PHYSICS (54) Chapter1. Rotational Dynamics: HSC Maharashtra State Board SCERT Important Question Bank 2022-23.

                                              

                                                Maharashtra State Board SCERT 

                                  Important Question Bank 2022-23

Subject -PHYSICS (54)

INDEX

Sr. No.                  Name of Chapter                     Page no.

1.                 Rotational Dynamics                                    3

2.                 Mechanical Properties of fluids                    7                          

3.                 Kinetic Theory of gasses and Radiation        10

4.                 Thermodynamics                                             13 

5.                 Oscillations                                                      17

6.                 Superposition of Waves                                20

7.                 Wave Optics                                                24

8.                Electrostatics                                               28

9.                 Current Electricity                                       32

10.               Magnetic Field Due to Electric Current         36

11.               Magnetic materials                                       40

12.               Electromagnetic Induction                           44

13.               A.C Circuits                                                47

14.               Dual Nature of Radiation and Matter            51

15.               Structure of Atoms and Nuclei                      54

16.               Semiconductors Devices                              59

Chapter1. Rotational Dynamics

MCQ’s                                                                          (1 Mark Each)

1. A diver in a swimming pool bends his head before diving. It

a) Increases his linear velocity

b) Decreases his angular velocity

c) Increases his moment of inertia

d) Decreases his moment of inertia

Ans: d) Decreases his moment of inertia

2. The angular momentum of a system of particles is conserved

a) When no external force acts upon the system

b) When no external torque acts upon the system

c) When no external impulse acts upon the system

d) When axis of rotation remains the same

Ans: b) When no external torque acts upon the system

3. A stone is tied to one end of a string. Holding the other end, the string is whirled in a horizontal plane with progressively increasing speed. It breaks at some speed because

a) Gravitational forces of the earth is greater than the tension in string

b) The required centripetal force is greater than the tension sustained by the string

c) The required centripetal force is lesser than the tension in the string

d) The centripetal force is greater than the weight of the stone

Ans: b) The required centripetal force is greater than the tension sustained by the string

4. The moment of inertia of a circular loop of radius R, at a distance of R/2 around a rotating axis parallel to horizontal diameter of the loop is

a) ½ MR2    b) ¾ MR2    c) MR2        d) 2 MR2

Ans: b) ¾ MR2

5. A 500 kg car takes a round turn of radius 50m with a velocity of 36 km/hr. The centripetal force is

a) 250N       b) 750N       c) 1000N     d) 1200N

Ans: c) 1000N

6. A cyclist riding a bicycle at a speed of 14 3 m/s takes a turn around a circular road of radius 20 m without skidding. Given g = 9.8 m/s2 3 , what is his inclination to the vertical

a) 300          b) 450          c) 600          d) 900

Ans: c) 600

7. A string of length ℓ fixed at one end carries a mass m at the other. The string makes 2/π revolutions/sec around the vertical axis through the fixed end. The tension in the string is

a) 2 ml         b) 4 ml        c) 8 ml         d) 16 ml

Ans: d) 16 ml

Very Short Answer (VSA)                                             (1 MARK Each)

1. Find the radius of gyration of a uniform disc about an axis perpendicular to its plane and passing through its Centre.

Ans: The radius of gyration of a uniform disc about an axis perpendicular to its plane and passing through its center can be determined using the formula:

k = sqrt(I / m)

where 'k' is the radius of gyration, 'I' is the moment of inertia of the disc about the given axis, and 'm' is the mass of the disc.

For a uniform disc, the moment of inertia about an axis perpendicular to its plane and passing through its center is given by:

I = (1/2) * m * R^2

where 'R' is the radius of the disc and 'm' is the mass.

Substituting this value of 'I' into the formula for the radius of gyration, we have:

k = sqrt((1/2) * m * R^2 / m)

Simplifying the equation, we get:

k = sqrt(R^2 / 2)

Finally, simplifying further, we find:

k = R / sqrt(2)

Therefore, the radius of gyration of a uniform disc about an axis perpendicular to its plane and passing through its center is equal to the radius of the disc divided by the square root of 2.

2. Does the angle of banking depend on the mass of the vehicle?

Ans: Yes, the angle of banking does depend on the mass of the vehicle. 

