Physics Test 229

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Physics Test 229

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Question 1
4 / -1

A particle is projected with velocity v₀ along x-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. ma = -ax². The distance at which the particle stops:

A

B

C

D
Solution
CONCEPT:
- The damping force is defined as the force which is proportional to the velocity of the mass but opposite to the motion of the mass.
- Newton's second law of motion, states that the force which is equal to the rate of change of momentum. For a constant mass, force is equal to mass times acceleration.
F = ma

Here, F is the force, m is the mass and a is the acceleration of the given body.

CALCULATION:

It is given that the damping force of the particle is proportional to the square of the distance and it is written as,
Question 2
4 / -1

Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.

Assertion(A) :

The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies.

Reason(R) :

The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal.

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true and R is NOT the correct explanation of A

C
A is true but R is false

D
Reason(R) :
Solution
Concept:

Photoelectric Effect: The phenomenon in which the electrons from the metal surface are ejected when the electromagnetic radiation is incident on it.

Characteristics of Photoelectric Effect :
- The threshold frequency varies with material, it is different for different materials.
- The photoelectric current is directly proportional to the light intensity.
- The kinetic energy of the photoelectrons is directly proportional to the light frequency.
- The stopping potential is directly proportional to the frequency and the process is instantaneous.
Explanation :
- Initially, the monochromatic light beam is incident on the metal surface.
- The electrons absorb all the energy from the photon.
- If this monochromatic light has frequency greater than that of the threshold frequency of the metal then the electrons get ejected out of the metals.
- Electrons are at different level inside the metal, so different electrons come out with different velocities.
- The deeper the electron in the material, the more it will lose its energy while coming out due to the collision with other atoms in the metal.
- Thus, there will be a spread in the kinetic energy of the photoelectrons as they come out with different velocities from different levels of the metal.
Hence, both the assertion and reason are true and reason is the correct explanation of the assertion.

Hence, the correct option is (1)
Question 3
4 / -1

A rod of length L revolves in a horizontal plane about the axis passing through its center and perpendicular to its length. The angular velocity of the rod is ω. If A is the area of cross-section of the rod and ρ is its density, then the rotational kinetic energy of the rod is

A

B

C

D

Solution

Concept:

Rotational energy or angular kinetic energy is kinetic energy due to the rotation of an object and is part of its total kinetic energy.
Question 4
4 / -1

Which of the following is not correct about the first law of thermodynamics:

A
First law does not explain about the direction of flow of heat

B
First law can't explain that the heat doesn't convert completely into work

C
The first law can explain whether the process is a spontaneous process or not

D
None of these
Solution
CONCEPT:

The first law of thermodynamics
- This law states that the total heat given or taken out of the system will always be equal to the work done and change in the internal energy of the system.
⇒ Q = W + ΔU

Limitations of the first law of thermodynamics:
1. The limitation of the first law of thermodynamics is that it does not say anything about the direction of the flow of heat.
2. It does not say anything whether the process is a spontaneous process or not.
3. It can't explain that why the reverse process is not possible. In actual practice, the heat doesn’t convert completely into work.
EXPLANATION:
- From the above, it is clear that the first law can't explain whether the process is a spontaneous process or not. Hence, option 3 is correct.
Question 5
4 / -1

Two particles of equal mass m have respective initial velocities They collide completely inelastically. The energy lost in the process is :

A

B

C

D

Solution

Concept:

Inelastic Collision is a collision in which kinetic energy is not conserved due to the action of internal friction.

The energy loss linked to the system is given by

Δ E = E_i - E_f

where E_i is the initial energy and E_f is the final energy.

Solution:

Let the two masses be m_A and m_B, and their initial velocity be u_A and u_Brespectively.

Now, according to the problem:

m_A = m_B = m

u_A = ui
Question 6
4 / -1

A particle of mass 'M' moves along a straight line and collides with a stationary point mass 'm' such that they follow the same straight-line path. The collision is elastic and the ratio of the final velocities of M and m is x. If m = 0.5M, the value of x is -

A
0.25

B
0.50

C
1.20

D
1
Solution
The correct answer is option 1) i.e. 0.25

CONCEPT:
- Elastic collision: Elastic collision is a phenomenon where the collision of objects takes place such that the total linear momentum and kinetic energy of the system are conserved.
- Collisions in one dimension:
Let m₁ and m₂ be the masses of two objects that undergo elastic collision.

From the principle of momentum conservation,

⇒ m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

where m₁, m₂ are the masses of the colliding bodies, v_1i, v_2i are the initial velocity and v_1f and v_2f are their final velocities.

