12th Physics 1 Mark Questions with Answers English Medium

Explanation: This problem deals with rotational kinematics under uniform angular acceleration starting from rest. When angular acceleration remains constant, angular displacement is not linear with time; instead, it follows a quadratic dependence on time. To analyse motion during a specific second, we focus on displacement differences between successive time intervals rather than total displacement alone. The standard approach involves expressing angular displacement as a function of time using rotational motion equations and then evaluating the change between t = 4 s and t = 5 s.

For uniformly accelerated rotational motion starting from rest, angular displacement is expressed in a time-dependent quadratic form. The displacement during a specific second is obtained by subtracting total displacement up to the previous second from the total displacement up to the current second. This method avoids direct substitution into instantaneous formulas and ensures accuracy for interval-based motion.

The key idea is that angular motion under constant acceleration behaves analogously to linear motion under constant acceleration, where displacement in successive seconds increases in a predictable pattern due to continuously increasing angular velocity. Careful application of interval subtraction is essential in such problems to correctly isolate motion during a particular second.

Explanation: This problem involves motion in two distinct stages: radial motion from the centre to the boundary of a circular track, followed by motion along the circular path. Average velocity depends only on NET displacement and total time, not on the actual path travelled. Therefore, understanding the geometry of displacement is more important than analysing the distance covered along the curve.

The displacement of the cyclist is determined by the straight-line Vector from the starting point at the centre to the final position on the boundary after completing the arc motion. Even though the cyclist travels along a curved path during the second stage, only the initial and final positions Matter for average velocity. The total time includes both the radial motion phase and the time spent along the circumference.

In such problems, the key step is to identify the resultant displacement Vector and divide it by the total time of travel. Since the second phase involves circular motion, it contributes significantly to time but not necessarily to displacement magnitude in a straightforward additive manner. Careful separation of path length and displacement is essential for correct interpretation of average velocity.

Explanation: This situation involves rotational motion with constant angular acceleration, where angular displacement increases non-linearly over time. The motion begins with an initial angular speed and continues to increase due to constant angular acceleration. To analyse such motion, angular displacement over a given time interval is expressed using standard rotational kinematic relations that combine initial angular velocity, angular acceleration, and time.

Since the motion spans a fixed duration, total angular displacement is determined over the entire interval rather than instantaneous values. Once angular displacement in radians is obtained, it is interpreted in terms of full circular rotations by comparing it with the angular measure of one complete revolution. This conversion is essential because rotational motion Questions often require interpreting angular displacement in more intuitive physical units.

A key idea is that both initial angular velocity and angular acceleration contribute to total displacement, with acceleration having an increasingly dominant effect as time progresses. The radius is not directly required for angular displacement calculation but becomes relevant when linking linear and angular quantities. Careful handling of time-dependent angular growth is crucial in such problems.

Explanation: This problem is based on uniformly accelerated rotational motion, where angular velocity increases steadily over time due to constant angular acceleration. Angular displacement depends on both the initial angular velocity and the continuous contribution from acceleration throughout the motion interval.

To understand this motion, angular displacement is expressed as a function of time using a standard quadratic relationship in time. The expression accounts for two parts: one arising from initial angular velocity and another arising from acceleration. The radius is not needed for angular displacement because angular quantities are independent of linear dimensions in rotational kinematics.

The key conceptual step is recognising that angular acceleration contributes increasingly over time, making the displacement grow faster than linear time dependence. After finding total angular displacement in radians, it can be interpreted in terms of revolutions if needed by comparing it with full rotational measure.

This type of problem highlights the importance of correctly combining initial motion conditions with acceleration-driven change, ensuring both contributions are included in the final angular displacement evaluation over the given time period.

Explanation: This is a reverse-engineering rotational kinematics problem where angular displacement and final angular velocity are given, and the initial conditions along with acceleration must be determined. The motion is uniformly accelerated, meaning angular velocity changes linearly with time.

The relationship between angular displacement, initial angular velocity, angular acceleration, and time is used to form one equation. A second equation comes from the relationship between final angular velocity and initial angular velocity under constant angular acceleration. Together, these two equations allow determination of the unknown parameters.

The key idea is that angular velocity changes steadily, so the final value is directly linked to the initial value through acceleration and time. Meanwhile, displacement reflects the cumulative effect of continuously changing velocity over the interval.