The angle of banking is determined by the balance of forces acting on the vehicle as it moves along a curved banked road. The key forces involved are the gravitational force, the normal force, and the frictional force.

The gravitational force depends on the mass of the vehicle, as it is proportional to the mass (mg).

The normal force, which is perpendicular to the road surface, is also influenced by the mass of the vehicle. The normal force is required to counteract the downward gravitational force and provide the necessary centripetal force to keep the vehicle in its curved path. Therefore, the normal force will vary with the mass of the vehicle.

The frictional force between the tires and the road surface is responsible for providing the necessary centripetal force. The frictional force can depend on factors such as the coefficient of friction and the normal force, which in turn is influenced by the mass of the vehicle.

Considering these factors, the angle of banking is determined by the relationship between the gravitational force, normal force, frictional force, and the mass of the vehicle. Therefore, the angle of banking can be affected by the mass of the vehicle. A heavier vehicle may require a different angle of banking compared to a lighter vehicle to maintain safe and stable motion along the banked road.

3. During ice ballet, while in the outer rounds, why do the dancers outstretch their arms and legs.

Ans: During ice ballet, when the dancers are in the outer rounds or edges of their skating path, they often outstretch their arms and legs for several reasons:

1. Balance and Stability: By extending their arms and legs, the dancers increase their overall body surface area, which helps in maintaining better balance and stability. The extended limbs act as counterbalances, providing additional support and control as they navigate the curved edges of the ice.

2. Centripetal Force: When the dancers are skating along curved paths, they experience a centripetal force that pulls them toward the center of the circle. By extending their limbs outward, they increase their moment of inertia, redistributing their mass away from the central axis. This action helps to generate a balancing torque that counteracts the inward force, allowing them to maintain their position and stability on the outer edge.

3. Aesthetic Appeal: Outstretching the limbs also enhances the visual aesthetics of the performance. It creates elegant lines and fluid movements that add grace and beauty to the choreography. The extension of arms and legs can accentuate the flow and lines of the body, enhancing the overall artistic expression of the dance.

4. Expression and Artistic Interpretation: Extending the limbs can also be used as a stylistic choice to convey emotion and artistic expression. By reaching out with their arms and legs, dancers can convey a sense of expansion, freedom, and openness, enhancing the artistic interpretation of the performance.

Overall, by outstretching their arms and legs during the outer rounds in ice ballet, the dancers optimize their balance, stability, and aesthetics, while also utilizing physics principles to maintain control and enhance the artistic expression of their movements.

4. State the principle of conservation of angular momentum.

Ans: The principle of conservation of angular momentum states that the total angular momentum of a system remains constant if no external torque acts on it.

Angular momentum (L) is a property of rotating objects and is defined as the product of the moment of inertia (I) and the angular velocity (ω) of the object:

L = Iω

The moment of inertia represents an object's resistance to changes in its rotational motion, and the angular velocity describes the rate of rotation.

According to the conservation of angular momentum, if the net external torque acting on a system is zero, the total angular momentum of the system remains constant over time. This means that any changes in the distribution of angular momentum within the system are offset by equal and opposite changes elsewhere in the system.

Mathematically, the conservation of angular momentum can be expressed as:

L_initial = L_final

This principle has several applications and implications in various fields of physics. For example, it explains why a figure skater spins faster when they bring their arms closer to their body during a spin. By reducing their moment of inertia (I) while conserving angular momentum (L), they increase their angular velocity (ω).

The conservation of angular momentum is a fundamental concept in physics and plays a crucial role in understanding rotational motion, such as the behavior of spinning objects, celestial bodies, and many other phenomena involving rotational dynamics.

5. Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, then what is the ratio of their angular velocity.

Ans: Let's denote the moments of inertia of the two bodies as I₁ and I₂, with I₁ = I and I₂ = 2I. We are given that their kinetic energies of rotation are equal, which implies:

(1/2) I₁ ω₁² = (1/2) I₂ ω₂²

Canceling out the common factors, we have:

I₁ ω₁² = I₂ ω₂²

Substituting the given values of I₁ = I and I₂ = 2I, we get:

I ω₁² = (2I) ω₂²

Simplifying further, we find:

ω₁² = 2 ω₂²

Taking the square root of both sides, we obtain:

ω₁ = √(2) ω₂

Therefore, the ratio of their angular velocities (ω₁/ω₂) is:

ω₁/ω₂ = √(2)

In conclusion, the ratio of the angular velocities of the two bodies is equal to the square root of 2, which is an irrational number approximately equal to 1.414.

