1967-05-31 — Page 14

育教僑

貞三第張四第鱨:

日三廿月四年未丁唇

育教僑華

英中會考物理科答案(續)

HONG KONG ENGLISH SCHOOL CERTIFICATE EXAM. 1967.)

Suggested Answers to Physics.

Section C.

7(a)(1) Snell's Law: When a ray of light passes from I'

one medium to another the ratio of the sine

of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium is a constant for any given pair of media.

(ii) Consider a ray passing from a medium to a legg

dense medium with a small angle of incidence. The refracted ray is diverged from the normal. If the angle of incidence is gradually increK- sed until for a certain critical angle of in- cidence, the angle of refraction is just 90°.

Let 1 be the critical angle. By Snell's Law

ein i

sin r

where n is the refractive

index from air to a par-

ticular medium.

Since r-90" when 1-1

(b)

white

light

Source

(0)

WAH KIU YAT PO

~STOMER TO B.

CITY HALL

(ii) T is not affected by a strong wind blowing

at right angles to the line AB.

(iii) T is not affected by an increase in the

frequency of the sound.

(iv) T becomes smaller by an increase in the

temperature of the air because the velocity,

· of sound increases.

(b) The velocity of sound is given by the formula

v = IX

If we can find the wavelen, th of the sound wave emitted by a standard tuning fork of frequency f, we may easily calculate the velocity v.

We use a resonance tube to accomplish this.

In the first position of resonance the length of the tube is 4, But the antinode at the top does not coincide exactly with the top of the tube 26 shown in the diagram. This ie called the end correction, To eliminate this, we can

measure a second position of

resonance. We eliminate the end correction by subtracting one from the other.

Length of tube for 1st position of resonance

L.

om.

Length or tube for 2nd position of resonance

.om.

air

J

then

glass

and

ne

sin90

Bin 1 1

(2) - (1)

sin 1.

y = fλ

I

સુ સસરા

。

(0)

E

国酱

(1)

(in)

(i)

(d) By Fleming's right hand rule, a magnetic field

is formed such that the magnetic flux passes from the bottom to the top inside the oil and back to the bottom from outside of the coil. The magnetic north pole was in the bottom of the coil while the magnetic south pole was at the top of the coil. Since the needle is unmagnetised ariginally and is placed at the the centre inside the coil, the position of the needle is unchanged. When the current is on, the steel needle is magnetised and the surface of the needle below the coil or facing the ground becomes south whereas the surface above becomes north. When the current is switched off The magnetism remains in the needle and the position of the needle is such that the north pole surface points to south and vice versa.

V 2f(L- 4.

(1)

(2)

(c: Let x yd. be the distance between the man and cliff riginally, the velocity of sound. Then roshe first cho

=

from the second echo

7=27 x + 220

put (1) into (ii) 3.7v

2.5vj± 440

v = 366---

(ii) x=

2-521003

yd./ses ft./sec.

(10(a)(1) Ohn's Law: The current passing through a wire

at constant temperature is proportional to the potential difference between its ends.

· Potential difference

Ne may write, (ii)

I, Ri

V

current

Resistance

R. and R. are two resistors in parallel, Suppose the tom. tal current is I amperes and the current passing through Ris, and the furrant passing through

R2 is 12

་ where V is the potential difference across the circuit. Then I I. +

V

Int 12

1

Where R is the combined resistance.

-

cliff

1

1

+

Solving for R, R =

or V

• 1 + 4/2

V

V

R1 R2

壕

+

Lens to focio

Right on alit

Prim

on

reim formsch

wate screen.

Production of a fairly pure spectrum,

Jem

Join PP

cutting BC at E. Let PE x cm.

PE is the apparent thickness of the block of glass. P is the real thickness.

From.

refractive index

9

1.5

I

x = 6

cm.

Real thickness Apparent thickness

Q

- 1375

yds.

2+220 -

'Answer:(i). The velocity of sound in air is

1100 per seo

(ii) The original distance of the man \

from the cliff is 1375 yds.

(d)(1) Tone or quality depends on the waveform and

overtones,

(11) Pitch depends on the frequency. (iii) Loudness depends on the amplitude.

(e) Let n be the -

n be the frequency of wire B

£ be the length of wire A.

