"In-Depth NEET Physics Review: 20 Critical Questions on Kinematics, Dynamics, Energy, Waves, and Optics to Enhance Knowledge and Boost Confidence for NEET Examination Success"

"NEET Physics Essentials: 20 Questions to Sharpen Your Skills"

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1. Radiation of wavelength 280 nm is used in an experiment of photoelectric effect with cathode of work function, 2.5 eV. The maximum kinetic energy of the photoelectrons is:

A. 0.5 eV

B. 1.93 eV

C. 2.5 eV

D. 3.0 eV

The correct answer is 1.93 eV. Using the photoelectric equation \( K.E_{max} = h\nu - \phi \), we first convert the wavelength to frequency and calculate the photon energy. The energy of the photon is 4.43 eV, and after subtracting the work function (2.5 eV), the maximum kinetic energy of the photoelectrons is 1.93 eV.

2. An electron and a photon possess the same de Broglie wavelength. If \( E_e \) is the kinetic energy of the electron and \( E_p \) is the energy of the photon, (Given speed of electron is \( v \)), then:

A. \( E_e = E_p \)

B. \( E_e > E_p \)

C. \( E_e < E_p \)

D. \( E_p = \frac{hc}{\lambda}, E_e = \frac{p^2}{2m} \) and \( E_p > E_e \)

The correct answer is \( E_p = \frac{hc}{\lambda} \) and \( E_e = \frac{p^2}{2m} \). Since the photon is massless, its energy \( E_p \) is greater than the kinetic energy \( E_e \) of the electron for the same wavelength.

3. A nucleus of mass \( M \) at rest emits a photon of frequency \( \nu \). Find the recoil energy of the nucleus.

A. \( \frac{h\nu}{M} \)

B. \( \frac{h\nu^2}{c^2} \)

C. \( \frac{h^2\nu^2}{2Mc^2} \)

D. \( \frac{h\nu^2}{M} \)

The correct answer is \( \frac{h^2\nu^2}{2Mc^2} \). When the photon is emitted, the nucleus recoils to conserve momentum. The energy of the photon is \( E = h\nu \), and by using conservation of momentum, we can derive the recoil energy as \( E_{\text{recoil}} = \frac{h^2\nu^2}{2Mc^2} \).

4. A bulb lamp emits light of mean wavelength 4500 Å. The lamp is rated at 150 W and 8% of the energy appears as emitted light. How many photons are emitted by the lamp per second?

A. \( 1.2 \times 10^{19} \)

B. \( 2.0 \times 10^{19} \)

C. \( 2.71 \times 10^{19} \)

D. \( 3.5 \times 10^{19} \)

The correct answer is \( 2.71 \times 10^{19} \) photons/second. The effective power for light emission is \( 12 \text{ W} \), and the energy of a single photon at 4500 Å is approximately \( 4.42 \times 10^{-19} \text{ J} \).

5. If 200 MeV energy is released in the fission of a single U nucleus, the number of fissions required per second to produce 1 kW power shall be:

A. \( 1.5 \times 10^{19} \)

B. \( 2.0 \times 10^{19} \)

C. \( 3.125 \times 10^{13} \)

D. \( 4.0 \times 10^{19} \)

The correct answer is \( 3.125 \times 10^{19} \) fissions/second. The energy released per fission is approximately \( 3.2 \times 10^{-11} \text{ J} \), and to produce 1 kW of power, approximately \( 3.125 \times 10^{13} \) fissions are needed per second.

6. The work done by the external forces on a system equals the change in:

A. total mechanical energy

B. kinetic energy

C. potential energy

D. none of these

The correct answer is kinetic energy. According to the work-energy theorem, the work done by external forces equals the change in kinetic energy of the system.

7. The potential energy for a force field \( F \) is given by \( U(x, y) = \sin(x + y) \). The magnitude of the force acting on a particle of mass \( m \) is:

A. \( 2 \, \text{N} \)

B. \( \sqrt{2} \, \text{N} \)

C. \( 1 \, \text{N} \)

D. none of these

The magnitude of the force \( F \) is given by \( F = -\nabla U \). Calculating the gradient, we find \( F_x = -\frac{\partial U}{\partial x} = -\cos(x + y) \) and \( F_y = -\frac{\partial U}{\partial y} = -\cos(x + y) \). Therefore, the magnitude is \( |F| = \sqrt{F_x^2 + F_y^2} = \sqrt{2 \cos^2(x + y)} = \sqrt{2} \).