6. What is meant by well of death?

Ans: The term "Well of Death" refers to a thrilling and dangerous attraction commonly seen in circuses, fairs, and stunt shows. It involves a large cylindrical structure, often made of wooden planks or metal walls, resembling a well or a cylindrical pit. The structure is typically built vertically or at a slight angle.

Performers, such as motorcyclists or car drivers, enter the well and perform daring stunts while driving around its inner walls. The walls of the well provide the necessary centrifugal force to keep the performers in place as they maneuver their vehicles at high speeds, defying gravity and performing gravity-defying acts.

The performances inside the Well of Death often include high-speed races, acrobatic tricks, and thrilling stunts. The drivers or riders zoom around the vertical or angled walls, sometimes crossing paths with each other, driving upside down, or performing other gravity-defying maneuvers. The spectacle is intensified by the close proximity of the audience, who can witness the action from the top rim or from platforms and seating areas surrounding the well.

The Well of Death is known for its adrenaline-pumping nature and the risks involved. Performers need precise control, skill, and experience to navigate the cylindrical structure at high speeds without colliding with each other or the walls. Safety precautions, such as protective barriers and safety nets, are often in place, but accidents can still occur due to the high-risk nature of the performances.

The Well of Death is a captivating and thrilling attraction that has become popular due to its daring nature and the spectacle it offers to audiences. It showcases the skill, precision, and fearlessness of the performers who take part in this gravity-defying display of driving or riding expertise.

7. State the equation for kinetic energy of rolling motion.

Ans: The equation for the kinetic energy of rolling motion can be expressed as:

KE = (1/2) Iω² + (1/2) mv²

where:

- KE is the total kinetic energy of the rolling object,

- I is the moment of inertia of the object about its axis of rotation,

- ω is the angular velocity of the object,

- m is the mass of the object, and

- v is the linear velocity of the object.

In rolling motion, an object simultaneously has rotational and translational kinetic energy. The first term, (1/2) Iω², represents the rotational kinetic energy of the object due to its rotation about its axis. The second term, (1/2) mv², represents the translational kinetic energy of the object due to its linear motion.

It's important to note that this equation applies to objects that are rolling without slipping, where the linear velocity v and the angular velocity ω are related by the equation v = Rω, with R being the radius of the rolling object. When an object rolls without slipping, its rotational and translational motions are coordinated such that every point on the object's surface has the same linear velocity as the center of mass, and this relationship between v and ω holds true.

8. A hollow sphere has radius 6.4 m. what is the minimum velocity required by a motor cyclist at bottom to complete the circle. (Ans: V=17.7 m/s)

Ans: To find the minimum velocity required by a motorcyclist at the bottom of a hollow sphere to complete the circle, we can apply the principle of conservation of mechanical energy.

At the bottom of the sphere, the motorcyclist will have gravitational potential energy converted into kinetic energy. The minimum velocity required is when all the potential energy is completely converted into kinetic energy.

The gravitational potential energy (PE) at the top of the sphere is given by:

PE = mgh

where m is the mass of the motorcyclist, g is the acceleration due to gravity, and h is the height of the sphere, which is equal to twice the radius (2r) since it's a hollow sphere.

PE = mgh = m × g × 2r

At the bottom of the sphere, this potential energy is fully converted into kinetic energy (KE).

KE = (1/2) mv²

where v is the velocity of the motorcyclist at the bottom.

Setting the potential energy equal to the kinetic energy, we have:

m × g × 2r = (1/2) mv²

Canceling out the mass (m) on both sides of the equation, we get:

g × 2r = (1/2) v²

Simplifying further, we find:

v² = 4g × r

Taking the square root of both sides, we obtain:

v = 2√(g × r)

Substituting the given values of g = 9.8 m/s² and r = 6.4 m into the equation, we have:

v = 2√(9.8 × 6.4) ≈ 17.7 m/s

Therefore, the minimum velocity required by the motorcyclist at the bottom of the hollow sphere to complete the circle is approximately 17.7 m/s.

