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Explanation:

Spring Potential Energy

Definition: The potential energy stored in a spring is due to its displacement from its equilibrium position. This energy is a form of mechanical energy and is given by the formula:

Potential Energy (U) = 1/2 * k * x²

where:

• k is the spring constant (a measure of the stiffness of the spring)
• x is the displacement of the spring from its equilibrium position

Working Principle: When a spring is displaced from its equilibrium position, it stores energy. This energy is due to the work done to compress or stretch the spring. The potential energy in the spring increases with the square of the displacement. This quadratic relationship means that even small changes in displacement can lead to significant changes in the potential energy stored in the spring.

Correct Option Analysis:

The correct option is:

Option 2: It decreases to 1/4 of its original value

To understand why this is the correct option, let’s delve into the details:

Using the formula for the potential energy of a spring:

U = 1/2 * k * x²

If the original displacement is x, the original potential energy (U_original) is:

U_original = 1/2 * k * x²

When the displacement is reduced to half its original value, the new displacement (x_new) is:

x_new = x / 2

The new potential energy (U_new) is:

U_new = 1/2 * k * (x / 2)²

U_new = 1/2 * k * (x² / 4)

U_new = (1/4) * (1/2 * k * x²)

U_new = (1/4) * U_original

Therefore, when the displacement is reduced to half its original value, the potential energy stored in the spring decreases to 1/4 of its original value.

The correct answer is option 1) i.e. 1 rad/s².

CONCEPT:

• Angular acceleration: It is the rate of change angular velocity of an object with respect to time i.e.

\(α = \frac{ω_{f}-ω_{i}}{t}\)

where,α = angular acceleration, ωf = final angular velocity, ωi = initial angular velocity, and t = time

CALCULATION:

Given,

initial angular velocity, ω_i = 10 rad/s

final angular velocity, ω_f = 20 rad/s

time, t = 10 second

Hence,

angular acceleration, α = (20-10)/10 rad/s²

⇒ α = 1 rad/s².

CONCEPT

For rounding along the level curve a body require some centripetal force.

Centripetal Force: It is a force required to move a body uniformly in a circular motion. This force acts along the radius and towards the centre of the circle.

• When a body moves in a circle, its direction of motion at any instant is along the tangent of a circle. But according to Newton’s first law of motion, A body cannot change its direction of itself an external force is required for this purpose. This external force is centripetal force

\({\bf{Centripetal}};{\bf{Force}};\left( {\bf{F}} \right) = \frac{{m{v^2}}}{r};\left[ {{\rm{m}} = {\rm{mass}},{\rm{;v}} = {\rm{velocity}},{\rm{;r}} = {\rm{radius}}} \right]\)

• The centripetal force required for circular motion along the surface of the road, towards the centre of the turn. The Static friction between tyre and road provides the necessary centripetal force.

• For the turn on-level curve of a moving object, maximum velocity is

\(\frac{{m{v^2}}}{r} \le F \Rightarrow \frac{{m{v^2}}}{r} \le {\mu _s};N\)

\(\therefore \frac{{m{v^2}}}{r} \le {\mu _s};mg\)

\(\Rightarrow {V_{maximum}} \le \sqrt {{\mu _s}rg}\)

Where, m = mass of the object, Vmaximum = maximum velocity of an object in a circular motion, r = radius of the circle, μs = static friction between object and surface, g = gravity of earth

Calculation

Given that,

Mass of car, M_car = 500 kg

Mass of truck, M_truck = 1000 kg

From the above explanation, we can see that the maximum permissible speed of any vehicle can be expressed as

\(\Rightarrow {V_{maximum}} \le \sqrt {{\mu _s}rg}\)

And from the above equation, we can conclude that the maximum safe speed is independent of mass, thus it will be same for any vehicle with the same coefficient of friction.

Hence option 3 is correct among all

Concept:

• We have a ball that is dropped from a height. The ball reaches the ground and rebounds. We need to find the height to which the ball rebounds. We know the equation for the velocity of a body falling from a height and the relation between velocity and coefficient of restitution. Substituting values in these equations we get the required solution.
• The ratio of the final velocity to the initial velocity between two objects after their collision is known as the coefficient of restitution.
• The restitution coefficient is denoted as 'e' and is a unitless quantity, and its values range between 0 and 1.

Explanation:

The velocity of a body falling from a height ‘H’ is given by,

\(v = { \sqrt{2gH} }\)

Coefficient of restitution and velocity,

v_f = ev ; v_f  =  Final velocity, v = initial velocity, e = coefficient of restitution

\(v = {v_f \over e} \)

\({\sqrt{2gh} } = {{\sqrt{2g\times 9}} \over 0.5}\)

\(\sqrt h= {3 \over0.5} = 6\)

h = 36 m

Concept:

• Acceleration due to gravity: The acceleration achieved by any object due to the gravitational force of attraction by any planet is called acceleration due to gravity by the earth.
• As each planet has a different mass and radius so the acceleration due to gravity will be different for a different planet.
• Weight: The weight(w) of an object is the force of gravity on the object and may be defined as the mass(m) times the acceleration of gravity(g).
• Weight is a force, the SI unit of weight is Newton.

Weight (W) = m g

Where m is mass and g is the acceleration due to gravity.

• Mass of the moon is 1/100 times and the radius of the moon is 1/4 times that of the earth.
• Since the acceleration due to the gravity of the moon is one-sixth of that of the earth.
• So Weight of any object on Moon = 1/6 × Weight of any object on the Earth surface.