8. A block of mass 5 kg is suspended by a light thread from an elevator. The elevator, initially at rest, is accelerating upward with a uniform acceleration of 2 m/s². Find the work done by tension on the block during the first 3 seconds.

A. \( 200 \, \text{J} \)

B. \( 531 \, \text{J} \)

C. \( 150 \, \text{J} \)

D. \( 400 \, \text{J} \)

The correct answer is \( 531 \, \text{J} \). The tension in the thread is \( 59 \, \text{N} \), and the displacement of the block during the first 3 seconds is \( 9 \, \text{m} \), resulting in work done by tension \( W = T \cdot s = 59 \times 9 = 531 \, \text{J} \).

9. A force is conservative, then correct statements are:- (i) work is path independent (ii) it must be central (iii) work done round a closed path is zero (iv) potential energy remains constant

A. (i, ii)

B. (ii, iii)

C. (i, iii)

D. (i, ii, iv)

The correct statements are (i) work is path independent and (iii) work done round a closed path is zero. Statement (ii) is false because not all conservative forces are central. Statement (iv) is also false as potential energy can change.

10. A string of mass \( m \) and length \( \ell \) rests over a frictionless table with \( \frac{1}{4} \) of its length hanging from a side. The work done in bringing the hanging part back on the table is:-

A. \( \frac{m g \ell}{16} \)

B. \( \frac{m g \ell}{8} \)

C. \( \frac{m g \ell}{32} \)

D. None of these

The work done in bringing the hanging part back onto the table is given by \( W = \frac{m g \ell}{32} \). This is calculated based on the mass of the hanging part and the distance it needs to be lifted.

11. A monkey of mass \( 20 \, \text{kg} \) is holding a vertical rope. The rope will not break when a mass of \( 25 \, \text{kg} \) is suspended from it but will break if the mass exceeds \( 25 \, \text{kg} \). What is the maximum acceleration with which the monkey can climb up along the rope? (Given \( g = 10 \, \text{m/s}^2 \))

A. \( 5 \, \text{m/s}^2 \)

B. \( 10 \, \text{m/s}^2 \)

C. \( 25 \, \text{m/s}^2 \)

D. \( 2.5 \, \text{m/s}^2 \)

The maximum acceleration with which the monkey can climb up the rope is \( 2.5 \, \text{m/s}^2 \). This is derived from the maximum tension the rope can support and the weight of the monkey.

12. A block of mass \( m \) is lying on an inclined plane at an angle \( \theta \) with the horizontal. The coefficient of friction between the plane and the block is \( \mu \). The minimum force \( F \) required to move the block up the inclined plane will be:

A. \( mg \sin \theta + \mu mg \cos \theta \)

B. \( mg \cos \theta - \mu mg \sin \theta \)

C. \( mg \sin \theta - \mu mg \cos \theta \)

D. \( mg \cos \theta + \mu mg \sin \theta \)

The minimum force required is the sum of the component of gravitational force along the plane, \( mg \sin \theta \), and the frictional force, \( \mu mg \cos \theta \). Therefore, the correct expression is \( F = mg \sin \theta + \mu mg \cos \theta \).

13. Find the maximum force \( F_{\text{max}} \) so that no sliding occurs between the blocks. Given \( \mu = 0.5 \), the mass of the upper block is 2 kg, and the mass of the lower block is 4 kg. The surface is smooth under the lower block.

A. 29.4 N

B. 19.6 N

C. 9.8 N

D. 39.2 N

The maximum frictional force between the two blocks is \( f_{\text{max}} = \mu N = 0.5 \times 19.6 = 9.8 \, \text{N} \). The acceleration of the upper block is \( a = 9.8 / 2 = 4.9 \, \text{m/s}^2 \). The total mass of the system is 6 kg, so the maximum applied force is \( F_{\text{max}} = 6 \times 4.9 = 29.4 \, \text{N} \).

14. If all conditions are ideal, what is the ratio \( \frac{m_1}{m_2} \) such that the block will not move? Given \( AO = 3 \, \text{m} \), \( BO = 4 \, \text{m} \), \( AB = 5 \, \text{m} \).