L be the length of wire B,

M

be the mass per unit length of wire A,

M be the mass per unit length of wire B,,

&

a

be the tension of wire A,

be the tension of wire. B,

be the diameter of wire A,

be the diameter of wire B.

k is a constant.

15 mA 75 mV

resistance of the meter.

R-155 ohms.

If the metter is to be modified to give full deflection for a current of 1.5 A, we have to} use a shunt in parallel with the meter, The Current passing through the shunt is

-0.015 1.485- amp. Remembering that the potential difference V across each is the same.

for the milliammeter V = 0.015 x 5 ...(1) V - 1.485 x R ...(2) -(2) 1.485R = 0.015 x 5

R * 0.0505

ohm.

Answer: The resistance of the meter is 5 ohms,

The shunt is 0.0505 ohm so as to give full scale deflection for a current

of 1.5 A.

LIBRARIES

Since wires A and B are made of the same materiai the dengity of them are the same.

(o) Resultant resistance

185+10+

10x10 10+30

-200

Current passing through the

V/R 2/200

185

-0.01 amp.

10 QA,

10 x 10

0.01 x

10+10

-0.05 V.

❤

Answer: The distance between P2 and Pa is.

9+6 = 15 cm.

(a) As shown in the diagram,

PQ is an incident ray through the centre of a ciroular glass digo com- posed of tyo similar semi“. circular glass blocks with a very narrow air space in. between. Starting A-50*-

N'

the combination is rotated in the clockwise di- rection. Let MN ba the normal,i.e. the straight Line perpendicular to the plane edges through the centre Than the angle of incidense will be 90-A. If this angle is smaller than the crit- ical angle, the ray will emerge into the air space

at an angle r to the normal. Then angle r wilI be the incident angle of the ray into the other glass block. Owing to the fact that the light raya are reversible, the angle of refraction in the second glass block will be 90-A which is the angle of incidence in the first glass block. Since the air space is very narrow, we can assume that there is no displacement between PQ and XY. Therefore PQXY is a straight line. Now we con- tinue rotating A to such a position that 90°-A is greater than the critical angle, total re- flection occurs, the ray bends along the air

spage. This valus of A lies between 90 MEILE

180. Suppose the total reflection occure at an angle of incidence which is above the ling FQXY. (Fig.2) Continuing rotating for further 96 and the ray emerges again along XY. This means - that the critical angle is now below PQXY(Fig.3] Comparing fig.2 and fig.3, we find that

2(908)

or 8 = 42°

96

zefractive index of glass

CNC 1.5

Fig (3)

sin.

Fig (2)

T is not affected by a strong wind blowing)

Answer. The ratio of the tensions in A and B)

is 15

Section D.

metal

Remisphere

insulator

-moving

Bilk belt

point conduct

battery

Van de Graaff generator.

(b)(1) When a pin is put upside down on the upper terminal of the Van de Graaff generator, all the charges on the metal hemisphere will ac- cumulate at the point of the pin, because charge is mostly concentrated at places where the surface is sharply curved. If the charge continues to accumulate, there will even be a electric wind away from the point, The positive charge is sprayed off quickly. (11) If the upper terminal of the generator was dusted with lycopodium powder, the powder was attracted to the hemispherical metal tex minal because the powder was induced a nezám tive charge.

milliammeter

Reading in the millivoltmeter

- 50

10.2

Answer: The reading or the milliammeter is

10 må. The reading of the millivolt- meter is 50 mV.

(á) Current flowing through the

resistance 573 amp.

Since each cell consumÓB

17 gma of zins, and the potential differenca of each Leckanohet cell is 1.5 V, there ara two çella in geries.

Total mass of zinc consumed 17 x 234 gms

0.8. - 0.00034 gms/coulomb: From Faraday's law or electrolysis, E.c.e. WIt FAL zack cell

17.

x 0.00034

x 10

78 hr. 2 min,

Bec.

Answer: The time required is 2 hr, 2 min,

Appendix,

Section B.

2(c) (211 The mechanical advantage of this pulley

Bystem is 6 x 33%

Effort

200

2.

-100 lbs.

Answer: The effort required to lift a load

of 200 lbs. is 100 lbs,

End