Mechanics

If some object is weighed when submerged in water, what will happen to its weight compared to its weight in air ?

[amp_mcq option1=”Increase” option2=”Decrease” option3=”Remain exactly the same” option4=”Increase or decrease cannot be predicted” correct=”option2″]

When an object is submerged in water, it experiences an upward force exerted by the water, known as the buoyant force. This buoyant force opposes the object’s weight. The apparent weight of the object when submerged is equal to its true weight (weight in air) minus the buoyant force. Since the buoyant force is a positive value for any object submerged in a fluid, the apparent weight will always be less than the weight in air. Therefore, the weight of the object appears to decrease when submerged in water.

– Buoyant force is an upward force exerted by a fluid on a submerged object.
– Apparent weight = True weight – Buoyant force.
– Buoyancy causes the apparent weight to be less than the weight in air.

Archimedes’ principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The magnitude of the decrease in apparent weight is exactly equal to this buoyant force.

The force acting on a particle of mass m moving along the x-axis is given by F(x) = Ax² – Bx. Which one of the following is the potential energy of the particle ?

[amp_mcq option1=”2Ax – B” option2=”$ -\frac{x^2}{6} (2Ax – 3B) $” option3=”$ Ax^3 – Bx^2 $” option4=”Zero” correct=”option2″]

The potential energy $U(x)$ is related to the conservative force $F(x)$ by the relation $F(x) = -\frac{dU}{dx}$. To find the potential energy, we need to integrate the force with respect to $x$ and negate the result: $U(x) = -\int F(x) dx + C$, where $C$ is the integration constant. Given $F(x) = Ax^2 – Bx$, we integrate:
$U(x) = -\int (Ax^2 – Bx) dx = – \left( A \int x^2 dx – B \int x dx \right) = – \left( A \frac{x^3}{3} – B \frac{x^2}{2} \right) + C$
$U(x) = -\frac{A}{3}x^3 + \frac{B}{2}x^2 + C$.
If we set the integration constant $C=0$, the potential energy is $U(x) = -\frac{A}{3}x^3 + \frac{B}{2}x^2$. Let’s rewrite Option B: $ -\frac{x^2}{6} (2Ax – 3B) = -\frac{2Ax^3}{6} + \frac{3Bx^2}{6} = -\frac{A}{3}x^3 + \frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$.

– The relationship between a conservative force $F(x)$ and potential energy $U(x)$ is $F(x) = -dU/dx$.
– To find $U(x)$ from $F(x)$, integrate: $U(x) = -\int F(x) dx$.

The integration constant $C$ represents the arbitrary choice of the zero point for potential energy. Different choices of $C$ result in different absolute values for potential energy, but the force (which depends on the *derivative* of potential energy, or the *difference* in potential energy between two points) remains the same. In multiple-choice questions involving potential energy derived from a force, the correct option often corresponds to setting the integration constant to zero or a value that makes the expression fit one of the choices.

An object moves in a circular path with a constant speed. Which one of the following statements is correct ?

[amp_mcq option1=”The centripetal acceleration of the object is smaller for a gentle curve (i.e., curve of larger radius) than that for a sharp curve (i.e., curve of smaller radius).” option2=”The centripetal acceleration is greater for a gentle curve than that for a sharp curve.” option3=”The centripetal acceleration is the same for both, the gentle and sharp curves.” option4=”The centripetal acceleration causes the object to slow down.” correct=”option1″]

When an object moves in a circular path with constant speed $v$, it experiences a centripetal acceleration $a_c$ directed towards the center of the circle. The formula for centripetal acceleration is $a_c = \frac{v^2}{r}$, where $v$ is the speed and $r$ is the radius of the circular path. For a constant speed $v$, the centripetal acceleration is inversely proportional to the radius $r$. A gentle curve corresponds to a larger radius ($r$), while a sharp curve corresponds to a smaller radius ($r$). Therefore, for a constant speed, a larger radius (gentle curve) results in a smaller centripetal acceleration, and a smaller radius (sharp curve) results in a larger centripetal acceleration. Option A correctly states that the centripetal acceleration is smaller for a gentle curve (larger radius) than for a sharp curve (smaller radius).

