Question 1. In a nuclear reaction, the β- particle is emitted
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Solution:
A β- particle is emitted when a neutron inside a nucleus is transformed into a proton.
Therefore, the correct option is (3).
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Question 1. In a nuclear reaction, the β- particle is emitted
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Solution:
A β- particle is emitted when a neutron inside a nucleus is transformed into a proton.
Therefore, the correct option is (3).
Question 2. What is the potential energy of the electron in an orbit of radius r in the hydrogen atom? (e = charge of electron)
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Solution:
For hydrogen, Z = 1.
Potential energy of the electron is U = -K(Ze)(e)/r.
Hence U = -Ke2/r.
Therefore, the correct option is (2).
Question 3. The binding energy of deuteron is 2.5 MeV and that of 4He is 26 MeV. If two deuterons are fused to form one 4He, then energy released in the process will be
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Solution:
Given reaction: 2H + 2H → 4He + Q.
Q = (B.E.)product - (B.E.)reactants.
Q = 26 - 2 × 2.5 = 26 - 5 = 21 MeV.
Therefore, the correct option is (3).
Question 4. The total energy of electron in the He+ atom in first excited state is -13.6 eV. The kinetic energy of this electron will be
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Solution:
For an electron in a Bohr orbit, kinetic energy is equal to the negative of total energy.
K.E. = -(-13.6 eV) = 13.6 eV.
Therefore, the correct option is (1).
Question 5. The electrical conductivity of pure silicon can be increased by
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Solution:
Conductivity of pure silicon increases when temperature is increased.
Doping with impurity atoms also increases the number of charge carriers.
Therefore, the correct option is (4).
Question 6. Match Column-I and Column-II.
| Column-I | Column-II |
|---|---|
| (i) Isotopes | (a) Sum of number of neutrons and protons |
| (ii) Isobars | (b) Same mass number but different atomic number |
| (iii) Isotones | (c) Same atomic number but different mass number |
| (iv) Nucleons | (d) Same number of neutrons but different atomic number |
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Solution:
Isotopes have the same atomic number but different mass numbers.
Isobars have the same mass number.
Isotones have the same number of neutrons.
Nucleons are the total protons and neutrons in a nucleus.
Therefore, the correct option is (3).
Question 7. Following diagram performs the logic function of
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Solution:
The first two gates act as NOT gates, giving Ā and B̄.
The final gate gives Y = Ā · B̄ = A + B by De Morgan theorem.
So the circuit performs OR operation.
Therefore, the correct option is (4).
Question 8. Which of the following statements is/are correct?
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Solution:
Pentavalent impurity in Ge produces an N-type semiconductor.
In an N-type semiconductor, majority carriers are electrons and minority carriers are holes.
Doped semiconductors remain electrically neutral overall, so N-type and P-type crystals are not negatively or positively charged as a whole.
Therefore, the correct option is (1).
Question 9. Which of the following is correct for Zener diode?
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Solution:
A Zener diode is heavily doped on both P and N sides.
It is designed to operate in reverse bias breakdown region and is used as a voltage regulator.
Therefore, the correct option is (3).
Question 10. In the given circuit, the current through the battery is (all diodes are ideal).
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Solution:
An ideal diode has zero resistance in forward bias and infinite resistance in reverse bias.
After replacing the ideal diodes by their conducting/open states, the equivalent resistance of the active network is 20 Ω.
I = V/R = 20/20 = 1 A.
Therefore, the correct option is (1).
Question 11. If kinetic energy of an electron in first orbit of hydrogen atom is K, then potential energy of an electron in first orbit of singly ionized helium atom is
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Solution:
For a Bohr orbit, potential energy is twice the total energy and equals -2 times kinetic energy.
Kinetic energy in the first orbit is proportional to Z2.
For He+, Z = 2, so kinetic energy becomes 4K.
Thus potential energy = -2 × 4K = -8K.
Therefore, the correct option is (3).
Question 12. The output current versus time curve of a half wave rectifier for sinusoidal input is shown in the figure. The average value of the output current in this case is
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Solution:
For a half-wave rectified sinusoidal current, only one half cycle contributes in each cycle.
The average output current is Iavg = i0/π.
Therefore, the correct option is (2).
Question 13. The shape of meniscus when contact angle is less than 90° is
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Solution:
When contact angle is less than 90°, the liquid wets the surface and the meniscus is concave upwards.
Therefore, the correct option is (2).
Question 14. Assertion (A): In general, heavier nuclides contain more number of neutrons than protons. Reason (R): To overcome the coulombic repulsion between protons, more number of neutrons are required. In the light of above statements, choose the correct option.
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Solution:
Heavier nuclides contain more protons, so neutron excess is needed to improve nuclear stability against proton-proton repulsion.