Such problems require careful algebraic substitution and elimination of variables rather than direct computation. The physical interpretation is that the system starts at some unknown rotational speed and uniformly speeds up or slows down to reach the given final state while covering the specified angular displacement.

Explanation: This is a direct application of uniformly accelerated rotational motion where angular velocity changes at a constant rate over time. The angular acceleration represents the rate at which angular velocity increases per second.

To understand the motion, angular velocity at any instant is expressed as a linear function of time, where the initial angular velocity is modified by the product of angular acceleration and elapsed time. This relationship shows that angular velocity grows uniformly, similar to linear velocity under constant acceleration.

The key idea is that acceleration continuously adds to the angular velocity, resulting in a steady increase throughout the motion interval. Time acts as a scaling factor determining how much change occurs in velocity.

Such problems emphasise the direct proportionality between time and velocity change under constant acceleration. The final value depends only on initial conditions and total time elapsed, making it one of the simplest forms of rotational kinematics analysis.

Explanation: This is a uniformly decelerated rotational motion problem where an initially rotating system gradually comes to rest due to friction. The key concept is that angular velocity decreases linearly with time under constant angular deceleration.

To analyse such motion, the initial angular speed must first be interpreted in consistent units, and then the total angular displacement is determined over the stopping interval. Since deceleration is uniform, the average angular velocity over the interval becomes a useful concept for determining total rotation.

The physical interpretation is that the system starts with a high rotational speed and gradually loses energy until it stops. During this process, the angular displacement depends on both the initial speed and the time taken to stop.

The radius or diameter is not required for angular displacement calculation, as the motion is purely angular. The key is recognising that total rotation depends on the average of initial and final angular velocities multiplied by time. Since final velocity becomes zero, the motion simplifies significantly.

Explanation: This problem involves motion starting from rest under constant angular acceleration, where displacement increases quadratically with time. In such cases, displacement during successive time intervals increases due to continuously rising angular velocity.

The first second displacement provides direct information about angular acceleration because initial angular velocity is zero. Once acceleration is understood implicitly, displacement for any time interval can be compared using time-squared dependence of angular motion.

The key idea is that motion under constant angular acceleration does not produce equal displacements in equal time intervals. Instead, each subsequent second covers more angular distance than the previous one due to increasing speed.

To analyse interval displacement, total angular displacement at different time instants is compared, and the difference between successive totals gives displacement for a specific second. This method is essential for correctly interpreting non-uniform interval motion in rotational systems.

Explanation: This is a rotational kinematics problem where angular velocity changes from an initial to a final value under constant angular acceleration. The goal is to determine the angular displacement during this change in speed.

Instead of using time explicitly, the relationship between angular velocity and displacement under constant acceleration is used. This eliminates time and directly connects velocity change with angular displacement.

The key idea is that acceleration represents how quickly angular velocity changes, so the difference between final and initial velocity reflects the total change accumulated over the motion. Angular displacement depends on how long the system spends transitioning between these velocities.

Such problems are typically solved using energy-like or kinematic elimination methods where time is removed from equations, leaving a direct relation between velocities and displacement. This approach simplifies analysis and highlights the connection between rotational speed change and total angular movement.

Explanation: This problem links linear motion under gravity with circular motion. The speed gained by a freely falling body is derived from gravitational acceleration over a given distance, which then becomes the tangential speed of circular motion.

Once the linear speed is determined from gravitational motion, it is used in the expression for centripetal acceleration in circular motion. Centripetal acceleration depends on the square of linear speed divided by the radius of the circular path.

The key concept is connecting energy gained in vertical free fall with kinetic energy in circular motion. This equivalence allows transformation between gravitational potential energy change and Rotational Dynamics.

The radius plays a crucial role in determining how strongly the body accelerates toward the center of the circular path. The relationship highlights how different domains of mechanics interact through shared energy and motion principles.

Explanation: This is a relative motion problem on a circular path where two objects move in opposite directions. Since they travel along the same circumference, their relative speed determines how quickly they meet again.

The key idea is that when two bodies move in opposite directions, their speeds add up to give an effective closing speed along the circular track. Collision occurs when the combined distance covered equals one complete circumference.

The radius determines the total length of the circular path, which must be matched by the total relative distance covered by both particles. Time of collision is then obtained by dividing total circumference by relative speed.

This problem highlights how circular motion simplifies into linear relative motion along the path when analysing meeting points. The geometry of the circle ensures Periodic encounters based on total path length and combined speed.