Physics questions and answers

Physics Questions and Answers

If you want to learn more about the nature and properties of matter and energy or you're simply preparing for a Physics exam, these Physics past questions and answers are ideal for you.

4,126.

The primary coil of a transformer has N turns and is connected to a 120 V a.c. power line. If the secondary coils has 1 000 turns and a terminal voltage of 1 200 volts, what is the value of N?

A.

120

B.

100

C.

1000

D.

1200

Correct answer is B

For the transformer, V ∝ ∩
=> V s ∝ ∩s; Vp ∝ ʌp
=> Vp/Vv = ∩p/∩s, therefore 120/1200 = ∩p/1000
therefore ∩p * 1200 = 120 * 1000
p = 120*1000/1200 = 100

4,127.

The current through a resistor in an a.c circuit is given as 2 sin wt. Determine the d.c. equivalent of the current

A.

A/√2

B.

2/√A

C.

2 A

D.

√2 A

Correct answer is B

If I = 2 Sin wt; but in a.c. circuit current I = I0 Sin wt; thus comparing the two equations, it implies that I0 = 2A; Again, the root means square (I(rms)) of an a.c. current is the value of the d.c. current that will dissipate the same amount of heat in a given resistance as the a.c.
and I(rms) = I0/√2 = 2.0A/√2

4,128.

I. Low pressure.
II. High pressure.
III. High p.d.
IV. Low p.d.
Which combination of the above is true of the conduction of electricity through gases?

A.

I and IV only

B.

I and III only

C.

II and IV only

D.

II and III only

Correct answer is B

Electricity conduction through gases is usually done under low pressure (I) and high voltage (III). Thus /B/

4,129.

The particles emitted when \( ^{39}_{19}K \text{ decay to } ^{39}_{19}K \) is

A.

gamma

B.

beta

C.

electron

D.

alpha

Correct answer is A

3919K 3919K.
Since neither the mass number (39), nor the atomic number (19) is affected by the emission,the particles emitted can only be gamma ray, which is not a charge particle, but simply a radiation.

4,130.

A cell can supply current of 0.4 A and 0.2 A through a 4.0Ω and 10.0Ω resistors respectively.
This internal resistance of the cell is

A.

2.0Ω

B.

1.0Ω

C.

2.5Ω

D.

1.5Ω

Correct answer is A

let the internal resistance of the cell = r,
therefore E = I (R+r)
(i) E = 0.4(4+r), (ii) E = 0.2 (10+r)
therefore 0.4 (4+r)= 0.2 (10+r)
1.6 + 0.4r = 2.0 * 0.2r
0.4r - 0.2r = 2.0 - 1.6
therefore 0.2r = 0.4 => r = 0.4/0.2 = 2Ω
therefore r = 2Ω