A. \( \frac{3}{4} \)

B. \( \frac{4}{3} \)

C. \( \frac{5}{4} \)

D. \( \frac{2}{3} \)

The system is in equilibrium when the forces along both inclines balance each other. The force pulling the block along the incline AO is \( m_1 g \sin \alpha \), where \( \sin \alpha = \frac{4}{5} \), and the force along BO is \( m_2 g \sin \beta \), where \( \sin \beta = \frac{3}{5} \). By setting the forces equal, we find \( \frac{m_1}{m_2} = \frac{3}{4} \).

15. A force acts on a block as shown in the figure. The force is given by \( F = 10t \, \text{N} \) and is applied at an angle of \( 37^\circ \) to the horizontal. The block has a mass of \( 10 \, \text{kg} \), and the coefficient of friction is \( \mu = 0.4 \). Find the time when the block loses contact with the surface.

A. 16.33 s

B. 12.5 s

C. 20.5 s

D. 14.5 s

The block will lose contact with the surface when the normal force \( N \) becomes zero. The normal force is given by \( N = mg - F_{\text{vertical}} \), where \( F_{\text{vertical}} = F \sin 37^\circ = 6t \, \text{N} \). Setting \( N = 0 \) and solving for time, we get \( t = 16.33 \, \text{seconds} \).

16. When a horse pulls a wagon, the force that causes the horse to move forward is the force:

A. He exerts on the wagon

B. The wagon exerts on him

C. The ground exerts on him

D. He exerts on the ground

The correct answer is C. When the horse pulls the wagon, it pushes backward against the ground. The ground responds by exerting an equal and opposite force that propels the horse (and the wagon) forward, according to Newton's third law of motion.

17. If the de-Broglie wavelengths for a proton and for an α-particle are equal, then the ratio of their velocities will be:

A. 4 : 1

B. 2 : 1

C. 1 : 2

D. 1 : 4

The correct answer is A. Since the de-Broglie wavelengths are equal, we can derive that the ratio of the velocities \( v_p : v_{\alpha} = 4 : 1 \), given that the mass of the α-particle is approximately four times that of the proton.

18. For the Bohr's first orbit of circumference \(2\pi r\), the de-Broglie wavelength of the revolving electron will be:

A. \(2\pi r\)

B. \(\frac{h}{mv}\)

C. \(\frac{2\pi r}{n}\)

D. \(h\)

The correct answer is A. According to Bohr's model, the de-Broglie wavelength of the electron in the first orbit is equal to the circumference of the orbit, which is \(2\pi r\).

19. When a spring is stretched by 3 cm, it stores 150 J of energy. If it is stretched further by 3 cm, the stored energy will be increased by:

A. 200 J

B. 450 J

C. 600 J

D. 150 J

The correct answer is B. The increase in stored energy when the spring is stretched from 3 cm to 6 cm is 450 J, as calculated using the potential energy formula for springs.

20. Ten blocks of 1 kg are placed one above another. The reaction force exerted by the 6th block from the bottom on the 7th block is (g = 10 m/s²):

A. 30 N

B. 40 N

C. 50 N

D. 60 N

The correct answer is B. The reaction force exerted by the 6th block on the 7th block is 40 N, as it must support the weight of the 7th, 8th, 9th, and 10th blocks.

21. S is a spring balance. Find the reading of the spring balance if two masses of 2 kg and 4 kg are connected over a pulley (g = 10 m/s²):

A. 8/3 kgf

B. 3/8 kgf

C. 3 kgf

D. 8 kgf

The correct answer is A. Let's calculate the tension in the string, which is the reading of the spring balance. The forces acting on the system are the gravitational forces on the 4 kg and 2 kg masses:

- For the 4 kg mass: Force (downward) = 4g = 40 N
- For the 2 kg mass: Force (downward) = 2g = 20 N

The net force causing acceleration is:

F_net = (4g - 2g) = 20 N

The total mass of the system is:

M_total = 4 kg + 2 kg = 6 kg

The acceleration \( a \) of the system is given by Newton's second law:
\( a = \frac{F_{\text{net}}}{M} = \frac{20}{6} = \frac{10}{3} \, \text{m/s}^2 \)

