Sunday, October 18, 2015

Physics 9702 Doubts | Help Page 213

  • Physics 9702 Doubts | Help Page 213


Question 1020: [Current of Electricity > Resistance of wire]
An electric shower unit is to be fitted in a house. The shower is rated as 10.5 kW, 230 V. The shower unit is connected to the 230 V mains supply by a cable of length 16 m, as shown in Fig. 6.1.

(a) Show that, for normal operation of the shower unit, the current is approximately 46 A.

(b) The resistance of the two wires in the cable causes the potential difference across the shower unit to be reduced. The potential difference across the shower unit must not be less than 225 V.
The wires in the cable are made of copper of resistivity 1.8 × 10–8 Ω m.
Assuming that the current in the wires is 46 A, calculate
(i) the maximum resistance of the cable,
(ii) the minimum area of cross-section of each wire in the cable.

(c) Connecting the shower unit to the mains supply by means of a cable having wires with too small a cross-sectional area would significantly reduce the power output of the shower unit.
(i) Assuming that the shower is operating at 210 V, rather than 230 V, and that its resistance is unchanged, determine the ratio
power dissipated by shower unit at 210 V / power dissipated by shower unit at 230 V

(ii) Suggest and explain one further disadvantage of using wires of small cross-sectional area in the cable.

Reference: Past Exam Paper – November 2007 Paper 2 Q6



Solution 1020:
(a)
Power = VI
Current, I = (10.5×103) / 230 = 45.7A

(b)
(i)
Potential difference across cable = (230 – 225 =) 5.0V
Resistance, R = (V/I =) 5.0 / 46
Resistance, R = 0.11Ω

(ii)
R = ρL / A
0.11 = [(1.8×10-8) × (16×2)] / A
Minimum area of cross-section of each wire, A = 5.3×10-6m2

(c)
(i)
EITHER Power = V2 / R        OR Power α V2
So, ratio = (210 / 230)2 = 0.83
{In this question, we are comparing the power dissipated when the unit is at 210V and 230V. It is given that the resistance is unchanged. So, we should use a formula that relates the power dissipated P to the resistance R and the voltage V. We cannot include other quantities that would also change when V is changed. This formula is P = V2 / R.
For example, we cannot use the formula P = VI because the current I will also change when V is changed. So, the comparison of the power dissipated depending on V will not be relevant because I is also changing.}

(ii) The resistance of the cable is greater. So, there is greater power loss / fire hazard / insulation may melt / wire may melt / cable gets hot









Question 1021: [Electromagnetism > Current-carrying wire]
The current in a long, straight vertical wire is in the direction XY, as shown in Fig. 6.1.

(a) On Fig. 6.1, sketch the pattern of the magnetic flux in the horizontal plane ABCD due to the current-carrying wire. Draw at least four flux lines.

(b) The current-carrying wire is within the Earth’s magnetic field. As a result, the pattern drawn in Fig. 6.1 is superposed with the horizontal component of the Earth’s magnetic field.
Fig. 6.2 shows a plan view of the plane ABCD with the current in the wire coming out of the plane.

The horizontal component of the Earth’s magnetic field is also shown.
(i) On Fig. 6.2, mark with the letter P a point where the magnetic field due to the current-carrying wire could be equal and opposite to that of the Earth.

(ii) For a long, straight wire carrying current I, the magnetic flux density B at distance r from the centre of the wire is given by the expression
B = μ0I / 2πr
where μ0 is the permeability of free space.
The point P in (i) is found to be 1.9 cm from the centre of the wire for a current of 1.7 A.
Calculate a value for the horizontal component of the Earth’s magnetic flux density.

(c) The current in the wire in (b)(ii) is increased. The point P is now found to be 2.8 cm from the wire.
Determine the new current in the wire.

Reference: Past Exam Paper – November 2009 Paper 41 Q6



Solution 1021:
(a) The flux lines should be concentric circles with increasing separation and a correct direction (anticlockwise) clear


(b)
(i) correct position to left of wire


(ii)
Magnetic flux density B = (4π × 10-7 × 1.7) / (2π × 1.9 × 10-2)
Magnetic flux density B = 1.8 × 10-5 T

(c)
{B = μ0I / 2πr. So, B ×r = μ0I. The distance r is proportional to the current I.}
distance current
{When current = 1.7A, distance of P from centre = 1.9cm
When distance of P from centre = 2.8cm, current = (2.8 / 1.9) × 1.7}
current = (2.8 / 1.9) × 1.7
current = 2.5 A









Question 1022: [Measurement > Uncertainty]
The volume V of liquid flowing in time t through a pipe of radius r is given by the equation
V / t = πPr4 / 8Cl
where P is the pressure difference between the ends of the pipe of length l, and C depends on the frictional effects of the liquid.
An experiment is performed to determine C. The measurements made are shown in Fig. 1.1.