Thus both statements are true and the reason explains the assertion.
Therefore, the correct option is (1).
Question 15. Consider the following statements regarding nuclear force and select the correct statement(s).
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Solution:
Nuclear forces are nearly charge independent.
They are not always attractive; at very short distances they become strongly repulsive.
They are non-central forces.
Therefore, the correct option is (3).
Question 16. A nucleus of 92U238 originally at rest emits an α-particle with speed v0. The speed of daughter nucleus will be
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Solution:
Linear momentum is conserved because the parent nucleus was initially at rest.
92U238 → 90Th234 + 2He4.
0 = 4v0 - 234vd, so vd = 4v0/234 = 2v0/117.
Therefore, the correct option is (2).
Question 17. If the metre scale shown below is in rotational equilibrium, then the unknown weight W will be equal to
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Solution:
At rotational equilibrium, clockwise moment equals anticlockwise moment about the support.
60 × 10 = W × (80 - 50).
600 = 30W, so W = 20 N.
Therefore, the correct option is (3).
Question 18. A potential barrier of 0.60 V exists across P-N junction. If the depletion region is 3 × 10-5 m wide, then the intensity of the electric field in this region is
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Solution:
Use |ΔV| = Ed.
E = ΔV/d = 0.60/(3 × 10-5) = 0.2 × 105 = 2 × 104 V/m.
Therefore, the correct option is (2).
Question 19. The mass of a 11B nucleus is 0.077 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of 11B nucleus is (given, 1 u × c2 = 930 MeV)
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Solution:
Total binding energy = Δm(u) × 930 MeV.
B.E. = 0.077 × 930 MeV.
Binding energy per nucleon = (0.077 × 930)/11 = 6.51 MeV.
Therefore, the correct option is (4).
Question 20. When radiation of wavelength λ is incident on a metallic surface, the stopping potential is 3 volt. If the same surface is irradiated with radiation of double wavelength, then the stopping potential becomes 1 volt. The work function of the metal is
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Solution:
Einstein photoelectric equation: hc/λ = φ0 + eVs.
For wavelength λ: hc/λ = φ0 + 3 eV.
For wavelength 2λ: hc/(2λ) = φ0 + 1 eV.
Half of the first equation equals the second: (φ0 + 3)/2 = φ0 + 1, giving φ0 = 1 eV.
Therefore, the correct option is (4).
Question 21. A particle of mass 9 × 10-25 kg has the same de-Broglie wavelength as an electron moving with a velocity of 4 × 106 m s-1. The velocity of the particle is (given, mass of electron = 9 × 10-31 kg)
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Solution:
For equal de-Broglie wavelength, momenta are equal: mpvp = meve.
vp = (9 × 10-31 × 4 × 106)/(9 × 10-25) = 4 m s-1.
Therefore, the correct option is (2).
Question 22. In a photoelectric experiment, the frequency and photon intensity of the light source are both quadrupled. Consider the following statements.
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Solution:
Saturation photocurrent is proportional to photon intensity, so quadrupling intensity changes it.
Maximum kinetic energy follows Kmax = hf - φ; quadrupling frequency does not generally make Kmax exactly four times.
Therefore, the correct option is (4).
Question 23. In a hypothetical atom, if transition from n = 5 to n = 3 produces visible light then the possible transition to obtain infrared radiation is
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Solution:
Infrared radiation has less energy than visible radiation.
The transition from n = 6 to n = 4 has smaller energy separation than n = 5 to n = 3.
Therefore, the correct option is (4).
Question 24. For an equilateral triangular prism, if angle of minimum deviation is δm, then refractive index μ is equal to
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Solution:
For a prism, μ = sin[(A + δm)/2] / sin(A/2).
For an equilateral prism, A = 60°.
μ = sin(30° + δm/2)/sin 30° = 2 sin(30° + δm/2).
Therefore, the correct option is (3).
Question 25. The ratio of radii of first shell of H-atom to that of third shell of Li++ ion is
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Solution:
Radius of Bohr orbit is proportional to n2/Z.
For H first shell: n = 1, Z = 1, so radius factor is 1.
For Li++ third shell: n = 3, Z = 3, so radius factor is 9/3 = 3.
Ratio = 1 : 3.
Therefore, the correct option is (3).
Question 26. An α-particle is bombarded on 7N14 nucleus and a proton is emitted. The product nucleus is
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Solution:
An α-particle is 2He4.
2He4 + 7N14 → 8O17 + 1H1.
Therefore, the correct option is (3).
Question 27. The speed of electron in second stable orbit of hydrogen atom is (c is speed of light in vacuum)
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Solution:
Speed of electron in the nth Bohr orbit is v = Zc/(137n).
For hydrogen, Z = 1 and for the second orbit n = 2.
v = c/(137 × 2) = c/274.
