# Physics 9702 Doubts | Help Page 208

__Question 1000:__

__[Current of Electricity]__
A uniform resistance wire AB has
length 50 cm and diameter 0.36 mm. The resistivity of the metal of the wire is
5.1 × 10

^{–7}Ω m.**(a)**Show that the resistance of the wire AB is 2.5 Ω.

**(b)**The wire AB is connected in series with a power supply E and a resistor R as shown in Fig. 5.1.

The electromotive force (e.m.f.) of
E is 6.0 V and its internal resistance is negligible.

The resistance of R is 2.5 Ω. A
second uniform wire CD is connected across the terminals of E. The wire CD has
length 100 cm, diameter 0.18 mm and is made of the same metal as wire AB.

Calculate

(i) the current supplied by E,

(ii) the power transformed in wire
AB,

(iii) the potential difference
(p.d.) between the midpoint M of wire AB and the midpoint N of wire CD.

**Reference:**

*Past Exam Paper – June 2015 Paper 23 Q5*

__Solution 1000:__**(a)**

Resistance R of wire = ρl / A

R = (5.1×10

^{−7}× 0.50) / π×(0.18×10^{−3})^{2}= 2.5 (2.51) Ω**(b)**

(i)

{Resistance R of wire AB =
ρ

*l*/ A
The length of CD is twice
that of AB and its diameter is half. Since the cross-sectional area depends on
(diameter)

^{2}, the cross-sectional area of CD is ¼ that of AB.
Resistance of wire CD = ρ(2

*l*) / (A/4) = 8(ρl/A) = 8R}
Resistance of CD = 8 × resistance of
AB = 20 (Ω)

{Resistance of AB = 2.5Ω. Total
resistance in the middle branch = 2.5 + 2.5 = 5.0 Ω. The middle branch is in
parallel with the bottom branch which contains the wire CD of wire 20 Ω, as calculated
above. So, we need to find the equivalent circuit resistance.}

Circuit resistance = [1/5.0 + 1/20]

^{−1}= 4.0 (Ω)
Current = V / R = 6.0 / 4.0

Current = 1.5 A

(ii)

Power in AB = I

^{2}R OR Power = V^{2}/ R
{

**For P = I**The current of 1.5A is the total current in the circuit. But, for a parallel combination, the current splits.^{2}R:
Current in wire CD = V / R
= 6.0 / 20 = 0.3 A

Thus, the current in the
middle branch = 1.5 – 0.3 = 1.2 A.

**For P = V**The p.d. across any of the 2 branches of the parallel combination is equal to the e.m.f. in the circuit. But in the middle branch, the wire AB is in series with the resistor R. So, the voltage is divided between them. As both have the same resistance, the voltage is divided equally, that is, 6.0 / 2 = 3.0V.}

^{2}/R:
Power in AB = (1.2)

^{2}× 2.5 = 3.6 W Power = (3.0)^{2}/ 2.5 = 3.6 W
(iii)

{The length wire AM is
half that of wire AB, so only half the resistance is considered (= 2.5 / 2 = 1.25
Ω).}

Ohm’s law: potential drop
V = IR}

potential drop A to M = 1.25 × 1.2 =
1.5 V

{Similarly, only half the
resistance is considered at point N. From Ohm’s law, the p.d. would also be
half that across the whole length of the wire (= 6.0 / 2 = 3.0 V).}

potential drop C to N = 3.0 V

p.d. MN {= 3.0
– 1.5} = 1.5 V

{Note that both point A
and C are connected to the ‘+’ terminal of the battery and are thus at a
potential of 6.0V. Normally, we would need to find the

**potential**at**point**M and**point**N and give the difference, but since we are considering the**potential drop**from the initial points having the same potentials, the p.d. between MN can be directly obtained by the difference in potential drop.}

__Question 1001:__

__[Current of Electricity > Capacitance]__
Three capacitors, each of
capacitance 48 μF, are connected as shown in Fig. 6.1.

**(a)**Calculate the total capacitance between points A and B.

**(b)**The maximum safe potential difference that can be applied across any one capacitor is 6 V.

Determine the maximum safe potential
difference that can be applied between points A and B.

**Reference:**

*Past Exam Paper – November 2014 Paper 43 Q6*

__Solution 1001:__**(a)**

{For capacitors in
parallel, total capacitance = C

_{1}+ C_{2}+ …}
For the two capacitors in parallel,
capacitance {= 48 + 48} = 96 μF

{For capacitors in series,
1 / total capacitance = 1/C

_{1}+ 1/C_{2}+ …}
For complete arrangement, 1 / C

_{T}= 1/96 + 1/48
C

_{T}= 32 μF**(b)**

The p.d. across the parallel
combination is one half the p.d. across single capacitor.

total p.d. {=
3V + 6V} = 9 V

__Question 1002: [Matter > Pressure > Liquids]__
A W-shaped tube contains two amounts of mercury, each open to the
atmosphere. Air at pressure P is trapped in between them. The diagram shows two
vertical distances x and y.

