Wednesday, February 4, 2015

Physics 9702 Doubts | Help Page 60

  • 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



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 3rd 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]





10 comments:

  1. 01/O/N/03 Q.31,34
    2/O/N/03 Q.5(c)
    01/M/J/05 Q.36
    01/M/J/06 Q.31,34
    01/O/N/06 Q.31

    ReplyDelete
    Replies
    1. one of them is explained at
      http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-61.html

      Delete
    2. June 2006 Paper 1 Q31 is explained at
      http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-62.html

      Delete
    3. June 2006 Paper 1 Q34 is explained at
      http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-66.html

      Delete
    4. November 2006 Paper 1 Q36 is also explained at
      http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-66.html

      Delete
  2. past paper 2006 paper 1 q36

    ReplyDelete
  3. I forgot to mention earlier its may june 2006 paper1 q36

    ReplyDelete
    Replies
    1. See solution 928 at
      http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-191.html

      Delete
  4. Replies
    1. For Q3, see solution 943 at
      http://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-194.html

      Delete

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