Therefore, the correct option is (2).
Question 28. Consider the following statements A and B and identify the correct answer.
A. Fission is the source of energy of all stars including our sun.
B. In fusion, lighter nuclei combine to form a larger nucleus.
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Solution:
Fusion of hydrogen nuclei into helium nuclei is the source of energy of stars including the Sun.
Fusion means lighter nuclei combine to form a heavier nucleus.
Thus A is incorrect and B is correct.
Therefore, the correct option is (2).
Question 29. 1 cm on the main scale of vernier callipers is divided into 10 equal parts. If 20 divisions of vernier scale coincide with 18 divisions of main scale, then least count of callipers is
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Solution:
Least count = 1 MSD - 1 VSD.
Given 20 VSD = 18 MSD, so 1 VSD = 0.9 MSD.
L.C. = 1 MSD - 0.9 MSD = 0.1 MSD.
Since 1 MSD = 0.1 cm, L.C. = 0.01 cm.
Therefore, the correct option is (3).
Question 30. The energy required to remove an electron from fourth orbit of hydrogen atom is
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Solution:
Energy of electron in nth orbit is En = -13.6Z2/n2 eV.
For hydrogen in fourth orbit, Z = 1, n = 4.
E4 = -13.6/16 = -0.85 eV.
Energy required to remove it is 0.85 eV.
Therefore, the correct option is (4).
Question 31. The energy of emitted photon when an electron jumps from third orbit to first orbit in a hydrogen atom is
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Solution:
Photon energy for transition to lower orbit is E = 13.6(1/n12 - 1/n22) eV.
For n2 = 3 to n1 = 1: E = 13.6(1 - 1/9) = 13.6 × 8/9 = 12.08 eV.
Therefore, the correct option is (1).
Question 32. The photoelectrons emitted from the surface of metal are such that they
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Solution:
Due to energy loss inside the metal, emitted photoelectrons come out with different kinetic energies.
Their speeds range from zero up to a maximum value fixed by the incident photon energy.
Therefore, the correct option is (4).
Question 33. An external voltage V is applied across a diode such that it is reverse biased. As the reverse bias voltage is gradually increased before it reaches breakdown voltage, then current
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Solution:
Before breakdown, reverse current is approximately the reverse saturation current and is nearly independent of reverse voltage.
At breakdown voltage it increases sharply, but that is beyond the condition asked.
Therefore, the correct option is (4).
Question 34. In an experiment to determine the resistance of galvanometer by half-deflection method, we use the following circuit where R = 600 Ω and G = 3 Ω. Then the value of resistance S is (where symbols have their usual meaning)
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Solution:
In half-deflection method, galvanometer resistance is G = RS/(R - S).
Putting G = 3 Ω and R = 600 Ω: 3 = 600S/(600 - S).
1800 - 3S = 600S, so S = 600/201 Ω.
Therefore, the correct option is (1).
Question 35. The ratio of velocities of deuteron and α-particle, if the de-Broglie wavelength of both the particles is same, will be
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Solution:
For same de-Broglie wavelength, mv is same.
mdvd = mαvα.
Since md ≈ 2mp and mα ≈ 4mp, vd/vα = 4/2 = 2.
Therefore, the correct option is (1).
Question 36. In a hypothetical fission reaction 92X236 → 56Y141 + 35Z93 + 0n1 + R, the identity of emitted particle R is
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Solution:
Let R = xRy.
Atomic number conservation: 92 = 56 + 35 + 0 + x, so x = 1.
Mass number conservation: 236 = 141 + 93 + 1 + y, so y = 1.
Thus R = 1H1, a proton.
Therefore, the correct option is (1).
Question 37. A beam of light consists of two wavelengths λ and 2λ, and the number of photons passing in unit time for the two wavelengths are 2n and 3n respectively. If the cross-sectional area of the beam is A, then intensity of the light beam will be
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Solution:
Intensity is energy crossing per unit area per unit time.
For wavelength λ: I1 = 2n(hc/λ)/A = 2nhc/(λA).
For wavelength 2λ: I2 = 3n(hc/2λ)/A = 3nhc/(2λA).
I = I1 + I2 = 7nhc/(2λA).
Therefore, the correct option is (2).
Question 38. A metal plate is irradiated with a certain wavelength of radiation. If all the electrons ejected from the metal plate can be stopped before travelling 2 m in the direction of uniform electric field of 3 N/C, then the possible value(s) of kinetic energy of the ejected electrons is/are
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Solution:
Maximum kinetic energy stopped by the field equals work done by the field.
Kmax = eEd = e × 3 × 2 = 6 eV.
Possible kinetic energies can be less than or equal to 6 eV, so 1 eV and 2 eV are possible, but 7 eV is not.