Important Points

Explanation:

We know that weight of the body changes at different planets but the mass all over the universe is the same.

Given the mass of the body on earth is 72 kg.

Therefore the mass of the body on the moon is the same as the earth i.e. 72 kg.

• Mass of an object remains the same on the earth and on the moon.
• Weight is the measurement of the pull of gravity on an object.
• Since the gravity is different on the Earth and on the moon, therefore the weight of an object is different on the earth and on the moon.

Mistake Points

The weight changes by 1/6 of original on moon not the mass of the body.

CONCEPT:

• Kepler's Law of area/second law: The line joining the sun to the planet sweeps out equal areas in equal interval of time i.e. areal velocity is constant.
• According to this law, the planet will move slowly when it is farthest from the sun and more rapidly, when it is nearest to the sun. it is similar to the law of conservation of angular momentum.

EXPLANATION:

• According to Kepler's second law, the planet will move slowly when it is farthest from the sun and more rapidly when nearer to the sun.
• When the earth is at perihelion then the distance is minimum so velocity will be maximum.
• And when the earth is at aphelion the distance is maximum so velocity is minimum.

⇒ Option 2 is correct.

Concept: 

The de Broglie wavelength is given by 

λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) = \(\frac{\mathrm{h}}{\mathrm{mv}} \) ​...(1)

The de Broglie wavelengths of the proton and alpha particles are λ and 2λ respectively. The mass of an alpha particle (m_α)  is 4 times the mass of a proton (m_p) 

Calculation: 

⇒ v = \(\frac{\mathrm{h}}{\mathrm{m} \lambda}\)....... (from 1)

\(\frac{v_p}{v_α}\) = \(\frac{m_α}{m_p} \) × \(\frac{\lambda_α}{\lambda_p}\)

⇒ 4 × 2 = 8

∴ the ratio of the velocities of the proton to the alpha particle is 8 : 1

Concept:

When a body is either wholly or partially immersed in a fluid, a lift is generated due to the net vertical component of hydrostatic pressure forces experienced by the body. This lift is called the buoyant force and the phenomenon is called buoyancy.

The Archimedes principle states that the buoyant force on a submerged body is equal to the weight of the liquid displaced by the body and acts vertically upward through the centroid of the displaced volume.

Thus, the net weight of the submerged body, (the net vertical downward force experienced by it) is reduced from its actual weight by an amount that equals the buoyant force.

FB = ρghA = ρgV

Weight of cube = buoyancy force

ρiceViceg = ρwVVDg

Calculation:

Given:

ρice = 920 kg/m3, ρw =1000 kg/m3

Let the x be the volume of iceberg floats above the water surface.

ρiceVg = ρwg(V - x)

920 × V × g = 1000 × g × (V - x)

0.92V = (V - x)

x = 8%

CONCEPT:

The efficiency of the Carnot cycle (η):

• It is defined as the ratio of net mechanical work done per cycle the gas (W) to the amount of heat energy absorbed per cycle from the source (Q1) i.e.,

\(η =\frac{W}{{{Q}_{1}}}~\)

As work done by the engine per cycle  is

⇒ W = Q1 – Q2

Where Q1 = amount of heat energy absorbed per cycle from the source and Q2 = energy absorbed per cycle from the sink.

\(⇒ η =\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}\)

As \(\frac{{{Q}_{2}}}{{{Q}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\)

\(⇒ η =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

Where T1 = temperature of the source and T2 = temperature of the sink.

• First, we need to find the sink temperature.
• after obtaining the sink temperature we have to put its value in the efficiency formula and find the new source temperature.

EXPLANATION:

Given - η = 40 % = 0.4 and temperature of the source (T₁) = 500 K

• The efficiency of the Carnot engine:

\(⇒ η =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

⇒ T2 = (1 - η) × T1

⇒ T2 = (1 - 0.4) × 500

⇒ T2 = 0.6 × 500

⇒ T2 = 300 K

• The temperature of the sink is 300 K.

 

Now Step: 2

Desired - η_D = 40 % = 0.4 and temperature of the sink (T2) = 833 K

Let's say T'₁ is the new source temperature.

• The efficiency of the Carnot engine:

\(⇒ η =1-\frac{{{T}_{2}}}{{{T}_{1}}}\)

\(⇒ T^{'}_1=\frac{T_2}{1-η}=\frac{300}{1-0.6}=\frac{300}{0.4}=750, K\)

• The temperature of the new source is 750 K.

Concept:

Simple Pendulum

• An ideal simple pendulum consists of a heavy point mass body (bob) suspended by a weightless, inextensible, and perfectly flexible string from rigid support about which it is free to oscillate.
• For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity,

\(\Rightarrow T=2\pi\sqrt{\frac{l}{g}},\)

• The above formula is only valid for small angular displacements.

Where, T = Time period of oscillation, l = Effective length of the pendulum, and g = gravitational acceleration

Explanation:

• The swinging girl can be considered as a simple pendulum.

The time period of oscillation of the simple pendulum is,

\(\Rightarrow T=2\pi\sqrt{\frac{l}{g}},\)

where l is the effective length of the simple pendulum.

• The length of the simple pendulum is equal to the distance from point of suspension to the centre of gravity (CG) of the oscillating body.
• When the girl stands up, the distance of CG from the point of suspension decreases.
• Therefore, the time period of oscillation also decreases.