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.

31.

A ball of mass 0.8 kg is dropped from a height, H. Just before hitting the ground, its velocity is 5.0 m/s. Determine H.
[g = 10 ms-2]

A.

1.25m

B.

4.00m

C.

6.25m

D.

1.60m

Correct answer is A

u = 0, g = 10 \(ms^2\), v = 5 m/s, h = ?
From \(v^2 = u^2 + 2gh\)
⇒ \(5^2 = 0^2 + 2 × 10 × h\)
25 = 0 + 20h
25 = 20h

Therefore, h = \(\frac{25}{20}\) = 1.25m


 

32.

The instrument used to measure relative humidity is a

A.

Hygrometer

B.

Barometer

C.

Hydrometer

D.

Manometer

Correct answer is A

The instrument used to measure relative humidity is a hygrometer. Relative humidity is the amount of moisture in the air compared to the maximum amount of moisture the air could hold at the same temperature. A hygrometer is a device used for measuring the moisture content in the atmosphere, soil, or confined spaces.

33.

Which of the following devices is used to compare the relative magnitudes of charges on two given bodies?

A.

Ebonite rod

B.

Proof Plane

C.

Gold Leaf Electroscope

D.

Electrophorus

Correct answer is C

The device used to compare the relative magnitudes of charges on two given bodies is a gold leaf electroscope.

A gold leaf electroscope consists of two thin gold leaves that are attached to a metal rod inside a glass or plastic enclosure. When a charged object is brought near the metal rod, it induces a charge on the leaves. The leaves, being like charges, repel each other and spread apart. By observing the degree to which the leaves separate, you can compare the relative magnitudes of charges on different bodies. This makes a gold leaf electroscope a valuable tool for demonstrating the presence and relative strength of electric charges.

34.

The following statements are advantages of a solid dielectric material between the plates of a capacitor except that

A.

the maximum potential difference the capacitor can withstand is increased

B.

it increases the electrical conduction of the capacitor.

C.

the capacitance of the capacitor of given dimensions is much greater than that of the plates in a vacuum.

D.

it keeps the plates at a fixed separation.

Correct answer is B

The statement that is not an advantage of a solid dielectric material between the plates of a capacitor is:

it increases the electrical conduction of the capacitor.

In fact, one of the primary advantages of using a solid dielectric material between the plates of a capacitor is that it reduces electrical conduction between the plates. Dielectric materials are insulators, and they prevent the flow of electric current between the plates, thus increasing the capacitor's ability to store charge.

35.

A solenoid is constructed by winding an insulated copper on a long test tube. A constant direct current is passed through the solenoid. Which of the following actions will not increase the strength of the magnetic field of the solenoid? Inserting a

A.

steel rod in the test tube

B.

Nickel rod in the test tube

C.

Brass rod in the test tube

D.

Soft iron rod in the test tube

Correct answer is C

Inserting a brass rod in the test tube will not increase the strength of the magnetic field of the solenoid.

The strength of the magnetic field inside a solenoid depends on factors such as the number of turns of wire, the current passing through it, and the core material within the solenoid. Materials with high permeability, like soft iron and steel, can enhance the strength of the magnetic field when inserted into the solenoid. Nickel, which also has some ferromagnetic properties, would have a similar effect.

However, brass is not a ferromagnetic material; it does not enhance the magnetic field strength when inserted into the solenoid. In fact, it might even slightly reduce the magnetic field strength due to its non-magnetic properties.

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.

An atom in an excited state is one whose

A.

electrons are in the conduction band.

B.

potential energy is minimum.

C.

potential energy is maximum.

D.

electrons have moved to a higher energy level.

Correct answer is D

An atom in an excited state is one whose electrons have moved to a higher energy level.

In an atom, electrons are typically found in specific energy levels or orbitals. When an electron absorbs energy, it can move from a lower energy level to a higher energy level, which is referred to as an excited state. In this higher energy state, the electron is farther from the nucleus and has more potential energy. When the electron returns to its lower energy level, it emits the excess energy as electromagnetic radiation, often in the form of visible light. This is what gives rise to the characteristic spectral lines observed in spectroscopy.

39.

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:


\(v_e = \frac{\sqrt{2GM}}{R}\)
 

40.

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.