JEE Main Previous Year Questions (2016- 2024): Modern Physics- 2 | Physics for Airmen Group X - Airforce X Y / Indian Navy SSR PDF Download

Q.37. A solution containing active cobalt 60/27 Co having activity of 0.8 µCi and decay constant λ is injected in an animal's body. If 1 cm³ of blood is drawn from the animal's body after 10 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? (1 Ci = 3.7 x 10¹⁰ decays per second and at t = 10 hrs e^–λt = 0.84)     [2018]
(1) 4 liters
(2) 5 liters
(3) 6 liters
(4) 7 liters

Ans: (2)
(dn/dt) = −λN_o
0.8 µ Ci = −λN_o
N₁ = N_o(0.84)
v → N₁
v cm³ → N₁/v


v = 3.7/5 × 0.84 × 0.84 × 10⁴ cm³
= 0.5 × 10⁴ cm³ = 5 × 10³ cm³ = 5 lit

Q.38. The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:      [2018]
(1) 109 eV
(2) 79 eV
(3) 20 eV
(4) 34 eV

Ans. (3)
E₁ = ionization energy of ionized He
E₂ = 2. 2E₂
E₁ = 13.6 eV
E₂ = 13.6/2.2 = 6.18 eV
Total = E₁ + E₂ = 13.6 ev + 6.18 eV
E_Total = 20 eV

Q.39. At some instant a radioactive sample S₁ having an activity 5 μCi has twice the number of nuclei as another sample S₂ which has an activity of 10 μCi. The half lives of S₁ and S₂ are:     [2018]
(1) 5 years and 20 years, respectively
(2) 20 years and 5 years, respectively
(3) 20 years and 10 years, respectively
(4) 10 years and 20 years, respectively

Ans: (1)
Given : N₁ = 2N

Hence,
The half life of S₁ is 5 years, and
The half life of S₂ is 20 years

Q.40. The de-Broglie wavelength (λ_B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state (λG) by:    [2018]
(1) λ_B = 3λ_G
(2) λ_B = 2λ_G
(3) λ_B = 3λ_G/3
(4) λ_B = 3λ_G/2

Ans: (1)

Q.41.  Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths λ_N,λ_A respectively. The ratio λ_N/λ_A  is closest to:     [2018]
(1) 10^–1
(2) 10^–6
(3) 10
(4) 10^–10

Ans: (2)

where Ea and EN are energies of photons from atom and nucleus respectively. EN is of the order of MeV and Ea in few eV.
So

Q.42. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λ_min is the smallest possible wavelength of X-ray in the spectrum, the variation of log λ_min with log V is correctly represented in    [2017]
(1)

(2)
(3)
(4)

Ans: (3)
In X-ray tube


Slope is negative
Intercept on y-axis is positive

Q.43. A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λ_A to λ_B after the collision is     [2017]
(1)
(2)
(3)
(4)

Ans: (4)

Q.44. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ₁/λ₂, is given by    [2017]

(1)
(2)
(3)
(4)

Ans: (2)
From energy level diagram

Q.45. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by    [2017]
(1) t =T log (1.3)
(2)
(3)
(4)

Ans: (4)

Q.46.  In a certain region static electric and magnetic fields exist. The magnetic field is given by  .  If a test charge moving with a velocity experiences no force in that region, then the electric field in the region, in SI units, is:    [2017]
(1)

(2)  
(3)
(4)  

Ans: (2)
F_e = F_n = 0 
F_e = –F_m

Ans: (3)
s = e (n_eμ_e + n_nμ_n)
= 1.6 × 10^-19 (5 × 10¹⁸ × 2 + 5 × 10¹⁹ × 0.01)
= 1.6 × 10^–19 (10¹⁹  + 0.05 × 10¹⁹)
= 1.6 × 1.05
= 1.68

Ans: (4)

=1/5 = 0.2
Frequency = 12 × 10³ kHz
F = 12.00 kHz
F₁ = 1200 – 20 = 1180 kHz
F₂ = 1200 + 20 = 1220 kHz

Ans: (1)

Ans: (2)
a = 0.1 mm = 10^–4 
λ = 6000 × 10^-10 
= 6 × 10^-7
D = 0.5 m
for 3^rd dark
a sin θ = 3λ

Q.51. Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be:    [2016]
(1)

(2)

(3)

(4)

Ans: A
.... (i)
 ...... (ii)
(ii) - (i) gives


.... (iv)
combining (iii) & (iv)

Ans: D
80 minutes = 4 half-lives of A = 2 half-lives of B
Let the initial number of nuclei in each sample be N
NA after 80 minutes = N/2⁴ 
⇒ Number of A nuclides decayed = 15/16N
NB after 80 minutes = N/2² 
⇒ Number of B nuclides decayed = 3/4N
Required ratio =

Ans: (1)
Energy of proton = 13.6 – 3.4 = 10.2 eV
For removal of electron

so minimum value of n = 4

Q.54. When photons of wavelength λ₁, are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength λ₂, are used , the corresponding stopping potential was thrice that of the above value. If light of wavelength λ₃. is used then find the stopping potential for this case     [2016]
(1)
(2)
(3)
(4)

Ans: (1)

using these equations

Ans. (2)
Energy of microwaves lie in range of vibration energy of water molecules.

Ans: (3)
Assuming perfectly inelastic collision for maximum loss in K.E.




⇒ K.E_i ≥ 20.4 eV

Ans: (4)
      ...(1)

    ...(2)

(Equating (i) & (ii))