– Centripetal acceleration $a_c = \frac{v^2}{r}$.
– For constant speed $v$, $a_c \propto \frac{1}{r}$.
– Gentle curve = larger radius; Sharp curve = smaller radius.

Centripetal acceleration is always perpendicular to the velocity vector in uniform circular motion and is responsible for changing the direction of velocity, thus keeping the object on the circular path, without changing its speed. Option D is incorrect because centripetal acceleration changes direction, not speed (in uniform circular motion).

A particle executes linear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is:

[amp_mcq option1=”2π” option2=”2π/√3″ option3=”√3/2π” option4=”2π√3″ correct=”option2″]

For a particle executing simple harmonic motion (SHM) with amplitude $A$ and angular frequency $\omega$, the velocity $v$ at a displacement $x$ from the mean position is given by $v = \pm \omega \sqrt{A^2 – x^2}$, and the acceleration $a$ is given by $a = -\omega^2 x$. The magnitudes are $|v| = \omega \sqrt{A^2 – x^2}$ and $|a| = \omega^2 |x|$. Given $A=2$ cm, and at $x=1$ cm, $|v| = |a|$.
Substituting the values: $\omega \sqrt{2^2 – 1^2} = \omega^2 |1|$
$\omega \sqrt{4 – 1} = \omega^2$
$\omega \sqrt{3} = \omega^2$
Since $\omega$ for SHM is non-zero, we can divide by $\omega$:
$\sqrt{3} = \omega$.
The time period $T$ is related to angular frequency $\omega$ by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{\sqrt{3}}$.

Formulas for velocity and acceleration in SHM are $|v| = \omega \sqrt{A^2 – x^2}$ and $|a| = \omega^2 |x|$. The time period is $T = 2\pi/\omega$.

The velocity is maximum at the mean position ($x=0$) and zero at the extreme positions ($x=\pm A$). The acceleration is maximum at the extreme positions and zero at the mean position. The relationship between velocity and displacement is elliptical in phase space, while the relationship between acceleration and displacement is linear.

Which one of the following four particles, whose displacement x and acceleration a, are related as follows, is executing simple harmonic motion?

[amp_mcq option1=”a = +3x” option2=”a = +3x²” option3=”a = -3″ option4=”a = -3x” correct=”option4″]

The particle executing simple harmonic motion is the one where acceleration ‘a’ and displacement ‘x’ are related as a = -3x.

Simple Harmonic Motion (SHM) is defined by a linear restoring force proportional to the displacement from the equilibrium position and directed towards it. According to Newton’s second law (F = ma), this implies that the acceleration is also proportional to the displacement and directed towards the equilibrium position. The general form of the acceleration-displacement relationship for SHM is a = -ω²x, where ω² is a positive constant (related to the system’s properties like mass and stiffness). Option D, a = -3x, fits this form with ω² = 3 (or ω = √3).

Option A, a = +3x, represents unstable equilibrium where the acceleration is away from the origin. Option B, a = +3x², is a non-linear relationship. Option C, a = -3, represents motion under constant acceleration, which is not SHM.

A particle is executing simple harmonic motion. Which one of the following statements about the acceleration of the oscillating particle is true ?

[amp_mcq option1=”It is always in the opposite direction to velocity” option2=”It is proportional to the frequency of oscillation” option3=”It is minimum when the speed is maximum” option4=”It decreases as the potential energy increases” correct=”option3″]

In simple harmonic motion, the acceleration of the oscillating particle is minimum when the speed is maximum.

Simple Harmonic Motion (SHM) is characterized by an acceleration that is directly proportional to the displacement from the equilibrium position and is always directed towards the equilibrium position (a = -ω²x, where x is displacement and ω is angular frequency). The equilibrium position is where the displacement x = 0. At this point, the restoring force and acceleration are zero, representing the minimum magnitude of acceleration. In SHM, the speed of the particle is maximum at the equilibrium position (x=0) and zero at the extreme positions (maximum |x|). Thus, acceleration (minimum at x=0) is minimum when speed (maximum at x=0) is maximum.