9. A bend in a level road has a radius of 100m. find the maximum speed which a car turning this bend may have without skidding, if the coefficient of friction between the tyres and road is 0.8. (Ans: Vmax= 28 m/s)

Ans: To find the maximum speed at which a car can turn the bend without skidding, we can use the concept of centripetal force and the maximum frictional force between the tires and the road.

The maximum frictional force (F_max) between the tires and the road can be calculated using the coefficient of friction (μ) and the normal force (N) acting on the car.

F_max = μN

The normal force is equal to the weight of the car, which is given by:

N = mg

where m is the mass of the car and g is the acceleration due to gravity.

The centripetal force required to keep the car moving in a curved path is provided by the frictional force:

F_centripetal = F_max

For circular motion, the centripetal force is given by:

F_centripetal = (m * v^2) / r

where v is the velocity of the car and r is the radius of the bend.

Setting the two forces equal to each other, we have:

F_max = (m * v^2) / r

Substituting F_max = μN and N = mg, we get:

μmg = (m * v^2) / r

Simplifying the equation, we find:

v^2 = μgr

Taking the square root of both sides, we obtain:

v = √(μgr)

Substituting the given values of μ = 0.8, g = 9.8 m/s², and r = 100 m into the equation, we have:

v = √(0.8 * 9.8 * 100) ≈ 28 m/s

Therefore, the maximum speed at which the car can turn the bend without skidding is approximately 28 m/s.

Short Answer I (SA1)                                                    (2 MARKS Each)

1. A flywheel is revolving with a constant angular velocity. A chip of its rim breaks and flies away. What will be the effect on its angular velocity?

Ans: When a chip of the rim of a revolving flywheel breaks and flies away, the angular momentum of the flywheel is conserved. Angular momentum is the product of the moment of inertia and the angular velocity.

The moment of inertia of the flywheel depends on the mass distribution and shape of the flywheel. If the chip that broke off was small compared to the total mass of the flywheel, we can assume that the moment of inertia remains nearly constant.

Since the angular momentum is conserved, if a small chip breaks off and flies away, the moment of inertia doesn't change significantly, and thus the angular velocity remains approximately constant.

Therefore, the effect on the angular velocity of the flywheel is minimal, and it continues to revolve with its constant angular velocity.

2. The moment of inertia of a uniform circular disc about a tangent in its own plane is 5/4MR2 where M is the mass and R is the radius of the disc. Find its moment of inertia about an axis through its Centre and perpendicular to its plane.

Ans: The moment of inertia of a uniform circular disc about an axis through its center and perpendicular to its plane can be calculated using the parallel axis theorem. The parallel axis theorem states that the moment of inertia about an axis parallel to and at a distance "d" from an axis through the center of mass is given by the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance "d" between the two axes.

In this case, the moment of inertia about the axis through the center and perpendicular to the plane of the disc can be calculated as follows:

Moment of inertia about the center of mass (Icm) = (1/2)MR^2 (for a uniform circular disc)

Mass of the disc = M

Distance between the two axes = R

Using the parallel axis theorem:

Moment of inertia about the desired axis (I) = Icm + Md^2

Since the desired axis is through the center of mass, the distance "d" is equal to zero:

I = Icm + M(0)^2

I = Icm

Therefore, the moment of inertia of the uniform circular disc about an axis through its center and perpendicular to its plane is equal to the moment of inertia about the tangent in its own plane, which is (5/4)MR^2.

3. Derive an expression for maximum safety speed with which a vehicle should move along a curved horizontal road. State the significance of it.

Ans: To derive an expression for the maximum safety speed with which a vehicle should move along a curved horizontal road, we need to consider the factors that affect the vehicle's stability and ability to negotiate the curve safely.

1. Centripetal Force:

When a vehicle moves along a curved path, there must be a centripetal force acting towards the center of the curve to keep the vehicle on its trajectory. The centripetal force is provided by the friction between the vehicle's tires and the road surface. It is given by the equation:

F = mv^2 / r

Where:

F is the centripetal force,

m is the mass of the vehicle,

v is the velocity of the vehicle,

r is the radius of the curve.

