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.

536.

The opposition offered to an alternating current in a C-R series circuit is known as

A.

impedance

B.

resistance

C.

reactance

D.

friction

Correct answer is A

No explanation has been provided for this answer.

537.

The diagram above illustrates two charged bodies, X and Y separated by a distance r. Each body has charge q and mass m. Which of the following statements about the electrostatic force \(F_{e}\) and gravitational force \(F_{g}\) acting between the bodies is correct?

A.

\(F_{e}\) is attraction while \(F_{g}\) is repulsion

B.

Both \(F_{e}\) and \(F_{g}\) are that of repulsion

C.

Both \(F_{e}\) and \(F_{g}\) is that of attraction

D.

\(F_{e}\) is repulsion while \(F_{g}\) is attraction

Correct answer is D

 electrostatic force can be repulsive or attractive and gravitational force is only attractive. 

538.

The main reason why a.c. is transmitted at very high voltage and low current is that

A.

flows faster through the conductor

B.

can be stepped down

C.

is cheaper to generate

D.

reduces heat loss

Correct answer is D

No explanation has been provided for this answer.

539.

A battery of e.m.f 3.0V is connected across a potentiometer wire AB of length 10\(\Omega m^{-1}\) as illustrated in the diagram above. If the jockey J makes contact at the 30cm mark, determine the resistance of length AJ and voltage across AJ.

A.

3.0\(\Omega\); 0.9V

B.

3.0\(\Omega\); 3.0V

C.

3.3\(\Omega\); 0.3V

D.

3.3\(\Omega\); 9.0V

Correct answer is A

E = 3.0V; \(\frac{R}{l} = 10\Omega m^{-1}\)

l = 30cm = 0.3m; R = \(10\Omega m^{-1} \times 0.3m = 3\Omega\)

\(\frac{E}{V} = \frac{l_{1}}{l_{2}} \implies \frac{3}{V} = \frac{1}{0.3}\)

= \(\frac{3\times 0.3}{1} = 0.9V\)

 

540.

A lamp rated 100W, 240V is lit for 5hours. Calculate the cost of lighting the lamp if 1kWh of electrical energy cost N5.

A.

N2.50

B.

N3.20

C.

N6.50

D.

N9.60

Correct answer is A

Power = 100W; Voltage = 240V;  Time = 5 hours

Total energy used = Power x time = 100 x 5 = 500wh

converting to kwh, we have

\(\frac{500}{1000} = 0.5kWh \times N5 = N2.50\)