• Physics 9702 Doubts | Help Page 60

Question 339: [Current of Electricity > Resistance]
Cylindrical piece of a soft, electrically-conducting material has resistance R. It is rolled out so that its length is doubled but its volume stays constant.
What is its new resistance?
A R / 2                        B R                              C 2R                            D 4R

Reference: Past Exam Paper – June 2005 Paper 1 Q33



Solution 339:

Answer: D.

Resistance R of the wire = ρL / A

Since the same material is used, the resistivity ρ is constant.

Volume V of cylinder = Length L x Cross-sectional Area A

If the length L is doubled while the volume V is kept constant, then the area A is be halved.

V = (2L) (A / 2)

New resistance = ρ (2L) / (A/2) = 4 (ρL / A) = 4R

Question 340: [Nuclear Physics]
Protons and neutrons are thought to consist of smaller particles called quarks.
The ‘up’ quark has charge of (2/3) e : a ‘down’ quark has charge of – (1/3) e, where e is the elementary charge (+1.6 x 10^–19 C).
How many up quarks and down quarks must a proton contain?

Reference: Past Exam Paper – June 2003 Paper 1 Q40



Solution 340:

Answer: D.

The proton has a charge of +e (+1.6 x 10^–19 C) while that of an electron is –e. So, we need a combination of ‘up’ and ‘down’ quarks such that the overall charge is +e.

Charge of ‘up’ quark: (2/3) e

Charge of ‘down’ quark: – (1/3) e

Choice A: Overall charge = 3 [– (1/3) e] = –e

Choice B: Overall charge = (2/3) e + [– (1/3) e] = (1/3) e

Choice C: Overall charge = (2/3) e + 2[– (1/3) e] = 0

Choice D: Overall charge = 2[(2/3) e] + [– (1/3) e] = e

Question 341: [Current of Electricity > Resistance]

A filament lamp operates normally at a potential difference (p.d.) of 6.0 V. The variation with p.d. V of the current I in the lamp is shown.

(a) Use Fig to determine, for lamp,

(i) resistance when it is operating at a p.d. of 6.0 V

(ii) change in resistance when p.d. increases from 6.0 V to 8.0 V

(b) Lamp is connected into circuit of Fig.

R is a fixed resistor of resistance 200 Ω. Battery has e.m.f. E and negligible internal resistance.

(i) On Fig, draw a line to show variation with p.d. V of the current I in the resistor R

(ii) Determine e.m.f. of the battery for the lamp to operate normally

Reference: Past Exam Paper – June 2003 Paper 2 Q5



Solution 341:

(a)

(i) Resistance = V / I = 6.0 = (40x10^-3) = 150 Ω

(ii)

At a p.d. of 8.0 V, resistance = 8.0 / (50x10^-3) = 160 Ω

Change in resistance (= 160 – 150) = 10 Ω

(b)

(i)

The graph is a straight line through the origin and passes through the point where I = 40 mA, V = 8.0 V

{When I = 40mA and V = 8.0V, the resistance is 8.0 / 0.040 = 200 Ω. The graph should be such that at all points, the resistance calculated is always 200 Ω. So, it is a straight line.}

(ii)

{Since this is a series connection,}

The current in both (the lamp and the resistor) must be 40 mA

{When current is 40mA in the lamp, the p.d. is 6.0V.}

e.m.f. of battery = 8.0 + 6.0V = 14.0V

Question 342: [Current of Electricity]
A wire carries current of 2.0 amperes for 1.0 hour.
How many electrons pass a point in wire in this time?
A 1.2 × 10^–15               B 7.2 × 10³                  C 1.3 × 10¹⁹                 D 4.5 × 10²²

Reference: Past Exam Paper – November 2003 Paper 1 Q30 & November 2016 Paper 12 Q33



Solution 342:

Answer: D.

1 hour = 60 minutes = 3600 seconds

Charge Q = It = 2.0 x 3600 = 7200 C

Charge, e of 1 electron = 1.6x10^-19 C

Charge Q = ne

Number of electrons, n = Q / e = 7200 / (1.6x10^-19) = 4.5x10²²

Question 343: [Dynamics > Newton’s laws of motion]
Ship of mass 8.4 × 10⁷ kg is approaching harbour with speed 16.4 m s^–1. By using reverse thrust it can maintain a constant total stopping force of 920 000 N.
How long will it take to stop?
A 15 seconds              B 150 seconds                         C 25 minutes               D 250 minutes

Reference: Past Exam Paper – June 2014 Paper 12 Q9



Solution 343:

Answer: C.

Force F = change in momentum  / time = Δp / t

The speed changes from 16.4 ms^-1 to zero (the ship stops).

Change in momentum, Δp = mΔv = m (16.4 – 0)

Time t = Δp / F = (8.4x10⁷ x 16.4) / 920 000 = 1497s = 25min

Question 344: [Radioactive decay]
Bombardment of a certain material with α-particles produces an emission which penetrated lead, ejects protons from paraffin wax, and travels at speed up to 5x10⁷ ms^-1. What does this emission consist of?
A X-rays         B α-particles    C β-particles    D neutrons      E ultraviolet light

Reference: Past Exam Paper – N82 / II / 36 & J84 / II / 34



Solution 344:
Answer: D.
The information given is:
1. penetrated lead,
2. ejects protons from paraffin wax,
3. and travels at speed up to 5x10⁷ ms^-1

X-rays and ultraviolet light are from the EM spectrum, so they travel at the speed of light (= 3.0x10⁸ms^-1). From their speed, they do not satisfy the 3^rd description. [A and E are incorrect]

It is known that α-particles have speed of about 1/200 that of light and they are easily by a piece of paper. From this, description 1 and 3 are not satisfied. [B is incorrect]

β-particles have high speed of up to2/3 the speed of light and they are stopped only by a few mm thickness of aluminium. From this, 1 and 3 are not satisfied. [C is incorrect]