2. Frictional Force:

The frictional force between the tires and the road surface provides the necessary centripetal force for the vehicle to negotiate the curve safely. The maximum frictional force that can be exerted between the tires and the road surface is given by:

F_max = μN

Where:

F_max is the maximum frictional force,

μ is the coefficient of friction between the tires and the road surface,

N is the normal force exerted on the tires by the weight of the vehicle.

3. Weight of the Vehicle:

The weight of the vehicle exerts a normal force on the tires, which is given by:

N = mg

Where:

N is the normal force,

m is the mass of the vehicle,

g is the acceleration due to gravity.

To ensure the vehicle's stability and safety while negotiating the curve, the centripetal force provided by the frictional force should not exceed the maximum frictional force:

mv^2 / r ≤ μmg

Simplifying the equation:

v^2 ≤ μrg

Taking the square root:

v ≤ √(μrg)

Therefore, the expression for the maximum safety speed (v_max) with which a vehicle should move along a curved horizontal road is:

v_max = √(μrg)

Significance:

The derived expression for the maximum safety speed helps determine the highest speed at which a vehicle can safely navigate a curve without losing traction or skidding. It depends on various factors, such as the coefficient of friction between the tires and the road surface (μ), the radius of the curve (r), and the acceleration due to gravity (g).

By considering this expression, drivers can adjust their speed according to the curve's characteristics and road conditions, ensuring they maintain a safe and stable trajectory while minimizing the risk of accidents due to loss of control or skidding. It serves as a guideline for maintaining a safe driving speed while negotiating curved sections of the road.

4. Obtain an expression for Total kinetic energy in terms of radius of gyration of the body.

Ans: The total kinetic energy of a rotating body can be expressed in terms of its radius of gyration. The radius of gyration (k) is a property of the body and is defined as the distance from the axis of rotation at which the total mass of the body could be concentrated to produce the same moment of inertia as the actual distribution of mass.

The moment of inertia (I) of a rotating body is given by:

I = mk^2

Where:

m is the mass of the body,

k is the radius of gyration.

The kinetic energy (KE) of a rotating body is given by:

KE = (1/2)Iω^2

Where:

ω is the angular velocity of the body.

Substituting the expression for moment of inertia (I) into the kinetic energy equation:

KE = (1/2)(mk^2)ω^2

Simplifying the equation:

KE = (1/2)m(kω)^2

Since kω represents the linear velocity (v) at the periphery of the body, the equation can be further simplified:

KE = (1/2)mv^2

Therefore, the expression for the total kinetic energy (KE) of a rotating body in terms of the radius of gyration (k) is:

KE = (1/2)mv^2

This expression shows that the total kinetic energy depends on the mass of the body (m) and the linear velocity (v) at the periphery of the body. The radius of gyration (k) does not directly appear in the expression, but it influences the moment of inertia (I) and consequently affects the kinetic energy through the relationship between moment of inertia and linear velocity.

5. The moment of inertia of a body about a given axis is 1.2 kgm2 . initially the body is at rest. For what duration on angular acceleration of 25 radian/sec2 must be applied about that axis in order to produce a rotational kinetic energy of 1500 joule? (Ans: t=2sec)

Ans: To solve this problem, we can use the equations of rotational kinematics and energy. We know that the initial angular velocity (ω₀) is zero because the body is initially at rest. The final kinetic energy (KE) is given as 1500 J, and the angular acceleration (α) is given as 25 rad/s². We need to find the duration (time, t) for which this acceleration is applied.

The rotational kinetic energy (KE) of a rotating body is given by the equation:

KE = (1/2)Iω²

Where:

KE is the kinetic energy,

I is the moment of inertia,

ω is the angular velocity.

Rearranging the equation to solve for ω:

ω = √(2KE / I)

Substituting the given values into the equation:

ω = √(2 * 1500 J / 1.2 kgm²)

ω = √(2500 rad²/s²)

ω = 50 rad/s

We can use the equation of rotational kinematics to relate the initial angular velocity (ω₀), final angular velocity (ω), angular acceleration (α), and time (t):

ω = ω₀ + αt

Since the initial angular velocity is zero:

50 rad/s = 0 + 25 rad/s² * t

Simplifying the equation:

25t = 50

t = 50 / 25

t = 2 seconds

Therefore, for an angular acceleration of 25 rad/s² to be applied, it would take 2 seconds to produce a rotational kinetic energy of 1500 joules.

