Mechanics

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]

[amp_mcq option1=”The pendulum has minimum potential energy at positions P and T.” option2=”The pendulum has minimum potential energy at positions Q and S.” option3=”The pendulum has minimum potential energy at position R.” option4=”The pendulum has same potential energy at all positions.” correct=”option3″]

In a simple pendulum, potential energy is related to its height above a reference point. The lowest point in the swing represents the minimum height, and therefore, the minimum potential energy. In the diagram, position R is the lowest point of the pendulum’s swing.

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.

At the extreme positions (P and T), the kinetic energy is minimum (zero) because the velocity is zero. At the mean position (R), the kinetic energy is maximum because the velocity is maximum. In the absence of air resistance and friction, the total mechanical energy (sum of potential and kinetic energy) of the pendulum remains conserved throughout its swing.

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 :

[amp_mcq option1=”reduce to half of its original value.” option2=”increase to two times of its original value.” option3=”remain the same as the original value.” option4=”increase to four times of its original value.” correct=”option4″]

The correct answer is 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.

This inverse square relationship means that if the distance is reduced by a factor, the force increases by the square of that factor. Conversely, if the distance were doubled, the force would reduce to (1/2)², or one-fourth, of its original value. This law governs gravitational interactions throughout the universe.

Consider a journey by a car represented by the graph given below in three parts A, B and C. The speed of the car in these parts is v_a, v_b and v_c, respectively :
[Image shows a Distance vs Time graph with segments O-A, A-B, B-C representing the journey parts]
Which one of the following is correct in this case ?

[amp_mcq option1=”v_a < v_b < v_c” option2=”v_b > v_a > v_c” option3=”v_a = v_b = v_c” option4=”v_a > v_b; v_a > v_c” correct=”option4″]

The correct answer is D) v_a > v_b; v_a > v_c.

In a distance vs. time graph, the speed is represented by the slope of the graph (speed = change in distance / change in time).
– Part A (O-A): The graph is a straight line with a positive slope, indicating constant positive speed. The slope is relatively steep.
– Part B (A-B): The graph is a horizontal line, meaning the distance is not changing over time. The slope is zero, indicating the car is stopped (v_b = 0).
– Part C (B-C): The graph is a straight line with a positive slope, indicating constant positive speed. The slope is less steep than in Part A.
Comparing the slopes: Slope of A > Slope of C > Slope of B (which is 0). Therefore, v_a > v_c > v_b.
Looking at the options, D) v_a > v_b; v_a > v_c is the only option that correctly reflects the relationship v_a > v_c and v_a > v_b (since v_b = 0).

A steeper slope on a distance-time graph indicates a higher speed, while a less steep slope indicates a lower speed. A horizontal line means the object is at rest, and a line with a negative slope would indicate movement back towards the starting point.

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

[amp_mcq option1=”100 N.s and 500 J” option2=”100 N.s and 1000 J” option3=”200 N.s and 500 J” option4=”200 N.s and 1000 J” correct=”option1″]

We are given the mass and uniform speed of an object and asked to calculate its linear momentum and kinetic energy.

– 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.

Linear momentum is a vector quantity (direction matters), while kinetic energy is a scalar quantity. The N.s unit for momentum comes from the impulse-momentum theorem, where Impulse (Force x time) equals the change in momentum. 1 Newton = 1 kg.m/s², so 1 N.s = (kg.m/s²) * s = kg.m/s.

A bus starting from a bus-stand and moving with uniform acceleration attains a speed of 20 km/h in 10 minutes. What is its acceleration?

[amp_mcq option1=”200 km/h²” option2=”120 km/h²” option3=”100 km/h²” option4=”240 km/h²” correct=”option2″]

The bus starts from rest, so its initial velocity (u) is 0 km/h. Its final velocity (v) is 20 km/h, attained in time (t) = 10 minutes. To find acceleration (a) in km/h², we need to convert time to hours: 10 minutes = 10/60 hours = 1/6 hours. Using the equation of motion v = u + at, we get 20 km/h = 0 km/h + a × (1/6) hours. Solving for a: a = 20 × 6 = 120 km/h².

Acceleration is the rate of change of velocity (a = (v-u)/t). Ensure that all units are consistent before performing calculations. If velocity is in km/h and time in hours, acceleration will be in km/h².

