Mechanics

A boy completes one round of a circular track of diameter 200 m in 30 s. What will be the displacement at the end of 3 minutes and 45 seconds ?

[amp_mcq option1=”50 m” option2=”100 m” option3=”200 m” option4=”236 m” correct=”option3″]

The boy completes 7.5 rounds in 3 minutes and 45 seconds. After 7 complete rounds, the displacement is zero. The displacement after the remaining 0.5 rounds (half a round) is the straight-line distance between the starting point and the diametrically opposite point. This distance is equal to the diameter of the circular track.

Displacement is a vector quantity defined as the shortest straight-line distance between the initial and final positions. In circular motion, after completing one or more full rounds, the displacement is zero because the final position is the same as the initial position. After half a round, the final position is diametrically opposite the initial position.

The diameter of the track is 200 m, so the radius is 100 m. The time taken for one round is 30 s. Total time = 3 minutes 45 seconds = 180 + 45 = 225 s. Number of rounds = Total time / Time per round = 225 s / 30 s = 7.5. Displacement after 7.5 rounds is the same as displacement after 0.5 rounds. The starting point and the point diametrically opposite are separated by the diameter, which is 200 m.

The weight of an object is due to

[amp_mcq option1=”the net force acting on it.” option2=”the total of all forces acting on it irrespective of their directions.” option3=”the force that it exerts on the ground.” option4=”its inert property.” correct=”option1″]

Weight is defined as the force of gravity acting on an object. In situations where gravity is the only force significantly influencing motion (like free fall), the net force acting on the object is equal to its weight. While weight is a specific type of force (gravitational force) and not the ‘net’ force in all scenarios (e.g., when supported), among the given options, ‘net force acting on it’ is the one most directly related to the force aspect of weight, particularly in simplified contexts where gravity is the primary force considered or its effect on motion is discussed.

Weight (W) is the force exerted by a gravitational field on a mass (m), usually expressed as W = mg, where g is the acceleration due to gravity. This gravitational force contributes to the net force acting on the object. In free fall, the net force acting on the object *is* its weight. The phrasing of option A is imprecise, as weight is a component of the net force in general, but in the context of forces acting on an object, it is the most relevant option compared to others that describe different concepts.

Mass is a measure of the amount of matter and inertia, an intrinsic property of an object. Weight is a force that depends on both the mass of the object and the strength of the gravitational field it is in. The force exerted by the object on a supporting surface (often called apparent weight) is related to, but not always equal to, the true weight. Options B, C, and D are incorrect definitions or concepts related to weight.

The direction of acceleration in uniform circular motion is along the

[amp_mcq option1=”direction of motion.” option2=”tangent to the circle at the point of observation.” option3=”direction of velocity.” option4=”direction perpendicular to velocity.” correct=”option4″]

In uniform circular motion, the speed of the object is constant, but its velocity is continuously changing direction as it moves along the circular path. Acceleration is the rate of change of velocity. For an object in uniform circular motion, there is an acceleration directed towards the center of the circle.

The velocity vector in circular motion is always tangent to the circle at the object’s position. The acceleration, known as centripetal acceleration, is always directed along the radius towards the center of the circle. A line tangent to a circle is always perpendicular to the radius at the point of tangency. Therefore, the direction of acceleration (towards the center) is always perpendicular to the direction of velocity (tangent to the circle).

The magnitude of the centripetal acceleration is given by a = v²/r, where v is the constant speed and r is the radius of the circle. Since there is no component of acceleration along the tangent (tangential acceleration), the magnitude of the velocity (speed) remains constant.

In spherical polar coordinates (γ, θ, α), θ denotes the polar angle around z-axis and α denotes the azimuthal angle raised from x-axis. Then the y-component of P⃗ is given by

[amp_mcq option1=”Psinθsinα” option2=”Psinθcosα” option3=”Pcosθsinα” option4=”Pcosθcosα” correct=”option1″]

In spherical polar coordinates (γ, θ, α), where γ is the magnitude of the vector P⃗ (let’s denote it as P), θ is the polar angle from the positive z-axis, and α is the azimuthal angle from the positive x-axis in the xy-plane, the Cartesian components (Px, Py, Pz) of the vector P⃗ are given by:
$P_x = P \sin\theta \cos\alpha$
$P_y = P \sin\theta \sin\alpha$
$P_z = P \cos\theta$
The question asks for the y-component of P⃗. According to the standard conversion from spherical to Cartesian coordinates, the y-component is $P\sin\theta\sin\alpha$.