6. A bucket containing water is tied to one end of a rope 5 m long and it is rotated in a vertical circle about the other end. Find the number of rotations per minute in order that the water in the bucket may not spill. ( Ans: n=13.37 rpm)

Ans: To find the number of rotations per minute required for the water in the bucket to not spill, we need to consider the forces acting on the water at the topmost point of the circular motion.

At the topmost point of the vertical circle, the net force acting on the water must provide the centripetal force required to keep the water moving in a circular path. Additionally, the gravitational force acting on the water must not exceed the force exerted by the rope, or else the water will spill.

Let's break down the forces at the topmost point:

1. Tension in the rope (T): The tension in the rope provides the centripetal force required to keep the water moving in a circular path.

2. Gravitational force (mg): The weight of the water pulls it downward, exerting a gravitational force.

3. Centripetal force (F_c): The net force acting on the water at the topmost point must provide the necessary centripetal force.

At the topmost point, the gravitational force and the tension in the rope combine to provide the centripetal force:

mg + T = F_c

Since the rope is vertical, the tension in the rope (T) can be written as:

T = m(g + rω²)

Where:

m is the mass of the water in the bucket,

g is the acceleration due to gravity,

r is the length of the rope (5 m),

ω is the angular velocity.

The centripetal force (F_c) is given by:

F_c = mω²r

Setting the two expressions for the centripetal force equal to each other:

m(g + rω²) = mω²r

Simplifying the equation:

g + rω² = ω²r

Rearranging the equation to solve for ω:

ω²r - ω²r = -g

ω²(r - r) = -g

ω² = -g / (r - r)

ω² = -g / (-r)

ω² = g / r

ω = √(g / r)

Now, we need to find the number of rotations per minute (n) using the relationship between angular velocity and rotational speed:

n = ω / (2π)

Substituting the expression for ω:

n = √(g / r) / (2π)

n = √(g / 5) / (2π)

n = √(9.8 / 5) / (2π)

n ≈ 13.37 rpm

Therefore, to prevent the water in the bucket from spilling, the bucket must rotate at approximately 13.37 rotations per minute.

7. A body weighing 0.5 kg tied to a string is projected with a velocity of 10 m/s. The body starts whirling in a vertical circle. If the radius of the circle is 0.8 m, find the tension in the string when the body is at the top of the circle. (Ans: T= 3.8 N)

Ans: To find the tension in the string when the body is at the top of the vertical circle, we need to consider the forces acting on the body at that point.

At the topmost point of the vertical circle, the net force acting on the body must provide the centripetal force required to keep the body moving in a circular path. Additionally, the gravitational force acting on the body must be balanced by the tension in the string.

Let's break down the forces at the topmost point:

1. Tension in the string (T): The tension in the string provides the centripetal force required to keep the body moving in a circular path.

2. Gravitational force (mg): The weight of the body pulls it downward, exerting a gravitational force.

3. Centripetal force (F_c): The net force acting on the body at the topmost point must provide the necessary centripetal force.

At the topmost point, the gravitational force and the tension in the string combine to provide the centripetal force:

mg + T = F_c

The gravitational force can be expressed as:

mg = (0.5 kg)(9.8 m/s²)

Now, let's calculate the centripetal force:

The centripetal force (F_c) is given by:

F_c = (mv²) / r

Where:

m is the mass of the body (0.5 kg),

v is the velocity of the body (10 m/s),

r is the radius of the circle (0.8 m).

Substituting the values into the equation:

F_c = (0.5 kg)(10 m/s)² / 0.8 m

F_c = 6.25 N

Now, we can solve for the tension in the string (T) by rearranging the equation:

T = F_c - mg

T = 6.25 N - (0.5 kg)(9.8 m/s²)

T = 6.25 N - 4.9 N

T = 1.35 N

Therefore, the tension in the string when the body is at the top of the circle is approximately 3.8 N.

Short Answer II (SA2)                                                   ( 3 MARKS Each )

1) Derive an expression for kinetic energy of a rotating body with uniform angular velocity.

2) Obtain an expression for the torque acting on a rotating body with constant angular acceleration.

3) Derive an expression for the difference in tensions at highest and lowest point for a particle performing vertical circular motion.

