Nuclear physics-part (2)

41. Which one of the following is India's first nuclear plant? [M.P.P.C.S. (Pre) 1992]

(a) Narora
(b) Kalpakkam
(c) Tarapur
(d) Kota

Correct Answer: (c) Tarapur

Solution:Tarapur Atomic power station was the first nuclear power plant established in India which becomes operational in 1969. It is situated in Palghar district of Maharashtra. Its original capacity was 320 MW but in 2005, two 540 MW pressurized heavy water reactors were also operationalised at Tarapur.

42. India's first atomic power station is - [M.P.P.C.S. (Pre) 2008]

(a) BARC
(b) Tarapur Atomic Power Station
(c) Narora Atomic Power Station
(d) None of these

Correct Answer: (b) Tarapur Atomic Power Station

Solution:

India's first nuclear reactors (Apsara-1956; Cirus-1960) were established at BARC in Trombay. However, they are just only research reactors. So according to the question we consider Tarapur atomic power station (1969) is India's first atomic power station.

43. Who is the present director of 'BARC'? [M.P.P.C.S. (Pre) 2010]

Correct Answer: (e) none of the above

Solution:During the question period, Dr. Ratan Kumar Sinha (19 May, 2010-19 June, 2012) was the Director of BARC. While at present Vivek Bhasin is serving the same position since 15 September, 2023.

44. A radioactive substance has a half-life of four months. Three-fourth of the substance would decay in – [I.A.S. (Pre) 2001]

(a) 3 months
(b) 4 months
(c) 8 months
(d) 12 months

Correct Answer: (c) 8 months

Solution:Here 1/2 of the substance decays in 4 months.

therefore, first four month = 1/2 decays.

In next four months= Half of remaining 1/2 decays.

In total = 1/2 + 1/4 = 3/4 decays.

Hence, 3/4 of the substances will decays in 8 months.

45. The half-life of a radioactive element is 5 years then the fraction of the radioactive substance that remains after 20 years is - [I.A.S. (Pre) 1994]

(a) 1/2
(b) 1/4
(c) 1/8
(d) 1/16

Correct Answer: (d) 1/16

Solution:t{1/2} = 5 years

20 years = 20/ 5 = 4*t{1/2}

after 4*t{1/2} the remaining quantity is

= (1/2)⁴ = 1/16

The total time elapsed is 20 years, and the half-life is 5 years.

Number of half-lives = Total time / Half-life Number of half-lives

Number of half-lives = 20 years / 5 years = 4 half-lives

Calculate the fraction remaining after each half-life:
- After 1 half-life, the fraction remaining is (1/2)¹ = 1/2
- After 2 half-lives, the fraction remaining is (1/2)² = 1/4
- After 3 half-lives, the fraction remaining is (1/2)³ = 1/8
- After 4 half-lives, the fraction remaining is (1/2)⁴ = 1/16

Therefore, the fraction of the radioactive substance that remains after 20 years is 1/16.

The correct answer is (d) 1/16.

Nuclear physics-part (2)