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.

36.

A 100 kg device is pulled up a plane inclined at 30° to the horizontal with a force of 1000 N. If the coefficient of friction between the device and the surface is 0.25, determine the total force opposing the motion.
[ g = 10 ms-2]

A.

283.5N

B.

975.0N

C.

216.5N

D.

716.5N

Correct answer is D

The forces opposing the motion of the device are "mgsin θ" and "μN" where N is the normal reaction
From the diagram, on resolving to components, N = mgcos θ
Total opposing force = mgsin θ + μmgcos θ = mg(sin θ + μcos θ)
= 100 × 10 × (sin 30° + 0.25 × cos 30°)
= 1000 × (0.5 + 0.2165)
= 1000 × 0.7165
= 716.5 N

37.

The most appropriate value of fuse for a kettle rated 220 V, 1.06 kW is

A.

3A

B.

6A

C.

5A

D.

4A

Correct answer is C

V = 220V, P = 1.60KW = 1060W, I = ?

P = IV

I = \(\frac{P}{V} = \frac{1060}{220}\)

 ≈ 5A.

38.

A planet has mass, M, and radius R, If the universal gravitational constant is G, what is the expression for the escape velocity of an object on the planet?

A.

\(\frac{\sqrt{GR}}{2}\)

B.

\(\frac{\sqrt{2GM}}{R}\)

C.

\(\frac{\sqrt{2G}}{R}\)

D.

\(\frac{\sqrt{GM}}{R}\)

Correct answer is B

The expression for the escape velocity \(v_e\) of an object from a planet of mass M and radius R using the universal gravitational constant G is:


\(v_e = \frac{\sqrt{2GM}}{R}\)
 

39.

Increasing the frequency of a sound wave produces a sound with

A.

higher pitch

B.

longer wavelength

C.

more overtones

D.

reduced loudness

Correct answer is A

The pitch of a sound is determined by its frequency. Higher frequencies correspond to higher pitches. For example, a high-pitched sound like a whistle has a higher frequency than a low-pitched sound like a drum.

40.

In a mass spectrometer, an ion of charge, q, and mass, m, moving in a path of radius, r, in a field of flux density, B, has a speed of

A.

\(\frac{qBr}{m}\)

B.

\(\frac{mr}{qB}\)

C.

\(\frac{qm}{Br}\)

D.

\(\frac{mq}{r}\)

Correct answer is A

F = qvB ( force on moving charge in a magnetic field) F = \(\frac{mv^2}{r}\)

equating F

qvB =  \(\frac{mv^2}{r}\)

qB =   \(\frac{mv}{r}\)

qBr = mv 

Therefore, v =   \(\frac{qBr}{m}\)