Therefore, the correct option is (4).
Question 39. Consider the following statements and choose the correct statement(s):
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Solution:
By Einstein equation, eVs = hf - φ0, so stopping potential increases with frequency.
Saturation photocurrent is proportional to light intensity.
Work function depends on the material and does not depend on incident frequency.
Therefore, the correct option is (3).
Question 40. The time of revolution of an electron around a nucleus of atomic number Z in nth Bohr’s orbit is directly proportional to
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Solution:
Time period T = 2πr/v.
For Bohr orbit, r ∝ n2/Z and v ∝ Z/n.
Therefore T ∝ (n2/Z)/(Z/n) = n3/Z2.
Therefore, the correct option is (2).
Question 41. The radius of nuclei X is measured to be twice the radius of nuclei Y. The ratio of nucleons in X to Y is
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Solution:
Nuclear radius R = R0A1/3.
Given RX = 2RY, so (AX/AY)1/3 = 2.
Hence AX/AY = 8.
Therefore, the correct option is (1).
Question 42. The ratio of the shortest wavelength of Paschen series to the shortest wavelength of Lyman series for hydrogen atom is
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Solution:
For shortest wavelength, final level is fixed and initial level is infinity.
Paschen series: n1 = 3, so 1/λP = R/9.
Lyman series: n1 = 1, so 1/λL = R.
Thus λP : λL = 9 : 1.
Therefore, the correct option is (2).
Question 43. The acceptor level of a p-type semiconductor is 5 eV. The maximum wavelength of light which can create a hole would be (given, hc = 12400 eV Å)
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Solution:
Use ΔE = hc/λ.
Here ΔE = 5 eV, so λmax = 12400/5 = 2480 Å.
Therefore, the correct option is (1).
Question 44. A screw gauge has 50 equal divisions marked on its circular scale and one full rotation of the circular scale advances the main scale by 0.02 cm. The least count of this screw gauge is
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Solution:
Least count = pitch / total number of circular scale divisions.
Pitch = 0.02 cm.
L.C. = 0.02/50 = 0.0004 cm = 4 × 10-4 cm.
Therefore, the correct option is (2).
Question 45. In a resonance tube experiment, first resonating length is 15 cm and second resonating length is 47 cm, then wavelength of the wave is
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Solution:
For resonance tube, difference between second and first resonating lengths is λ/2.
λ/2 = 47 - 15 = 32 cm.
λ = 64 cm.
Therefore, the correct option is (3).
Question 46. Hydrogen gas sample in ground state is given an energy of 12.75 eV. How many spectral lines will be emitted due to transition of electrons?
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Solution:
Energy 12.75 eV excites hydrogen from n = 1 to n = 4.
Number of spectral lines emitted from level n is n(n - 1)/2.
N = 4(4 - 1)/2 = 6.
Therefore, the correct option is (3).
Question 47. In the circuit shown, if the potential drop in forward bias condition across Si and Ge diodes are 0.7 V and 0.3 V respectively, then the potential difference across 2 kΩ resistor is
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Solution:
Total diode drop = 0.3 V + 0.7 V = 1.0 V.
Voltage across resistors = 20 - 1 = 19 V.
Total resistance = 2 kΩ + 3 kΩ = 5 kΩ, so i = 19/5 mA = 3.8 mA.
Across 2 kΩ: V = iR = 3.8 × 10-3 × 2 × 103 = 7.6 V.
Therefore, the correct option is (2).
Question 48. If M0 is the mass of nucleus 12C, Mp and Mn are the masses of proton and neutron, then nuclear binding energy of nucleus 12C is
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Solution:
For 12C, there are 6 protons and 6 neutrons.
Mass defect Δm = 6Mp + 6Mn - M0.
Binding energy = Δmc2 = (6Mp + 6Mn - M0)c2.
Therefore, the correct option is (1).
Question 49. A proton, a deuteron and an electron have the same kinetic energies. Their de-Broglie wavelengths will be compared as
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Solution:
For same kinetic energy, λ = h/√(2mK), so λ ∝ 1/√m.
Mass order is md > mp > me.
Therefore wavelength order is λe > λp > λd.
Therefore, the correct option is (4).
Question 50. The de-Broglie wavelength of a particle accelerating with 130 V potential is 1 Å. If the accelerating potential is increased to 520 V, then its de-Broglie wavelength will be
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Solution:
For a particle accelerated through potential V, λ = h/√(2mqV), so λ ∝ 1/√V.
Here the potential increases from 130 V to 520 V, i.e. by a factor of 4.
Thus wavelength becomes 1/√4 = 1/2 times the original.
λ = 0.5 Å.
Therefore, the correct option is (3).
For one-to-one online Physics classes, contact Kumar Sir.