Option A is incorrect; acceleration is in the opposite direction to velocity only when the object is slowing down (moving away from equilibrium). When moving towards equilibrium, velocity and acceleration are in the same direction. Option B is incorrect; acceleration is proportional to the *square* of the angular frequency (ω²), not frequency directly, though frequency is proportional to ω. Option D is incorrect; potential energy is maximum at the extreme positions where |x| is maximum, and acceleration magnitude |a| = ω²|x| is also maximum there. So, as potential energy increases, acceleration magnitude increases, not decreases.

How is the kinetic energy of a moving object affected if the net work done on it is positive ?

[amp_mcq option1=”Decreases” option2=”Increases” option3=”Remains constant” option4=”Becomes zero” correct=”option2″]

If the net work done on a moving object is positive, its kinetic energy increases.

This is directly explained by the Work-Energy Theorem, which states that the net work (W_net) done on an object by external forces is equal to the change in its kinetic energy (ΔKE): W_net = ΔKE = KE_final – KE_initial. If W_net is positive, then ΔKE must be positive, meaning KE_final > KE_initial. Therefore, the kinetic energy increases.

Conversely, if the net work done is negative, the kinetic energy decreases. If the net work done is zero, the kinetic energy remains constant (although the velocity vector might change direction, the speed and thus KE remain the same).

Along a streamline flow of fluid

[amp_mcq option1=”the velocity of all fluid particles at a given instant is constant” option2=”the speed of a fluid particle remains constant” option3=”the velocity of all fluid particles crossing a given position is constant” option4=”the velocity of a fluid particle remains constant” correct=”option3″]

Along a streamline flow of fluid, the velocity of all fluid particles crossing a given position is constant.

Streamline flow is a characteristic of steady flow. In steady flow, the velocity of the fluid at any fixed point in space does not change over time. Therefore, any fluid particle that passes through a particular point will have the same velocity as any other particle that passes through that same point at a different time. Streamlines are the paths followed by fluid particles in steady flow.

Option A is incorrect because the velocity can vary from one streamline to another at a given instant. Option B is incorrect because the speed of a particle can change along a streamline if the cross-sectional area of the flow changes (e.g., in a converging or diverging pipe, based on the continuity equation). Option D is incorrect because the velocity of a specific fluid particle changes as it moves along a streamline, unless the flow is uniform (velocity is the same everywhere). Option C accurately describes a key property of steady flow: the velocity field is constant with respect to time.

When a force of 1 newton acts on a mass of 1 kg which is able to move freely, the object moves in the direction of force with a/an

[amp_mcq option1=”speed of 1 km/s” option2=”acceleration of 1 m/s²” option3=”speed of 1 m/s” option4=”acceleration of 1 km/s²” correct=”option2″]

The correct answer is B. According to Newton’s Second Law of Motion, the force (F) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a): F = m × a.

Given F = 1 Newton (N) and m = 1 kg, we can calculate the acceleration:
a = F / m
a = 1 N / 1 kg
The unit of Newton is defined as 1 kg·m/s².
So, a = (1 kg·m/s²) / 1 kg = 1 m/s².
The object will move in the direction of the force with an acceleration of 1 m/s².

Acceleration is the rate of change of velocity. A constant force acting on a mass produces a constant acceleration. The object’s speed will increase by 1 meter per second every second, starting from rest or its initial velocity. Options A and C refer to speed, not acceleration, and D uses incorrect units (km/s²).

Which one of the following is not a contact force ?

[amp_mcq option1=”Push force” option2=”Gravitational force” option3=”Frictional force” option4=”Strain force” correct=”option2″]

The correct answer is B. A contact force is a force that acts only when there is direct physical contact between two interacting objects.

Gravitational force is a non-contact force (also known as a field force). It acts between any two objects possessing mass, even when they are separated by a distance, without any physical contact between them (e.g., the force between the Earth and the Sun, or the force between the Earth and an object in free fall). Push force, frictional force, and strain force (which arises from interactions between atoms/molecules within a material under deformation) are all examples of contact forces.

Other common examples of non-contact forces include electrostatic force (between charges) and magnetic force (between magnets or currents). The concept of forces acting at a distance was initially puzzling but is now understood in terms of fields (gravitational field, electric field, magnetic field) which mediate the interaction.