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**Q 31. 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 cm ^{3} 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 10^{10} decays per second and at t = 10 hrs e^{–λt} = 0.84) [2018]** (2)

(1) 4 liters

(2) 5 liters

(3) 6 liters

(4) 7 liters

Ans:

(dn/dt) = −λN

0.8 µ Ci = −λN

N

v → N

v cm

v = 3.7/5 × 0.84 × 0.84 × 10

= 0.5 × 10

Answer:

E

E

E

E

Total = E

E

(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:

Given : N

Hence,

The half life of S

The half life of S

(1) λ

(2) λ

(3) λ

(4) λ

Ans:

Ans:

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

(1)

**(2)(3) (4)Ans:** (3)

Solution:

In X-ray tube

Slope is negative

Intercept on y-axis is positive

(1)

(2)

(3)

(4)

Ans

(1)

(2)

(3)

(4)

Ans:

Solution:

From energy level diagram

(1) t =T log (1.3)

(2)

(3)

(4)

Ans

(1)

**(2) **

**(3) **

**(4) Ans:** (2)

(Take charge of electron as 1.6×10

A: 0.59 (Ω - m)

B: 1.83 (Ω - m)

C: 1.68 (Ω - m)

D: 1.20 (Ω - m)

Ans:

s = e (n

= 1.6 × 10

= 1.6 × 10

= 1.6 × 1.05

= 1.68

**Q 42. A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement. [2017](1) Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz(2) Modulation index = 5, side frequency bands are at 21.2 kHz and 18.8 kHz(3) Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz(4) Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHzAns:** (4)

=1/5 = 0.2

Frequency = 12 × 10

F = 12.00 kHz

F

F

(1) away from the wire

(2) towards the wire

(3) parallel to the wire along the current

(4) parallel to the wire opposite to the current

Ans:

(1) 3 mm

(2) 9 mm

(3) 4.5 mm

(4) 1.5 mm

Ans:

a = 0.1 mm = 10

λ = 6000 × 10

= 6 × 10

D = 0.5 m

for 3

a sin θ = 3λ

(1)

**(2) **

**(3) **

**(4) Ans:** A

.... (i)

...... (ii)

(ii) - (i) gives

.... (iv)

combining (iii) & (iv)

(1) 1 : 16

(2) 4 : 1

(3) 1 : 4

(4) 5 : 4

Ans:

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/2

⇒ Number of A nuclides decayed = 15/16N

NB after 80 minutes = N/2

⇒ Number of B nuclides decayed = 3/4N

Required ratio =

(1) 4

(2) 5

(3) 2

(4) 3

Ans:

Energy of proton = 13.6 – 3.4 = 10.2 eV

For removal of electron

so minimum value of n = 4

(1)

(2)

(3)

(4)

Ans:

using these equations

(1) giving rotational energy to water molecules

(2) giving vibrational energy to water molecules

(3) giving translational energy to water molecules

(4) transferring electrons from lower to higher energy levels in water molecule

Ans.

(1) 10.2 eV

(2) 12.1 eV

(3) 20.4 eV

(4) 16.8eV

Ans:

⇒ K.E

(1) 3hc/λ

(2) hc/λ

(3) hc/3λ

(4) hc/2λ

Ans:

Solution:

...(1)

...(2)

(Equating (i) & (ii))

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