From the principle of kinetic energy conservation,

On solving (1) and (2),
Question 7
4 / -1

One end of a metal wire is fixed at the centre of a uniform disc of radius 4.0 cm and mass 100 g. The other end of the wire is fixed with a clamp. The hanging disc is rotated about the wire through a small angle and is released. If the disc makes torsional oscillations with time period 0.20 s, the torsional constant of wire is (Given π² = 10)

A
4 × 10^-2 kg m²s^-2

B
8 × 10^-2 kg m²s^-2

C
1.2 × 10^-2 kg m²s^-2

D
8 × 10^-1 kg m²s^-2

Solution

Here we have " I" as the moment of inertia, and C is the torsional constant.

CALCULATION:

Given: Mass, m = 100 g

radius. r = 4.0 cm

time period, T = 0.20 s

The moment of inertia of a disc is written as;
Question 8
4 / -1

The equation of motion of a particle is x = a cos (α t)².

The motion is

A
periodic but not oscillatory.

B
periodic and oscillatory.

C
oscillatory but not periodic.

D
neither periodic nor oscillatory.

Solution

Concept:

Periodic motion:

→A motion that repeats itself after an equal interval of time is known as periodic motion. For example, A minute hand of a clock returning to its original state every 60 minutes is a great example of periodic motion.
→Similarly, the function which repeats its value after a fixed interval of time is termed a periodic function.
→Oscillation is going back and forth repeatedly between two positions or states. Oscillation can be a periodic or aperiodic motion (irregular).

Simple harmonic motion:

Simple harmonic motion (S.H.M) is a type of periodic oscillation where the restoring force is directly proportional to the displacement.

→The motion of a simple pendulum is a good example of simple harmonic motion in which the bob oscillates freely along with the lowest and highest position under the influence of gravity.

Explanation:

Given:

The equation of motion of a particle is

x = a cos (α t)²

= a cos α² t²

The equation of motion involves the cosine function and t².

∴ the motion is oscillatory but not periodic.

Hence, option (3) is the correct answer.
Question 9
4 / -1

Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively.

If T_A and T_B are the time periods of A and B respectively then the value of T_B - T_A :

[Given : radius of earth = 6400 km, mass of earth = 6 × 10²⁴ kg]

A
1.33 × 10³ s

B
4.24 × 10³ s

C
4.24 × 10² s

D
3.33 × 10² s

Solution

CONCEPT:

According to Kepler's law, the square of the time period of the satellites is proportional to the cube of the radius and it is written as;

Here T is the time period, r is the radius, the Gravitational constant and M is the mass of the earth.

CALCULATIONS:

Given: Mass of the satellite A = 200 kg

Mass of the satellite B = 400 kg

Height of satellite A = 600 Km

Height of the satellite B = 1600 Km

The radius of satellite r_A = 6400 + 600

= 7000 Km = 7000 × 10³ m

and the radius of satellite r_B = 6400 + 1600

= 8000 Km = 8000 ×10³ m

Now, by using equation (1) we have;

⇒ T_B - T_A = 1.33 × 10³ s

Hence, option 1) is the correct answer.
Question 10
4 / -1

In an experiment on the specific heat of a metal, a 0.30 kg block of the metal at 120 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 100 cm³ of water at 27 °C. The final temperature is 35 °C. Calculate the specific heat of the metal?

A
0.164 Jg^-1 K^-1

B
0.434 Jg^-1 K^-1

C
0.282 Jg^-1 K^-1

D
0.636 Jg^-1 K^-1
Solution
Concept:
- In this question heat lost by the metal will be responsible for the rise in temperature of the calorimeter water.
- Thermal energy stored in a body can be calculated as:
Q = mc_pdT

Where m is the mass of the body, c_p is the heat capacity of the body at a constant temperature, dT is the temperature difference between the environment and the temperature of the body.
- But remember this heat totally cannot be extracted for work only the useful work obtained by Gibbs free energy.
Calculation:

Given: Mass of water m_w = 0.025 kg = 25 g ,

Mass of metal M = 0.3 kg = 300g , T_i = 120^oC,

Final temperature, T_f = 35^oC,

Density of the water = 1 g/cm³

So, the mass of the water of 100 cm³ volume = 100 g

Specific heat of water, C_w = 4.186 Jg^-1 K

⇒ Mc_pT_w = (M+ m) × C_w× ( T_wi - T_f)

Heat lost by metal = Heat gained by the water and calorimeter system.

⇒ mCΔT_m = (M + m) × C_w × T_w

⇒ 300 × C × (120 - 35) = (100 + 25)× 4.186 × (35 -27)

⇒ C × 25500 = 10883.6

⇒ C = 0.164 J g^-1K^-1

Note: Here, the answer was not matching with the official options so we modified the option which was close to the correct answer.