# Physics 9702 Doubts | Help Page 121

__Question 611: [Attenuation]__**(a)**State what is meant by

*attenuation*of a signal.

**(b)**Transmission cable has length of 30 km. Attenuation per unit length of the cable is 2.4 dB km

^{−1}.

Calculate, for a signal being
transmitted along the cable,

(i) total attenuation, in dB

(ii)

**ratio**of input power of signal**to**output power of signal**(c)**By reference to answers in (b), suggest why the attenuation of transmitted signals is usually expressed in dB

**Reference:**

*Past Exam Paper – June 2014 Paper 41 & 43 Q14*

__Solution 611:__**(a)**The attenuation of a signal is the reduction in power of the signal

**(b)**

(i)

{The attenuation per unit
length of the cable is 2.4 dB km

^{−1}. So, in 1km of cable, the signal is attenuated by 2.4 dB. For 30km of cable, attenuation = 30 (2.4) dB.}
Total attenuation = 30 (2.4) = 72dB

(ii)

{The total attenuation is
72 dB. So, the power of the output signal has been reduced by 72 dB when
compared to the power of the input signal. [This is the meaning of attenuation
as stated before.]

But attenuation is also
given by Gain / attenuation / dB = 10 lg(P

_{2}/P_{1}).}
Gain / attenuation / dB = 10 lg(P

_{2}/P_{1})
{The general formula is Gain / dB = 10 lg(P

_{2}/P_{1}). Note that this is only a formula for comparing 2 quantities, here P_{2}and P_{1}.
If there is a gain, P

_{OUT}should be greater than P_{IN}. (This is not the case here)
If there is an
attenuation, P

_{OUT}is less than P_{IN}(this is the case here). So, the ratio P_{OUT}/ P_{IN}would be a fraction with a value between 0 and 1. The log of such a value is negative. (What I have been explaining now is maths, not physics.) So, we should also take the value of 72 with a negative sign.
Like I said, the formula
is only comparing 2 quantities. Let’s say we say want the ratio P

_{IN}/ P_{OUT}. The value of this ratio would be greater than 1 since P_{IN}is greater than P_{OUT}. The log would give a positive value. That’s why we take 72 with a positive sign here.}
72 = 10 lg(P

_{IN}/P_{OUT}) or – 72 = 10 lg(P_{OUT}/P_{IN})
{Consider 72 = 10 lg(P

_{IN}/P_{OUT})
lg(P

_{IN}/P_{OUT}) = 72 / 10}
P

_{IN}/P_{OUT}= 10^{7.2}= 1.6x10^{7}}
Ratio (= P

_{IN}/P_{OUT}) = 1.6x10^{7}**(c)**Example

It enables smaller / more manageable
numbers to be used

The gains in dB for series
amplifiers are added, not multiplied

__Question 612: [Current of Electricity]__
Cell has electromotive force
(e.m.f.) E and internal resistance r. It is connected in series with a variable
resistor R, as shown in Fig.1.

**(a)**Define electromotive force (e.m.f.)

**(b)**Variable resistor R has resistance X. Show that

Power dissipated in resistor R /
power produced in cell = X / X+R

**(c)**Variation with resistance X of the power P

_{R}dissipated in R is shown in Fig.2.

(i) Use Fig.2 to state, for maximum
power dissipation in resistor R, magnitude of this power and the resistance of
R.

(ii) Cell has e.m.f. 1.5 V.

Use answers in (i) to calculate
internal resistance of the cell.

**(d)**In Fig.2, it can be seen that, for larger values of X, the power dissipation decreases. Use relationship in (b) to suggest one advantage, despite the lower power output, of using the cell in a circuit where the resistance X is larger than the internal resistance of the cell.

**Reference:**

*Past Exam Paper – November 2009 Paper 21 Q6*

__Solution 612:__**(a)**Electromotive force (e.m.f.) is the energy transferred from a source / changed from some form to electrical per unit charge (to drive a charge round a complete circuit).

