Physics Test 233

Concept:

• Centrifugal Force: As the rod rotates, the bead experiences an outward centrifugal force, 
• F_centrifugal = mLω²
  • m = the mass of the bead,
  • L = the distance of the bead from the axis of rotation,
  • ω = the angular velocity of the rod.
• Frictional Force: The frictional force resists slipping and is given by,
•  F_friction = μN
  • μ= the coefficient of friction,
  • N= the normal force acting on the bead, which is provided by the centrifugal force.

Calculation:

Centrifugal Force,

The angular velocity ω increases with time due to the constant angular acceleration α,  ω = αt

The centrifugal force at time t is, F_centrifugal = mL(αt)² = mLα²t² 

The normal force N acting on the bead is equal to the centrifugal force because gravity is neglected,

N = mLα²t² 

The maximum frictional force is,F_friction = μN = μmLα²t² 

The bead will start slipping when the centrifugal force equals the maximum frictional force,

 F_centrifugal = F_friction 

The centrifugal and frictional forces,  mLα²t² = μmL

Concept:

Radioactive Decay Chain:

• The radioactive decay chain describes the series of successive radioactive decays that certain isotopes undergo until a stable isotope is formed.
• During each decay, the nucleus emits either an alpha particle (α) or a beta particle (β).

Alpha (α) Particle Emission:

• An alpha particle is composed of 2 protons and 2 neutrons.
• When an alpha particle is emitted, the mass number of the nucleus decreases by 4 and the atomic number decreases by 2.

Beta (β) Particle Emission:

• A beta particle is an electron (β-) emitted from the nucleus when a neutron decays into a proton.
• When a beta particle is emitted, the atomic number increases by 1, while the mass number remains unchanged.

Calculate the change in mass number and atomic number:

Change in mass number: 226 - 206 = 20

Change in atomic number: 88 - 82 = 6

For each alpha particle emission:

⇒ The mass number decreases by 4.

⇒ The atomic number decreases by 2.

Let the number of alpha particles emitted be n.

⇒ 4n = 20

⇒ n = 5

For each beta particle emission:

⇒ The atomic number increases by 1.

Let the number of beta particles emitted be m.

Since the net change in atomic number is 6 (decrease):

⇒ 2n - m = 6

⇒ 2×5 - m = 6

⇒ 10 - m = 6

⇒ m = 4

∴ The number of alpha particles emitted is 5 and the number of beta particles emitted is 4.

Concept:

• Projected with an initial velocity u₁ at an angle 30° to the horizontal.
• At its highest point, P has zero vertical velocity and a constant horizontal velocity component.
• Thrown vertically upwards with an initial velocity u₂ from a point directly below the highest point of P.
• To collide with P at its highest point, Q must reach the same vertical height as P and do so in the same time.

Calculation:

Vertical Component of Initial Velocity:

u₁y = u₁ sin 30° = u₁ × 1/2

Time to Reach Highest Point:

t_max = u₁y /g = (u₁/2)/g = u₁ / 2g

Horizontal Distance Covered:

x_max = u₁x × t_max = (u₁ cos 30°) × (u₁ / 2g)

xmax = (u₁ × √3/2) × (u₁/ 2g)

xmax = u₁₂ √3/4g

Vertical Height at Highest Point:

y_max = (u²_1y) / (2g) = ((u₁ / 2)²) / (2g) = u₁² / 8g

Vertical Distance to Highest Point of P:

Particle Q must reach the vertical height y_max at the same time P reaches its highest point.

y_max = u₂ t - (1/2) g t²

For a collision to occur, Q must reach y_max in time tmax, so:

u₂ = u × (3/2)^1/2

This condition ensures that both particles meet at the highest point of P.

∴ The correct option is 2

Concept:

Tilting of Water Surface in an Accelerating Tank:

• Horizontal Acceleration: When a tank filled with water is accelerated horizontally, the surface of the water tilts due to the combined effect of gravitational acceleration and the horizontal acceleration.
• The tilt of the water surface is determined by the ratio of the horizontal acceleration to the gravitational acceleration.
• The angle of tilt is given by: 

• • (a): Horizontal acceleration of the tank (SI unit: m/s²)
  • (g): Gravitational acceleration (SI unit: m/s²)
• As the tank accelerates, the water level rises on one side and drops on the other, creating a height difference.
• Once the angle is determined, the amount of water that spills can be calculated based on the geometry of the tank.

Calculation:

Given,

• Acceleration of the tank, a = 2 ms²
• Gravitational acceleration,g = 9.8 ms²
• Side of the cubical tank, L = 1m

Concept:

The coefficient of restitution (e) is the ratio of the relative speed after collision to the relative speed before collision in the vertical direction,

Concept:

To determine the power radiated by a spherical black body when its radius and temperature are changed, we use the Stefan Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its temperature and the surface area of the body.

Stefan Boltzmann Law:

The power radiated by a black body is given by:

The power radiated when the radius is halved and the temperature is doubled is 18000 W.

The correct option is (4).

Concept:

A flywheel rotates about an axis and experiences angular retardation proportional to its angular velocity due to friction.

The angular retardation is given as,

α = -kω

Where:

• ω is the angular velocity
• k is the constant of proportionality
• α is the angular acceleration (retardation)

We are tasked with finding how many more rotations the flywheel will make before coming to rest after its angular velocity falls to half of its initial value, while making n rotations.

Calculation:

We know that angular retardation is proportional to the angular velocity. Thus, we use the following relation:

dω/(-kω) = dθ

By integrating this equation, we find,

∫(ω₀ to ω₀/2) (1/ω) dω = -k ∫(0 to θ₁) dθ

Which leads to,

ln(1/2) = -kθ₁

So, the angular displacement when the angular velocity falls to half its initial value is,

θ₁ = ln(2)/k

Given that the flywheel makes n rotations during this time, the displacement is,

θ₁ = 2πn

This gives us the constant k as,

k = ln(2)/(2πn)

At this point, the angular velocity has fallen to half of its initial value. We repeat the integration for the time until rest,

∫(ω₀/2 to 0) (1/ω) dω = -k ∫(θ₁ to θ₂) dθ

This leads to,

ln(1/2) = -kθ₂

Thus, the additional angular displacement until the flywheel comes to rest is,

θ₂ = ln(2)/k

We already know that k from earlier calculations. Thus, we get,

θ₂ = 2πn

This shows that the flywheel will make an additional n rotations before coming to rest.

∴ The correct option is 2

Concept:

The process in which the change in the heat is zero is known as the adiabatic process. The exchange from the surroundings to the system is prohibited in the process. All the work done on or by the system will alter the internal energy of the system.