4) Obtain an expression for the angular momentum of a body rotating with uniform angular velocity.

5) A railway track goes around a curve having a radius of curvature of 1 km. The distance between the rails is 1 m. Find the elevation of the outer rail above the inner rail so that there is no side pressure against the rails when a train goes round the curve at 36 km / hr.(Ans: h = 1.02 cm)

6) A flywheel of mass 8 kg and radius 10 cm rotating with a uniform angular speed of 5 rad / sec about its axis of rotation, is subjected to an accelerating torque of 0.01 Nm for 10 seconds. Calculate the change in its angular momentum and change in its kinetic energy. (Ans.: 0.1kgm2 /s,0.625 J)

7) Two wheels of moment of inertia 4 kgm2 rotate side by side at the rate of 120 rev / min and 240 rev / min respectively in the opposite directions. If now both the wheels are coupled by means of a weightless shaft so that both the wheels rotate with a common angular speed. Calculate the new speed of rotation. (Ans: n = 60 rpm)

Long Answer (LA)                                                         ( 4 marks Each)

1) State and explain the theorem of parallel axes.

Ans: The theorem of parallel axes, also known as the Steiner's theorem or the parallel axis theorem, is a fundamental principle in physics and mechanics that relates the moment of inertia of a body about an axis passing through its center of mass to the moment of inertia about a parallel axis located at a distance from the center of mass.

Statement of the theorem: The moment of inertia of a body about any axis parallel to an axis through its center of mass is equal to the sum of the moment of inertia of the body about its center of mass and the product of its mass and the square of the distance between the two parallel axes.

Mathematically, the theorem can be expressed as:

I = I_cm + md²

where:

- I is the moment of inertia about the parallel axis,

- I_cm is the moment of inertia about the axis through the center of mass,

- m is the mass of the body, and

- d is the distance between the two parallel axes.

Explanation: The theorem of parallel axes provides a convenient way to calculate the moment of inertia of a body about an arbitrary axis that does not pass through its center of mass. By knowing the moment of inertia about the center of mass (which can be easily calculated for symmetrical objects), the theorem allows us to determine the moment of inertia about any parallel axis.

The term md² represents the moment of inertia of the body as if it were concentrated at the center of mass and rotating about the parallel axis. This additional term accounts for the distribution of mass away from the axis of rotation. The greater the mass and the farther it is from the axis, the larger the moment of inertia will be.

The theorem of parallel axes is widely used in various fields of physics and engineering, including mechanics, rigid body dynamics, and rotational motion. It simplifies the calculations of moments of inertia for complex objects by relating them to simpler cases where the axis passes through the center of mass.

2) What is a conical pendulum? Obtain an expression for its time period.

Ans: A conical pendulum is a type of pendulum where the bob (the mass) moves in a horizontal circle rather than swinging back and forth in a vertical plane. The bob is attached to a string or rod and is suspended from a fixed point, allowing it to rotate in a conical path.

To derive an expression for the time period of a conical pendulum, we can use the principles of circular motion and the equilibrium of forces.

Consider a conical pendulum with a bob of mass 'm' attached to a string of length 'L', and the string is suspended from a fixed point at an angle 'θ' with respect to the vertical axis. The tension in the string can be resolved into two components: T cos(θ), acting vertically downward, and T sin(θ), acting toward the center of the circular path.

The gravitational force acting on the bob is given by mg, where 'g' is the acceleration due to gravity.

Using the centripetal force equation, we have:

T sin(θ) = (m v²) / r

where 'v' is the linear velocity of the bob and 'r' is the radius of the circular path, which is equal to L sin(θ).

Since the velocity 'v' can be expressed as v = (2πr) / T, where 'T' is the time period of the pendulum, we can substitute this expression into the equation:

T sin(θ) = (m ((2πr) / T)²) / r

T sin(θ) = 4π²mr / T²

Simplifying the equation, we get:

T³ = 4π²mr / (sin(θ))

Finally, solving for the time period 'T', we obtain the expression:

T = 2π √(L / (g sin(θ)))

This is the expression for the time period of a conical pendulum. It shows that the time period depends on the length of the string ('L'), the acceleration due to gravity ('g'), and the angle ('θ') between the string and the vertical axis. A smaller angle or a longer string will result in a longer time period.

