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.

4,196.

Gamma rays are produced when

A.

high velocity electrons are abruptly stopped in metals

B.

energy changes occur within the nucleus of atoms

C.

energy changes occur within the electronic structure of atoms

D.

electrons are deflected in very strong magnetic fields

Correct answer is B

Gamma rays are electromagnetic radiation of nuclear origin; they are produced from energy changes which occur within the nucleus of ato

4,197.

The core of an efficient transformer should consist of laminated pieces of metal in order to

A.

increase the heat produced by increasing the eddy current

B.

increase the heat produced by reducing the eddy current

C.

reduce the heat produced by increasing the eddy current

D.

reduce the heat produced by reducing the eddy current

Correct answer is D

No explanation has been provided for this answer.

4,198.

At what frequency would a 10H inductor have a reactance of 2000Ω?

A.

π/200Hz

B.

π/100Hz

C.

100/πHz

D.

100πHz

Correct answer is E

Inductive reactance XL = 2πfL
∴ 2πfL = 2000Ω
then f = 2000Ω/2 x π x 10
f = 100/πHz

4,199.

In fleming's right-hand rule, the thumb, the forefinger and the middle finger if held mutually at right angles represent respectively, the

A.

motion, the field and the induced current

B.

induced current, the motion and the field

C.

field, the induced current and the motion

D.

induced current, the field and the motion

Correct answer is A

No explanation has been provided for this answer.

4,200.

A galvanometer has a resistance of 5Ω. By using a shunt wire of resistance 0.05Ω, the galvanometer could be converted to an ammeter capable of reading 2Amp. What is the current through the galvanometer?

A.

2mA

B.

10mA

C.

20mA

D.

25mA

Correct answer is C

Shunt resistance is usually given as
R = igrg/(I - ig)
Where R = 0.05Ω
rg = 5Ω
I = 2M
∴ ig = IR/(rg + R)
= (2 x 0.05)/(5 + 0.05)
= 0.10/5.05
= 0.0198A
= 19.8mA
= 20mA