Now, we calculate the tension in the string, which is the force the spring balance reads. For the 2 kg mass, the equation is:
\( T = 2g - 2a \)
Substituting \( a = \frac{10}{3} \, \text{m/s}^2 \):
\( T = 20 - 2 \times \frac{10}{3} = 20 - \frac{20}{3} = \frac{60}{3} - \frac{20}{3} = \frac{40}{3} \, \text{N} \)

Converting the tension to kgf (kilogram-force):
\( T = \frac{40}{3g} = \frac{40}{3 \times 10} = \frac{8}{3} \, \text{kgf} \)

Therefore, the reading of the spring balance is \( \frac{8}{3} \, \text{kgf} \).

22. For the given system, when the string is cut, what will be the accelerations of blocks A and B? (Assume both blocks have mass \( m \) and the spring is ideal):

A. \( g \) and \( g \)

B. \( 2g \) and \( 0 \)

C. \( g/2 \) and \( g/2 \)

D. \( 3g \) and \( g/2 \)

The correct answer is B. Let's analyze the system:

When the string is cut, block A is no longer supported and accelerates downward. Since it is attached to block B via the spring, the spring initially stretches but does not immediately affect block B's motion. Thus, block A is in free fall with an acceleration of \( 2g \) (since it is pulling block B downward too), while block B remains stationary for a moment (acceleration = 0) due to the spring.



- For block A: Block A falls freely under gravity, accelerating at \( g \). Additionally, because it pulls block B via the spring, it experiences another \( g \), resulting in a total acceleration of \( 2g \).

- **For block B**: Block B is connected to block A by the spring. Initially, block B does not move, as the spring stretches and has not yet exerted a significant force on it. Hence, its acceleration is zero at the moment the string is cut.

Thus, the accelerations are:
Block A: \( a_A = 2g \)
Block B: \( a_B = 0 \)

23. Statement-1: A person walking on a horizontal road with a load on his head does no work on the load against gravity.
Statement-2: No work is said to be done if directions of force and displacement of load are perpendicular to each other.

A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT explanation for Statement-1

C. Statement-1 is True, Statement-2 is False

D. Statement-1 is False, Statement-2 is True

The correct answer is A.

- **Statement-1** is true because when the person walks horizontally, no work is done on the load against gravity, as there is no displacement in the vertical direction (the direction of the force).
- **Statement-2** is also true because work is defined as \( W = F \cdot d \cdot \cos\theta \), and when the force and displacement are perpendicular (\( \theta = 90^\circ \)), the work done becomes zero due to \( \cos(90^\circ) = 0 \).
Thus, Statement-2 correctly explains Statement-1.

24. The mass defect for the nucleus of helium is 0.0303 amu. What is the binding energy per nucleon for helium in MeV?

A. 28

B. 7

C. 4

D. 1

The correct answer is B.

The total binding energy is calculated using the mass defect: \[ E_b = \Delta m \times 931.5 \, \text{MeV} \] Substituting the given mass defect of 0.0303 amu: \[ E_b = 0.0303 \times 931.5 = 28.22 \, \text{MeV} \] Since helium has 4 nucleons, the binding energy per nucleon is: \[ E_b \, \text{per nucleon} = \frac{28.22}{4} = 7.06 \, \text{MeV} \] Therefore, the binding energy per nucleon for helium is approximately 7 MeV.

25. Statement-1: Graph between potential energy of a spring versus the extension or compression of the spring is a parabola.
Statement-2: Potential energy of a stretched or compressed spring is proportional to the square of extension or compression. Select the correct option:

A. Statement-1 is True; Statement-2 is True; Statement-2 is NOT explanation for Statement-1

B. Statement-1 is True; Statement-2 is True; Statement-2 is a correct explanation for Statement-1

C. Statement-1 is True, Statement-2 is False

D. Statement-1 is False, Statement-2 is True

The correct answer is B.

The graph of potential energy (PE) of a spring against extension (x) or compression is a parabola, represented by the formula: \[ PE = \frac{1}{2} k x^2 \] This shows that the potential energy is proportional to the square of the extension or compression. Therefore, Statement-2 correctly explains Statement-1.