The problem specifies uniform acceleration, allowing the use of standard kinematic equations. Other equations of motion like s = ut + ½at² and v² = u² + 2as can also be used depending on the given and required variables.

An athlete completes one round of a circular track of diameter 100 m in 20 s. What will be the displacements after 1 minute and 10 s, respectively ?

[amp_mcq option1=”0 m, 50 m” option2=”300 m, 100 m” option3=”300 m, 50 m” option4=”0 m, 100 m” correct=”option4″]

The athlete completes one round of a circular track (diameter 100 m, radius 50 m) in 20 seconds.
After 1 minute (60 seconds), the athlete has completed 60 s / 20 s/round = 3 full rounds. In circular motion, after completing a full circle or multiple full circles, the displacement is zero because the final position is the same as the starting position. So, displacement after 1 minute is 0 m.
After 1 minute and 10 seconds (70 seconds), the athlete has completed 70 s / 20 s/round = 3.5 rounds. This is 3 full rounds plus half a round. After 3 full rounds, the displacement is zero. The remaining half-round takes the athlete from the starting point to the diametrically opposite point on the track. The displacement in this case is the straight-line distance between the starting point and the diametrically opposite point, which is equal to the diameter of the circle. The diameter is given as 100 m. So, displacement after 1 minute and 10 seconds is 100 m.

– Displacement is the shortest distance between the initial and final positions.
– For motion along a closed loop (like a full circle), the displacement is zero if the motion starts and ends at the same point.
– For motion along half a circle, the displacement is equal to the diameter.

Distance covered is the total path length traced. In this case, distance after 60s would be 3 * circumference, and distance after 70s would be 3.5 * circumference. Displacement is a vector quantity and depends only on the initial and final positions.

What is the effect of pressure of a human body on sand ?

[amp_mcq option1=”Larger while standing than while lying” option2=”Smaller while standing than while lying” option3=”Same while standing or lying” option4=”Larger while standing during the daytime and smaller during the night time while lying” correct=”option1″]

Pressure is defined as force per unit area (Pressure = Force / Area). The force exerted by the human body on the sand is equal to its weight, which remains the same whether standing or lying down. However, the area of contact with the sand is significantly smaller when standing (area of feet) compared to when lying down (area of the entire body in contact). Since the force is constant, a smaller area results in larger pressure. Therefore, the pressure exerted on the sand is larger while standing than while lying down.

Pressure is inversely proportional to the area of contact for a constant force.

This principle explains why sharp objects exert more pressure than blunt objects under the same force, and why wide tires are used on vehicles intended for soft ground.

If the distance between two objects is increased by two times, the gravitational force between them will

[amp_mcq option1=”remain same” option2=”increase by two times” option3=”decrease by two times” option4=”decrease by four times” correct=”option4″]

The gravitational force between the two objects will 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.

The inverse square law is a common relationship in physics, appearing in gravitation, electrostatics, and light intensity. It implies that the effect of a source diminishes rapidly with increasing distance.

If the linear momentum of a moving object gets doubled due to application of a force, then its kinetic energy will

[amp_mcq option1=”remain same” option2=”increase by four times” option3=”increase by two times” option4=”increase by eight times” correct=”option2″]

The kinetic energy of the moving object will 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.

This relationship shows that kinetic energy is proportional to the square of momentum when mass is constant. Doubling the momentum results in quadrupling the kinetic energy.

If an object is at rest, then the time (X-axis) versus distance (Y-axis) graph

[amp_mcq option1=”is vertical” option2=”is horizontal” option3=”has 45° positive slope” option4=”has 45° negative slope” correct=”option2″]

If an object is at rest, its distance from a reference point does not change over time. In a time (X-axis) versus distance (Y-axis) graph, the distance (Y-value) remains constant regardless of how time (X-value) increases. This is represented by a horizontal line parallel to the X-axis.

A horizontal line on a distance-time graph signifies zero velocity, meaning the object is stationary or at rest.

A vertical line would imply infinite velocity, which is physically impossible. A graph with a positive slope (like 45°) indicates movement away from the origin at a constant positive velocity. A graph with a negative slope (like 45°) indicates movement towards the origin at a constant negative velocity.