Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If three $^{4} H e$ nuclei combine to form a $^{12} C$ nucleus, energy is liberated. Why can't helium nuclei combine on their own and minimize the energy?

A. He gas is rare, so collisions between nuclei are also rare;

B. It happens and the

^{12} C

we see in nature was formed by this method only in Earth's atmosphere;

C. The strong nuclear force between nucleons is attractive but short range; the Coulomb force between protons is long range, but repulsive;

D. Helium nucleus is more stable than carbon.

SOLUTION

Solution : C

Nuclear interactions are interesting battle grounds of different forces - mainly the Coulomb force, which is electrostatic in nature, and the strong nuclear force.
The strong nuclear force is a strong attractive force, but only activates when you bring the nucleons closer to each other than a few femtometers. At those length scales, no force in the universe is stronger than the strong nuclear force.
The problem is, $^{4} H e$ nuclei are positively charged, because of the two protons. To even bring them to a close distance, we will need to overcome the Coulomb repulsion between them, which increases as separation decreases. Only when they have reached a separation of few fermi the strong nuclear attraction becomes large enough to hold the nucleons together against the Coulomb repulsion.
$^{4} H e$ nuclei don't automatically fuse to make $^{12} C$ because of the energy that needs to be supplied to make them overcome the Coulomb repulsion, before the strong nuclear force even comes to the scene.

Question 2

1 g of hydrogen is converted into 0.993 g of helium in a thermonuclear reaction. The energy released

63 \times 10^{7} J

63 \times 10^{10} J

63 \times 10^{14} J

63 \times 10^{20} J

SOLUTION

Solution : B

$Δ m = 1 - 0.993 = 0.007 g m$
$∴$ $E = Δ m c^{2} = 0.007 \times 10^{- 3} \times (3 \times 10^{8})^{2} = 63 \times 10^{10} J$

Question 3

The binding energy per nucleon of deuteron $(_{1}^{2} H)$ and helium nucleus $(_{2}^{4} H e)$ is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

A. 13.9 MeV

B. 26.9 MeV

C. 23.6 MeV

D. 19.2 MeV

SOLUTION

Solution : C

$_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + Q$
Total binding energy of helium nucleus = $4 \times 7 = 28 M e V$
Total binding energy of each deutron = $2 \times 1.1 = 2.2 M e V$
Hence energy released = $28 - 2 \times 2.2 = 23.6 M e V$

Question 4

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV]

A. 28.4 MeV

B. 20.8 MeV

C. 27.3 MeV

D. 14.2 MeV

SOLUTION

Solution : A

Helium nucleus consist of two neutrons and two protons.
So binding energy $E = Δ m \times 931 M e V$
$\Rightarrow E = (2 \times m_{p} + 2 m_{n} - M) \times 931 M e V = (2 \times 1.0073 + 2 \times 1.0087 - 4.0015) \times 931 = 28.4 M e V$

Question 5

A atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of uranium atom $U^{238}$ is 170 MeV. The number of uranium atoms fission per hour will be

5 \times 10^{15}

10 \times 10^{22}

40 \times 10^{21}

30 \times 10^{25}

SOLUTION

Solution : C

By using $P = \frac{W}{t} = \frac{n \times E}{t}$ where n = Number of uranium atom fission and E = Energy released due to each fission so $300 \times 10^{6} = \frac{n \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{3600} \Rightarrow n = 40 \times 10^{21}$

Question 6

The binding energy per nucleon of $O^{16}$ is 7.97 MeV and that of $O^{17}$ is 7.75 MeV. The energy (in MeV) required to remove a neutron from $O^{17}$ is

A. 3.52

B. 3.64

C. 4.23

D. 7.86

SOLUTION

Solution : C

$O^{17} \to O^{16} +_{O} n^{1}$
$∴$ Energy required = Binding of $O^{17}$ - binding energy of $O^{16} = 17 \times 7.75 - 16 \times 7.97 = 4.23 M e V$

Question 7

$M_{n}$ and $M_{p}$ represent mass of neutron and proton respectively. If an element having nuclear mass M has N-neutron and Z-proton, then the correct relation will be

M < [N M_{n} + Z M_{p}]

M > [N M_{n} + Z M_{p}]

M = [N M_{n} + Z M_{p}]

M = N [M_{n} + M_{p}]

SOLUTION

Solution : A

Actual mass of the nucleus is always less than total mass of nucleons so $M < [N M_{n} + Z M_{p}]$

Question 8

If a Hydrogen nucleus is completely converted into energy, the energy produced will be around

A. 1 MeV

B. 931 MeV

C. 9.38 MeV

D. 238 MeV

SOLUTION

Solution : B

Mass of Hydrogen nucleus =mass of proton =1 amu energy equivalent to 1 amu is 931 MeV so correct option is (b).

Question 9

The radius of a nucleus of a mass number A is directly proportional to

A^{3}

A

A^{\frac{2}{3}}

A^{\frac{1}{3}}

SOLUTION

Solution : D

$R = R_{0} A^{\frac{1}{3}} \Rightarrow R \propto A^{\frac{1}{3}}$ .

Question 10

Two protons exerts a nuclear force on each other, the distance between them is

10^{- 14} m

10^{- 10} m

10^{- 12} m

10^{- 8} m

SOLUTION

Solution : A

From the options, only at a distance of $10^{- 14} m$ the nuclear force will act. This is equal to ten Fermis, as 1 Fermi = $10^{- 15} m$ , and this falls within the short range within which it acts.

Question 11

If m, $m_{n}$ and $m_{p}$ are the masses of $_{Z} X^{A}$ nucleus, neutron and proton respectively

m < (A - Z) m_{n} + Z m_{p}

m = (A - Z) m_{n} + Z m_{p}

m = (A - Z) m_{p} + Z m_{n}

m > (A - Z) m_{n} + Z m_{p}

SOLUTION

Solution : A

The mass of nucleus formed is always less than the sum of the masses of the constituent protons and neutrons i.e. $m < (A - Z) m_{n} + Z m_{p}$ .

Question 12

A radioactive substance has a half life of 60 minutes. After 3 hours, the percentage of atom that have decayed would be

A. 12.5%

B. 87.5%

C. 8.5%

D. 25.1%

SOLUTION

Solution : B

$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$ . Hence fraction of atoms decayed
$= 1 - \frac{N}{N_{0}} = 1 - {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} = 1 - {(\frac{1}{2})}^{\frac{3 \times 60}{60}} = \frac{7}{8}$
In percentage it is $\frac{7}{8} \times 100 = 87.5 %$

Question 13

The radioactive element which is used to find the age of the fossil is

A. C-14

B. U-234

C. U-238

D. Po-94

SOLUTION

Solution : A

C-14 is carbon dating substance.

Question 14

After two hours, one- sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is

A. 15 minutes

B. 30 minutes

C. 45 minutes

D. 1 hour

SOLUTION

Solution : B

$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T}} \Rightarrow (\frac{1}{16}) = {(\frac{1}{2})}^{\frac{2}{T}} \Rightarrow {(\frac{1}{2})}^{4} = {(\frac{1}{2})}^{\frac{2}{T}}$
$\Rightarrow$ T = 0.5 hour = 30 minutes.

Question 15

A radioactive nucleus decays from $_{92} X^{235}$ to $_{91} Y^{231}$ . Which of the following particles are emitted

A. One alpha and one electron

B. Two deuterons and one positron

C. One alpha and one proton

D. One proton and four neutrons

SOLUTION

Solution : A

$_{92} X^{235} α ⟶_{90} X^{231} - 1^{e^{0}} ⟶_{91} Y^{231}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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