Physics Test 278

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Physics Test 278

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Weekly Quiz Competition

Question 1
4 / -1

The energy released by the fission of one uranium atom is 200 MeV. The number of fissions required per second to produce 3.2 W of power is

A
10⁷

B
10¹⁰

C
10¹⁵

D
10¹¹

Solution

The energy released by the fission of one uranium atom, E = 200 MeV

E = 200 × 10⁶× 1.6 × 10^-19 J

The number of fissions required per second, n/t = P/E.
Question 2
4 / -1

If in a nuclear fission, a piece of uranium of mass 0.5 g is lost, then the energy obtained in kWh is

A
1.25 × 10⁷

B
2.25 × 10⁷

C
3.25 × 10⁷

D
0.25 × 10⁷

Solution

E = mc² = (0.5/1000) × (3 × 10⁸)² = 4.5 × 10¹³ Ws

1000 W = 1 kW

3600 seconds = 1 hour

So, by unit conversion, E = 1.25 × 10⁷ kWh
Question 3
4 / -1

What is the rest mass energy of an electron?

A
1 eV

B
0.51 MeV

C
931 MeV

D
None of these

Solution
Question 4
4 / -1

Which of the following are not emitted by a radioactive substance?

A
Electrons

B
Electromagnetic radiations

C
Helium nuclei with a charge equal to that of two protons

D
Neutrons

Solution

Neutrons are not emitted by a radioactive substance.
Question 5
4 / -1

During a negative beta decay,

A
An atomic electron is ejected

B
An electron which is already present within the nucleus is ejected

C
A neutron in the nucleus decays, emitting an electron.

D
A part of the binding energy of the nucleus is converted into an electron

Solution
Question 6
4 / -1

The half-life of Pa-218 is 3 minutes. What mass of a 16 g sample of Pa-218 will remain after 15 minutes?

A
3.2 g

B
2.0 g

C
1.6 g

D
0.5 g

Solution

Since 15 minutes = 5 x 3 minutes = 5 half-lives, the number of nuclei left after 15 minutes = 1/2⁵ = 1/32 of the original number

Therefore, the mass of 16 g sample left after 15 minutes = 16/32 = 0.5 g

Hence, the correct choice is (d).
Question 7
4 / -1

M_p denotes the mass of a proton and M_n that of a neutron. A given nucleus of binding energy B contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given by

A
M (N, Z) = NM_n + ZM_p - Bc²

B
M (N, Z) = NM_n + ZM_p + Bc²

C
M (N, Z) = NM_n + ZM_p - B/c²

D
M (N, Z) = NM_n + ZM_p + B/c²

Solution
Question 8
4 / -1

The circuit shown in the figure contains two diodes D₁ and D₂, each with a forward resistance of 50 ohms and infinite backward resistance. If the battery voltage is 6 V, the current (in amperes) through the 100 ohm resistance is

A
Zero

B
0.02

C
0.03

D
0.036

Solution

In the given circuit only diode D₁ will allow the current to pass through as it is forward-biased.

Hence, the current through the 100 ohm resistance is (∵ resistance of D₁ = 50 Ω)
Question 9
4 / -1

I is the current passing through a circular wire. If the radius of the wire is changed to twice, then what will be the current passing through the wire?

A
2I

B
4I

C
I/2

D
I/4

Solution

Current, I = V/R

If the radius becomes twice, the area becomes 4 times and resistance becomes R/4, as resistance is inversely proportional to the area of wire.

New current = 4V/R = 4I
Question 10
4 / -1

To obtain a p-type semiconductor germanium crystal, it must be doped with foreign atoms whose valency is

A
2

B
3

C
4

D
5

Solution

To obtain a p-type semiconductor the impurity must be trivalent, i.e. it must be doped with foreign atoms whose valency is 3. In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. P-type semiconductors are created by doping an intrinsic semiconductor with acceptor impurities. A common p-type dopant for silicon is boron.

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Re-Attempt Weekly Quiz Competition

Physics Test 278

Result

Physics Test 278

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Question 1

107

1010

1015

1011

Solution

Question 2

1.25 × 107

2.25 × 107

3.25 × 107

0.25 × 107

Solution

Question 3

1 eV

0.51 MeV

931 MeV

None of these

Solution

Question 4

Electrons

Electromagnetic radiations

Helium nuclei with a charge equal to that of two protons

Neutrons

Solution

Question 5

An atomic electron is ejected

An electron which is already present within the nucleus is ejected

A neutron in the nucleus decays, emitting an electron.

A part of the binding energy of the nucleus is converted into an electron

Solution

Question 6

3.2 g

2.0 g

1.6 g

0.5 g

Solution

Question 7

M (N, Z) = NMn + ZMp - Bc2

M (N, Z) = NMn + ZMp + Bc2

M (N, Z) = NMn + ZMp - B/c2

M (N, Z) = NMn + ZMp + B/c2

Solution

Question 8

Zero

0.02

0.03

0.036

Solution

Question 9

2I

4I

I/2

I/4

Solution

Question 10

2

3

4

5

Solution

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10⁷

10¹⁰

10¹⁵

10¹¹

1.25 × 10⁷

2.25 × 10⁷

3.25 × 10⁷

0.25 × 10⁷

M (N, Z) = NM_n + ZM_p - Bc²

M (N, Z) = NM_n + ZM_p + Bc²

M (N, Z) = NM_n + ZM_p - B/c²

M (N, Z) = NM_n + ZM_p + B/c²