Atmospheric pressure is equal to the pressure that would be exerted by a
column of mercury of height 760 mm. The pressure P is expressed in this way.

Which values of x, y and P are possible?

x / mm y
/ mm P / mm of mercury

A 20 20 780

B 20 30 780

C 30 20 810

D 30 30 790

**Reference:**

*Past Exam Paper – June 2015 Paper 13 Q21*

__Solution 1002:__**Answer: B.**

This is a relatively complex
situation.

Atmosphere pressure P

_{A}= column of mercury of height 760mm
Compare the left part of the diagram
with the middle (left) part.

Since the position of the mercury
column is lower at the middle, the pressure P is greater than the atmospheric
pressure acting at the left ‘open’ tube.

P – P

_{A}= x
P – 760 = x (1)

Now, compare the right part of the diagram
with the middle (right) part.

P – P

_{A}= 50 – y
P – 760 = 50 – y

P + y = 810 (2)

The values form the table should
satisfy both equation (1) and (2).

Choice A:

Put x = 20 and P = 780 in eqn (1):
780 – 760 = 20 [correct]

Put y = 20 and P = 780 in eqn (2):
780 + 20 = 810 [incorrect]

Choice B:

Put x = 20 and P = 780 in eqn (1):
780 – 760 = 20 [correct]

Put y = 30 and P = 780 in eqn (2):
780 + 30 = 810 [correct]

Choice C:

Put x = 30 and P = 810 in eqn (2): 810
– 760 = 30 [incorrect]

Choice D:

Put x = 30 and P = 790 in eqn (1):
790 – 760 = 30 [correct]

Put y = 30 and P = 790 in eqn (2):
790 + 30 = 810 [incorrect]

__Question 1003: [Work, Energy and Power]__**(a)**Define power.

**(b)**A cyclist travels along a horizontal road. The variation with time t of speed v is shown in Fig. 3.1.

The cyclist maintains a constant
power and after some time reaches a constant speed of 12 m s

^{–1}.
(i) Describe and explain the motion
of the cyclist.

(ii) When the cyclist is moving at a
constant speed of 12 m s

^{–1}the resistive force is 48 N. Show that the power of the cyclist is about 600 W. Explain your working.
(iii) Use Fig. 3.1 to show that the
acceleration of the cyclist when his speed is 8.0 m s

^{–1}is about 0.5 m s^{–2}.
(iv) The total mass of the cyclist
and bicycle is 80 kg. Calculate the resistive force R acting on the cyclist
when his speed is 8.0 m s

^{–1}. Use the value for the acceleration given in (iii).
(v) Use the information given in
(ii) and your answer to (iv) to show that, in this situation, the resistive
force R is proportional to the speed v of the cyclist.

**Reference:**

*Past Exam Paper – November 2012 Paper 23 Q3*

__Solution 1003:__**(a)**Power is defined as the rate of doing work OR power = work done / time (taken) OR power = energy transferred / time (taken)

**(b)**

(i) As the speed increases drag /
air resistance increases. The resultant force reduces and hence the acceleration
is less. A constant speed is reached when the resultant force is zero.

(ii)

{For constant speed, the
resultant force is zero.}

Force F from cyclist = drag force /
resistive force

Power P {=
Fv} = 12 × 48

Power P = 576 W

(iii)

A tangent should be drawn at speed =
8.0 m s

^{–1}
{The gradient of the
speed-time graph gives the acceleration.}

Gradient values that show the acceleration
to be between 0.44 to 0.48 m s

^{–2}should be obtained
(iv)

{Resultant force = Force F
by cyclist – Resistive force R = ma.}

F – R = ma

{Power = Force × Speed. Force F = Power / Speed}

600/8 – R = 80 × 0.5 [using
P = 576] 576/8 – R = 80 × 0.5

Resistive force R = 75 – 40 = 35 N R = 72 – 40 = 32 N

(v)

At 12 m s

^{–1}, the resistive (drag) force R is 48 N, and at 8 m s^{–1}, the resistive (drag) force is 35 or 32 N
The ratio R / v can be calculated to
be 4 and 4 or 4.4 for both cases {48 / 12 = 4 and 32 /
8 = 4}

Thus, the resistive force R is
proportional to the speed v.

Please consider answering ALL of the following questions:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.5(b),Q.9(b)

41/O/N/09 Q.6(a),(b)(i),Q.10

41/M/J/10 Q.7(a)

51/M/J/10 Q.2(d)

For 41/M/J/10 Q.7(a), check solution 1005 at

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j15 p13 Ques 26, 29 ,37

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thank u in advance

For Q26, see solution 1014 at

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thank you very very much!!!!

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