Correct Answer: C. 2 x 10^-3

Explanation

Strain = \(\frac{extension}{\text{original length}}\)

\(\frac{2 {\times} 10^{-5}}{0.01}\)

= 2 x 10^-3

Explanation

(a) Conditions necessary for:

Constructive interference of waves

i. lt must involve at least two waves.

ii. Waves must vibrate in phase.

iii. The distance between the sources of the wave must be of the order of the wavelength.

Total internal reflection

i. The waves must travel from a denser medium to a less dense medium.

ii. The angle of incidence (in the denser medium) must be greater than the critical angle.

iii. It must involve two media of different optical densities.

Production of beats

1. The amplitude of the two interfering waves should be the same.

ii. The difference between the frequencies of interfering waves should be small.

iii. The two sources of waves must be sounded together.

(b) Calculation of velocity of sound in air

V = f\(\lambda\)

V = 2f(l\(_{2}\) - l\(_{1}\))

= 2 x 256(100 - 35)

= 33280 cm s\(^{-1}\)

= 332.80 m s\(^{-1}\)

(c)(i) Classification of musical instruments

i. Wind

ii. Stringed

iii. Percussion

(ii) Examples of classified instruments saxophone

I. Wind: flute, trumpet, pipe, organ,

iii. Stringed: piano, guitar, violin

iii. Percussion: piano, drum, bell, gong. xylophone

(d) Calculation of critical angle

\(\eta\) = \(\frac{1}{\text {Sin C}}\)

1.5 = \(\frac{1}{\text {Sin C}}\)

Sin C = 0.667

C = 41.80º

(e) Calculation of temperature

\(\frac{V_{1}}{\sqrt T_{1}} = \frac{V_{2}}{\sqrt T_{2}}\)

\(\frac{240} {\sqrt 375} = \frac {310} {\sqrt T_{2}}\)

240 \(\sqrt T_{2}\) = 310 x 19.36

\(\sqrt T_{2}\) = 625 K = 352 ºC

Correct Answer: B. Decreasing the distance between the particles

Explanation

(i) It is not bulky

(ii) It does not need time to warm up

(iii) It is cheaper to manufacture in large quantities.

(iv) It needs low voltage to operate.

Correct Answer: C) the clock losses time

Explanation

(a)(i) Fusion: Fusion is the coming together of two light nuclei to form a heavier nucleus with the release of energy.

Example: \(^{2} _{1} H + ^{2} _{1} H \to ^{4} _{2} He + Energy\)

(ii) Fission: Fission is the breaking up of heavy nucleus into two nuclei of comparable masses when bombarded with fast neutrons or energetic particles releasing energy in the process.

Example: \(^{235} _{92} U + ^{1} _{0} N \to ^{90} _{39} Kr + ^{144} _{50} Ba + ^{1} _{0} N + ^{1} _{0} N\)

(iii) Radiation Hazards. Radioactive substances can be dangerous if one is exposed to their radiations for a long time. Some of the radiations are highly penetrating and if the body is over-exposed to them, they can destroy the cells in tissues and upset the natural chemistry of the body. Examples include: – Leukaemia – Skin burns

(b) Advantages of fusion over fission include: (1) Easily achieved with lightest element so that nuclear repulsion is easily overcome as nuclei approach each other (2) Raw materials are cheaply available. (3) Produces less dangerous by products. (4) There is no upper limit to the mass of hydrogen that can be obtained. Very high temperature is required to overcome the coulomb repulsive forces between the two light nuclei. This poses severe technological problem due to the fact that materials to withstand this high temperature are difficult to come by.

(c) \(I = I_{m} \sin 2 \pi ft ... (1)\)

I - Instantaneous value of current

\(I_{m}\) - Peak value of current

F - frequency of current

\(I = 30 \sin 100 \pi t ... (2)\)

Comparing (1) and (2),

(i) \(2 \pi f = 100\pi \therefore f = 50 Hz\)

(ii) \(I_{m} = 30A\)

(iii) \(I_{r.m.s} = \frac{I_{m}}{\sqrt{2}}\)

= \(\frac{30}{\sqrt{2}} = 21.2A\)

Correct Answer: A) 3.47 x 10¹⁸

Explanation

(a) — The sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point

— The sum of the forces in one recticn is equal to the sum of the forces in the opposite direction.

b)

Method: Suspend the metre rule on a thread loop from clamp. Adjust its position until the metre rule balances horizontally and obtain its centre of gravity G. Suspend known mass M at a point on one side of the rule and just the position of the loop until the rule balances horizontally. Read and record the distances y and x in the diagram above. sing the principle of moments Mgx = Mgy

m = \(\frac{My}{x}\)

Therefore the mass M of the metre rule is calculated.

ii) Precautions:

— Firmly fixed the loop

— Avoid draught

— Avoid parallax when reading metre rule.

.

c)(i) Declaration a = \(\frac{v - u}{t} = \frac{0 — 20}{0.1} = — 200m/s^2\)

Resistance F = Ma = 0.12 x 200 = 24N

ii) Distance S = \(\frac{v^2 - u^2}{2a} = \frac{0^2 - 20^0}{2 \times 200}\)

= \(\frac{400}{400} = 1.0m\)

Explanation

(a)(i) An echo is a reflected sound.

(ii) Echoes can be used to:

- Determine the speed of sound in air.

- Determine the depth of a sea bed.

- Locate the positions of shoals of fish.

- Echo sounding is used in mineral prospecting.

(iii) Padding absorbs sound and minimizes reflections at the walls, floors or ceilings. This helps to minimize the effects of echoes.

(b)(i) Timbre is the characteristic quality of a sound produced by a particular voice or instrument. Timbre of the note is pure and it comes out unaccompanied by undesired notes. Overtones are sound notes which accompany the fundamental notes.

(ii) Resonance is said to occur whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other system which is vibrating with the same frequency.

(c) Let S\(_1\) = distance i between the ship and the cliff in the first instance

\(\frac{2S}{12}\) = 340; S\(_1\) = 2040m

In the second instance \(\frac{2S_2}{8}\) = 340

S\(_2\) = 1360m

Distance travelled by ship in 2.1 minute S\(_1\) - S\(_2\)

= 2040 - 1360 = 680m

Ship velocity V = \(\frac{680 }{2.1 \times 60}\) = 5.4ms\(^{-1}\)

Explanation

(a) A magnetic field is a region or space where a magnetic force is experienced.

(b)

Connect a straight wire to the circuit as shown above, sprinkle iron fillings on the cardboard paper round the straight wire; and tap gently. Then close the key and gently tap again, the iron fillings form a concentric circles round the wire. This shows that there is magnetic field round a current carrying conductor.

(c)(i)

\(x_c = \frac{1}{2 \pi fc} = \frac{1}{2 \pi \times 50(40 \times 10^{-6})}\)

= 79.567\(\Omega\)

z = \(\sqrt{R^2 + X^2_c} = \sqrt{40^2 + 79,580^2}\)

(ii) I = \(\frac{v}{2} = \frac{100}{89.1} = 1.12A\)

(iii) V = IX\(_c\) = 1.12 x 79.567

= 89.1V