(a) Calculate the value of C.

(b) Calculate the uncertainty in C.

(c) State the value of C and its uncertainty to the appropriate number of significant figures.

Reference: Past Exam Paper – June 2012 Paper 22 Q1



Solution 1022:
(a)
V / t = πPr4 / 8Cl
C = πPr4t / 8Vl
C = [π × 2.5×103 × (0.75×10-3)4] / (8 × 1.2×10-6 × 0.25)
C = 1.04×10-3 Nsm-2

(b)
ΔC/C = ΔP/P + 4(Δr/r) + Δ(V/t)/(V/t) + Δl/l
Percentage uncertainty in C, %C = ΔC/C × 100%
%C = %P + 4(%r) + %V/t + %l
%C = 2% + 5.3% + 0.83% + 0.4% = 8.6%
ΔC = (8.6/100) × 1.04×10-3 = ± 0.089×10-3 Nsm-2

(c)
C = (1.04 ± 0.09) × 10-3 Nsm-2                                                [A1]
{The actual uncertainty should always be given to 1sf. Then, the number of significant figures in the actual value of C is given to the same number of decimal place as in the uncertainty.
Now, consider the following example (as in November 2014 Paper 23 Q2)
Frequency = 530 ± 30 Hz
This is also correct as it could be written as (5.3 ± 0.3) × 102 Hz. In other words, for uncertainties, the ‘zeros’ are not considered as significant figures.}









Question 1023: [Kinematics > Linear motion]
A stone is thrown vertically upwards. The variation with time t of the displacement s of the stone is shown in Fig. 2.1.

(a) Use Fig. 2.1 to describe, without calculation, the speed of the stone from t = 0 to t = 3.0 s.

(b) Assume air resistance is negligible and therefore the stone has constant acceleration.
Calculate, for the stone,
(i) the speed at 3.0 s,
(ii) the distance travelled from t = 0 to t = 3.0 s,
(iii) the displacement from t = 0 to t = 3.0 s. 

(c) On Fig. 2.2, draw the variation with time t of the velocity v of the stone from t = 0 to t = 3.0 s.

Reference: Past Exam Paper – June 2015 Paper 22 Q2



Solution 1023:
(a) The speed decreases/stone decelerates to rest/zero at 1.25 s. The speed then increases/stone accelerates (in opposite direction)

(b)
(i)
Equation for uniformly accelerated motion:
v = u + at         (or s = ut + ½at2 and v2 = u2 + 2as)
{Consider the motion starting at time 1.25s. The ball is at its highest position where its speed is zero (this is taken as the initial speed u). The acceleration (due to gravity) is downwards. The duration of the motion until the time is 3.0s is given by (3.00 – 1.25) s.}
v = 0 + (3.00 – 1.25) × 9.81
v = 17.2 (17.17) m s–1

(ii)
s = ut + ½at2
From t = 0 to t = 1.25s: distance s = ½ × 9.81 × (1.25)2 [= 7.66]
From t = 1.25s to t = 3.0s (time taken = 3.0 – 1.25 = 1.75s)
From t = 1.25s to t = 3.0s: distance s = ½ × 9.81 × (1.75)2 [= 15.02]
(distance = 7.66 + 15.02)
Distance = 22.7 (22.69 or 23) m

(iii)
{From t = 0 to t = 1.25s: distance moved up = 7.66m
From t = 1.25s to t = 3.0s: distance moved down = 15.02m}
Displacement s = 15.02 – 7.66 = 7.4 (7.36) m
Displacement is down

(c) It is a straight line from positive value of v to t axis. The same straight line crosses t axis at t = 1.25 s. The same straight line continues with same gradient to t = 3.0 s.
{Since the acceleration is constant, the gradient of the velocity-time graph should be constant. We do not know the initial value of velocity at t = 0. It is taken to be a positive value (since the initial upwards displacement is taken as positive in the graph above). Since the acceleration of free fall opposes the initial motion of the stone, it would have a negative value. Thus, the gradient is negative.}


4 comments:

  1. For the question solution 1023, question (ii). the distance travelled from t=0 to t=1.25s, why is the u=0? The initial velocity at t=0 is not zero as the gradient of the graph at t=0 is not zero.. so why you use s=1/2at^2 instead of s=ut + 1/2at^2?

    ReplyDelete
    Replies
    1. we are considering the motion to start at 1.25s. This is at the highest position and at this point, the speed is zero.

      Delete
  2. Question1020 (b) (i) why is the length multiplied by 2? why isn't the area multiplied by 2 instead? (I mean, the wires are parallel so shouldn't it be that way instead)

    ReplyDelete
    Replies
    1. It's in series, not parallel.

      Try to see it like this: current flows through one wire, goes to the unit, then flow back through the other wire to the mains.
      So, they are in series.

      The resistance calculated previously was for both wires. So, we need to consider twice the length or half the area.

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 213