26. Assertion: On stationary body acting friction force is always known as limiting friction.
Reason: Kinetic friction is more than that of limiting friction.
Select the correct option:

A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

C. Assertion is true but Reason is false.

D. Both Assertion and Reason are false.

The correct answer is D.

The friction force acting on a stationary body is called limiting friction, which is false. There is no friction.
Kinetic friction is actually less than limiting friction, making the reason false. Actually it is greater than limiting friction.

27. Assertion: Mass is a measure of inertia of the body in linear motion.
Reason: Greater the mass, greater is the force required to change its state of rest or of uniform motion.
Select the correct option:

A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

C. Assertion is true but Reason is false.

D. Both Assertion and Reason are false.

The correct answer is A.

Mass measures inertia, indicating how much an object resists changes in its motion. A greater mass requires greater force to change its state, confirming that both the assertion and reason are true, and the reason correctly explains the assertion.

28. Nuclear Force is -

A. Central force

B. Non-Central force

C. Short range force

D. Both 2 and 3

The correct answer is D.

Nuclear forces are non-central, as they do not act along the line connecting the centers of nucleons, and they are short-range forces, effective only at distances on the order of 1 femtometer (10^-15 meters).

29. The binding energies per nucleon for a deuteron and an α-particle are x and x respectively. What will be the energy Q released in the reaction

A. 4(x + x)

B. 4(x - x)

C. 2(x + x)

D. 2(x - x)

The correct answer is B.

In this nuclear reaction, two hydrogen nuclei (protons) combine to form a helium nucleus (α-particle). The binding energy per nucleon for the α-particle is greater than that of the individual protons.

The total binding energy before the reaction (for two hydrogen nuclei) is 0, as they are not bound together. After the reaction, the binding energy for the helium nucleus (with 4 nucleons) is \(4x\). Therefore, the energy released \(Q\) can be calculated as:

\[ Q = \text{Total Binding Energy of products} - \text{Total Binding Energy of reactants} = 4x - 0 = 4x. \]
Hence, since we express this in terms of binding energies, the energy released is given by \(4(x - x)\).

30. Statement-1: The instantaneous power of an agent is measured as the dot product of instantaneous velocity and the force (only one force applied by agent) on it at that instant.
Statement-2: The unit of instantaneous power is watt-hour.
Select the correct option:

A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT an explanation for Statement-1.

C. Statement-1 is True, Statement-2 is False.

D. Statement-1 is False, Statement-2 is True.

E. Both statements are false.

The correct answer is C.

Statement-1 is true as instantaneous power is calculated using the dot product of velocity and force. Statement-2 is false because the unit of instantaneous power is watt (W), not watt-hour, which is a unit of energy.

31. Statement-1: Action and reaction forces are always found in pairs.
Statement-2: Action and reaction forces do not need any physical contact.
Select the correct option:

A. Statement-1 is true, Statement-2 is true.

B. Statement-1 is false, Statement-2 is true.

C. Statement-1 is true, Statement-2 is false.

D. Statement-1 is false, Statement-2 is false.

The correct answer is A.

Statement-1 is true as action and reaction forces always occur in pairs. Statement-2 is also true because these forces can act at a distance without physical contact, such as gravitational forces.

32. In a Wheatstone's bridge, three resistances P, Q, and R are connected in three arms and the fourth arm is formed by two resistances S and S connected in parallel. What is the condition for the bridge to be balanced?

A.   P/Q = 2R/S

B.   P/Q = R/S

C.   P/Q = R/2S

D.   P/Q = 2S/R

The correct answer is A.

In a Wheatstone bridge, when three resistances P, Q, and R are connected in three arms, and the fourth arm is made up of two equal resistances S and S connected in parallel, the bridge will be balanced when:

1. The equivalent resistance of the parallel combination of S and S is given by:
   \[ S_{eq} = \frac{S \cdot S}{S + S} = \frac{S}{2} \]

2. The balance condition for the Wheatstone bridge is expressed as:
   \[ \frac{P}{Q} = \frac{R}{S_{eq}} \]
Substituting \( S_{eq} \) in the equation gives:
   \[ \frac{P}{Q} = \frac{R}{\frac{S}{2}} \implies \frac{P}{Q} = \frac{2R}{S} \]
Therefore, the correct condition for the Wheatstone bridge to be balanced is:
   \[ \frac{P}{Q} = \frac{2R}{S} \]