 2 | 3 | 4
——- | – | – | – | – | –
Speed (m/s) | 0 | 2 | 4 | 6 | 8

Which of the following is/are correct? The car travels

1. with a uniform acceleration of 2 m/s².
2. 16 m in 4 s.
3. with an average speed of 4 m/s.

Select the correct answer using the code given below :

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

This question was previously asked in

UPSC NDA-1 – 2017

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Let’s analyze each statement based on the given speed data:
1. with a uniform acceleration of 2 m/s²: The speed increases by 2 m/s every second (0 to 2, 2 to 4, 4 to 6, 6 to 8). Acceleration = change in speed / change in time = 2 m/s / 1 s = 2 m/s². Since this value is constant, the acceleration is uniform. Statement 1 is correct.
2. 16 m in 4 s: Assuming the car starts from rest (speed = 0 at time = 0) and moves with uniform acceleration (a = 2 m/s²), the distance covered in time t is given by s = ut + (1/2)at², where u is the initial speed (0 m/s). For t = 4s, s = (0)(4) + (1/2)(2)(4)² = 0 + 1 * 16 = 16 m. Statement 2 is correct.
3. with an average speed of 4 m/s: Average speed is total distance / total time. Total distance covered is 16 m (from statement 2) and total time is 4 s. Average speed = 16 m / 4 s = 4 m/s. Statement 3 is correct.
All three statements are correct.

– Uniform acceleration means the velocity changes by the same amount in equal intervals of time.
– For uniformly accelerated motion, cinematic equations can be used to calculate distance, speed, and time.
– Average speed is calculated as total distance traveled divided by the total time taken.

In uniformly accelerated motion starting from rest, the speed at time t is v = at, and the distance covered is s = (1/2)at². The average speed over a time interval is (initial speed + final speed) / 2 if acceleration is uniform. In this case, average speed over 4s is (0 + 8)/2 = 4 m/s, confirming statement 3.