– Spherical coordinates typically use (r, θ, φ) or (ρ, θ, φ). The question uses (γ, θ, α) with meanings specified.
– γ (or P) is the magnitude.
– θ is the angle from the z-axis (polar angle).
– α is the angle from the x-axis in the xy-plane (azimuthal angle).
– The projection onto the xy-plane has length $P\sin\theta$.
– This projection is resolved into x and y components using the azimuthal angle α.

The formulas for converting spherical coordinates (P, θ, α) to Cartesian coordinates (Px, Py, Pz) are derived from trigonometry. The projection of the vector onto the z-axis is $P\cos\theta$, giving the z-component. The projection onto the xy-plane has length $P\sin\theta$. This projection forms a right triangle in the xy-plane with the x and y axes, where the hypotenuse is $P\sin\theta$ and the angle with the x-axis is α. The x-component is $(P\sin\theta)\cos\alpha$ and the y-component is $(P\sin\theta)\sin\alpha$.

If two vectors $\vec{A}$ and $\vec{B}$ are at an angle $\theta \neq 0^\circ$, then

[amp_mcq option1=”$|\vec{A}|+|\vec{B}|=|\vec{A}+\vec{B}|$” option2=”$|\vec{A}|+|\vec{B}|>|\vec{A}+\vec{B}|$” option3=”$|\vec{A}|+|\vec{B}|” option4=”$|\vec{A}|+|\vec{B}|=|\vec{A}-\vec{B}|$” correct=”option2″]

The correct option is B.

For any two vectors $\vec{A}$ and $\vec{B}$, the triangle inequality states that the magnitude of their sum is less than or equal to the sum of their magnitudes: $|\vec{A}+\vec{B}| \le |\vec{A}|+|\vec{B}|$. Equality holds only when the vectors are collinear and point in the same direction, which corresponds to the angle between them $\theta = 0^\circ$. If $\theta \neq 0^\circ$, the vectors form two sides of a triangle, and the sum vector forms the third side. In a triangle, the length of any side is strictly less than the sum of the lengths of the other two sides. Therefore, if $\theta \neq 0^\circ$, $|\vec{A}+\vec{B}| < |\vec{A}|+|\vec{B}|$, or equivalently, $|\vec{A}|+|\vec{B}| > |\vec{A}+\vec{B}|$.

The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}$. Since $|\vec{A}|+|\vec{B}|$ is always positive, we can compare their squares: $(|\vec{A}|+|\vec{B}|)^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|$. Comparing this with $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta$, we see that $(|\vec{A}|+|\vec{B}|)^2 > |\vec{A}+\vec{B}|^2$ when $\cos\theta < 1$, which is true for $\theta \neq 0^\circ$.

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)

[amp_mcq option1=”the gold piece will weigh more” option2=”the silver piece will weigh more” option3=”both silver and gold pieces weigh equal” option4=”weighing will depend on their masses” correct=”option2″]

The correct option is B.

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 < V_silver. To determine which one weighs more in air, consider the relation derived from the apparent weight equation: W_air,gold - W_air,silver = (V_gold - V_silver) * density_water * g. Since V_gold < V_silver, (V_gold - V_silver) is negative. Therefore, W_air,gold - W_air,silver is negative, meaning W_air,gold < W_air,silver. The silver piece weighs more in air.

This problem highlights the effect of buoyancy, which is dependent on the volume of the object. Although gold is denser, the condition of equal apparent weight in water necessitates that the less dense silver piece has a larger volume, which compensates for its lower density when determining its weight in air.

