# 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 = (10.5×10

^{3}) / 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

^{-6}m^{2}**(c)**

(i)

EITHER Power = V

^{2}/ R OR Power α V^{2}
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 = V

^{2}/ 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 = μ

_{0}I / 2πr
where μ

_{0}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 = μ

_{0}I / 2πr. So, B × 2π**r**= μ_{0}**I**. 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 = πPr

^{4}/ 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 = πPr

^{4}/ 8Cl
C = πPr

^{4}t / 8Vl
C = [π × 2.5×10

^{3}× (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.

Frequency = 530 ± 30 Hz

This is also correct as it could be written as (5.3 ± 0.3)
× 10

^{2}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 + ½at

^{2}**and**v^{2}= u^{2}+ 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 + ½at

^{2}
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.

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?

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

DeleteNo Ut is zero as the Time t is zero, U is 12.26.

DeleteQuestion1020 (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)

ReplyDeleteIt's in series, not parallel.

DeleteTry 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.

in the solution of 1023 why isnt the velocity time graph starting from the origin. in the d-t graph at t=0 d also is 0

ReplyDeletethe stone is thrown: so it already had some velocity at time t = 0.

DeleteIf at time t = 0 the velocity is zero, it means that the stone would not move.

A sprinter runs a 100 m race in a straight line. He accelerates from the starting block at a constant acceleration of 2.5 m s–2 to reach his maximum speed of 10 m s–1. He maintains this speed until he crosses the finish line. Which time does it take the sprinter to run the race? A 4s B 10s C 12s D 20s. Answer is C, how?

ReplyDeleteGo to

Deletehttp://physics-ref.blogspot.com/2017/12/9702-june-2015-paper-12-worked.html