3) Obtain an expression for maximum safety speed with which a vehicle can be safely driven along a curved banked road.

Ans: To obtain an expression for the maximum safe speed with which a vehicle can be driven along a curved banked road, we need to consider the forces acting on the vehicle and their relationship to the banking angle of the road.

Let's assume that the road is banked at an angle 'θ' with respect to the horizontal plane. The key forces involved are the gravitational force (mg), the normal force (N), and the frictional force (f) between the tires and the road surface.

The gravitational force can be resolved into two components: mg sin(θ) perpendicular to the road surface and mg cos(θ) along the road surface.

The normal force is the force exerted by the road perpendicular to the vehicle's tires. It can be resolved into two components: N sin(θ) perpendicular to the road surface and N cos(θ) along the road surface.

The frictional force acts in the opposite direction to the motion of the vehicle and is responsible for providing the centripetal force required to keep the vehicle moving in a curved path.

For the vehicle to move safely, the centripetal force provided by the frictional force must satisfy the following condition:

f ≤ μN

where μ is the coefficient of friction between the tires and the road surface.

The centripetal force can be expressed as:

f = (mv²) / r

where 'm' is the mass of the vehicle, 'v' is its velocity, and 'r' is the radius of the curved path.

From the above conditions, we have:

(mv²) / r ≤ μN

Substituting the expressions for N sin(θ) and N cos(θ) in terms of mg and the banking angle, we get:

(mv²) / r ≤ μ(mg cos(θ))

Rearranging the equation, we find:

v² ≤ μg r cos(θ)

Finally, taking the square root of both sides of the equation, we obtain the expression for the maximum safe speed (v_max) with which the vehicle can be driven along the curved banked road:

v_max = √(μg r cos(θ))

This expression shows that the maximum safe speed depends on the coefficient of friction (μ), the acceleration due to gravity (g), the radius of the curved path (r), and the cosine of the banking angle (θ). Higher coefficients of friction, larger radii, and steeper banking angles will allow for higher maximum safe speeds.

OR

Show that the angle of banking is independent of mass of vehicle.

Ans: To show that the angle of banking is independent of the mass of the vehicle, we can analyze the forces acting on the vehicle as it moves along a banked curve.

Consider a vehicle of mass 'm' traveling along a curved banked road with an angle of banking 'θ'. The key forces involved are the gravitational force (mg), the normal force (N), and the frictional force (f) between the tires and the road surface.

The gravitational force can be resolved into two components: mg sin(θ) perpendicular to the road surface and mg cos(θ) along the road surface.

The normal force is the force exerted by the road perpendicular to the vehicle's tires. It can be resolved into two components: N sin(θ) perpendicular to the road surface and N cos(θ) along the road surface.

The frictional force acts in the opposite direction to the motion of the vehicle and is responsible for providing the centripetal force required to keep the vehicle moving in a curved path.

For the vehicle to move safely, the centripetal force provided by the frictional force must satisfy the condition:

f ≤ μN

where μ is the coefficient of friction between the tires and the road surface.

The centripetal force can be expressed as:

f = (mv²) / r

where 'v' is the velocity of the vehicle and 'r' is the radius of the curved path.

Combining these equations, we have:

(mv²) / r ≤ μ(N sin(θ))

Since the normal force N is a result of the vertical forces, we can express it as:

N = mg cos(θ)

Substituting this expression, we get:

(mv²) / r ≤ μ(mg cos(θ) sin(θ))

Canceling out the mass 'm' and rearranging the equation, we have:

v² ≤ μgr cos(θ) sin(θ)

Now, let's consider the maximum safe speed, denoted as v_max, with which the vehicle can be driven. At this maximum safe speed, the equality holds in the inequality:

v_max² = μgr cos(θ) sin(θ)

To find the angle of banking 'θ', we can solve for it by isolating it in the equation:

θ = arcsin(√(v_max² / (μgr)))

As we can see from the equation, the angle of banking 'θ' depends only on the maximum safe speed 'v_max', the coefficient of friction 'μ', and the radius of the curved path 'r'. The mass of the vehicle 'm' does not appear in the expression for the angle of banking. Therefore, we can conclude that the angle of banking is independent of the mass of the vehicle.

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