The pressure of a fluid varies with depth h as P = P₀ + pgh, where ρ is the fluid density. This expression is associated with

[amp_mcq option1=”Pascal’s law” option2=”Newton’s law” option3=”Bernoulli’s principle” option4=”Archimedes’ principle” correct=”option1″]

The expression P = P₀ + ρgh gives the total pressure at a depth h in a fluid. P₀ is the pressure at the surface (e.g., atmospheric pressure), ρ is the fluid density, g is the acceleration due to gravity, and h is the depth. This formula describes the hydrostatic pressure at a given depth. This concept and formula are fundamental to hydrostatics, a field significantly contributed to by Blaise Pascal. While the formula is derived from basic principles of force and pressure (considering the weight of the fluid column), it is most directly associated with Pascal’s principles of hydrostatics, which deal with the pressure exerted by fluids at rest and its transmission.

– 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.

Newton’s laws relate to motion and force. Bernoulli’s principle relates pressure, velocity, and height in fluid flow (fluid dynamics). Archimedes’ principle relates to buoyancy and displaced fluid. While all are important concepts in fluid mechanics, the specific formula for pressure variation with depth in a static fluid is most closely linked to the principles and studies of hydrostatics pioneered by Pascal.

Consider the following statements:

• There is no net moment on a body which is in equilibrium.
• The momentum of a body is always conserved.
• The kinetic energy of an object is always conserved.

Which of the statements given above is/are correct?

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

Let’s evaluate each statement:
1. There is no net moment on a body which is in equilibrium. This is correct. A body is in rotational equilibrium if the net torque (moment) acting on it is zero. For complete equilibrium (translational and rotational), both the net force and net moment must be zero.
2. The momentum of a body is always conserved. This is incorrect. The momentum of a body is conserved only if the net external force acting on it is zero. If a net force acts on a body, its momentum changes according to Newton’s second law (F = dp/dt).
3. The kinetic energy of an object is always conserved. This is incorrect. Kinetic energy is conserved only in specific situations, such as elastic collisions where no energy is lost as heat, sound, or deformation, and when no non-conservative forces (like friction) do work on the object, and no external forces change its speed. For example, when an object falls under gravity, its kinetic energy increases.
Only statement 1 is correct.

– Equilibrium implies zero net force (translational equilibrium) and zero net torque or moment (rotational equilibrium).
– Momentum of a system is conserved in the absence of external forces. Momentum of a single body is conserved only if the net force on it is zero.
– Kinetic energy of an object is conserved only if no net work is done on it by non-conservative forces or if the system is isolated and interactions are perfectly elastic.

Conservation laws (momentum, energy, angular momentum) are fundamental in physics, but their application requires careful definition of the system and consideration of external interactions (forces, torques). “Always conserved” statements for a single body or general process are usually incorrect unless specific conditions are met.

For which one of the following does the centre of mass lie outside the body?

[amp_mcq option1=”A fountain pen” option2=”A cricket ball” option3=”A ring” option4=”A book” correct=”option3″]

The correct answer is 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.

For a fountain pen, cricket ball, and book, the mass is distributed throughout their volume, and their center of mass will typically be located somewhere within the material of the object itself. Examples of other objects where the center of mass is outside the physical body include a hollow sphere or a boomerang.

The rate of change of momentum of a body is equal to the resultant :

[amp_mcq option1=”energy” option2=”power” option3=”force” option4=”impulse” correct=”option3″]

The correct option is C) force.

According to Newton’s Second Law of Motion, the rate of change of momentum of a body is directly proportional to the net force applied to it and occurs in the direction of the applied force. Mathematically, this is expressed as F = dp/dt, where F is the net force, p is the momentum, and t is time.
Momentum (p) is defined as the product of mass (m) and velocity (v), i.e., p = mv.
If mass is constant, the rate of change of momentum is d(mv)/dt = m(dv/dt) = ma, where a is acceleration. This is the more commonly known form of Newton’s Second Law, F = ma.

Energy is related to work and the capacity to do work. Power is the rate at which work is done or energy is transferred. Impulse is the change in momentum, calculated as the product of force and the time interval over which it acts (Impulse = F * Δt = Δp). The question specifically asks for the *rate* of change of momentum, which is defined as force.