**(b)**

Power in resistor R = I

^{2}X
{Ohm’s law: V = IR. Total
resistance = X+r}

__And__E = I (X+r)

Power in cell = EI

__and__algebra clear leading to ratio = X / (X+r)
{Power in R / power in
cell = I

^{2}X / EI = IX / E
From E = I (X+r), current
I = E / (X+r)

Power in R / power in cell
= [E / (X+r)] X / E = X / (X+r)}

**(c)**

(i)

Maximum
power = 1.4W

Resistance
= 0.40Î© (allow ± 0.05Î©)

(ii)

{Power, P = I

^{2}R so, I = √(P/R)}
Current
in the circuit = (√(P/R) =) √(1.4/0.4) = 1.87A

{E = I (X+r)}

1.5
= 1.87 (0.40 + r)

Internal
resistance, r = 0.40Î©

**(d)**

{Power dissipated = I

^{2}X = [E / (X+r)]^{2}X = E^{2}X / (X+r)^{2}.
Power in the cell = EI = E

^{2}/ (X+r).
From the equation obtained
for the power dissipated, for a greater value of X, the power dissipated is
less since the denominator is squared. So, less power is lost (there is greater
efficiency).}

EITHER Less power is lost / energy
wasted / lost OR There is greater efficiency (of energy transfer)

__Question 613: [Dynamics > Resultant force]__
Graph shows how total resistive
force acting on a train varies with its speed.

Part of this force is due to wheel
friction, which is constant. The rest is due to wind resistance.

What is the ratio (wind resistance /
wheel friction) at a speed of 200 km h

^{–1}?
A 4 B
5 C 8 D 10

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q16*

__Solution 613:__**Answer: A.**

The wheel friction is constant and
the wind resistance increases with speed.

When the speed is zero, the wind
resistance is also zero. So, wheel friction = 8 kN.

At a speed of 200 km h

^{–1}, the total resistive force is 40 kN.
Wheel friction = 8 kN

Wind resistance = 40 – 8 = 32 kN

Ratio = wind resistance / wheel
friction = 32 / 8 = 4

__Question 614: [Dynamics]__**(a)**State Newton’s first law.

**(b)**A log of mass 450 kg is pulled up a slope by a wire attached to a motor, as shown in Fig.1.

Angle that the slope makes with the
horizontal is 12°. Frictional force acting on the log is 650 N. The log travels
with constant velocity.

(i) With reference to motion of the log,
discuss whether the log is in equilibrium

(ii) Calculate tension in wire

(iii) State and explain whether the gain
in potential energy per unit time of the log is equal to output power of the
motor

**Reference:**

*Past Exam Paper – June 2012 Paper 22 Q3*

__Solution 614:__**(a)**Newton’s first law of motion states that a body continues at rest or constant velocity unless acted on by a resultant (external) force.

**(b)**

(i) The log moves with constant
velocity (zero acceleration), so no resultant force acts on it. Since there is
no resultant force (and no resultant torque), the log is in equilibrium.

(ii)

{Weight W of the log = mg
= 450 (9.81)

From the laws of angles,
the weight (which is vertically downwards) is at an angle of 12° to the line
perpendicular to the slope.

So, the component of the
weight along the slope is W sin12.}

__Component of weight of the log acting down the slope__= (450)(9.81)sin(12) (=917.8)

{That component of the weight
acts down along the slope. The log is being pulled up along the slope by the
motor. So, the direction of motion of the log is up along the slope while the
component of the weight discussed above is down along the slope. Friction,
which always opposes motion, will also be down along the slope since the log is
moving up along the slope.

For zero acceleration, the
resultant force along the slope should be zero.

Upwards forces along slope
= Downwards forces along slope

Tension = Friction +
Component of weight}

Tension = 650 + 450gsin(12) = (650 +
917.8)

Tension = 1600 (1570) N

(iii) Work is done against frictional force
(friction between log and slope). So, the output power is greater than the gain
in potential energy per unit time.

__Question 615: [Ultrasound]__**(a)**By reference to ultrasound waves, state what is meant by

*specific acoustic impedance*of a medium.

**(b)**A parallel beam of ultrasound of intensity I is incident normally on a muscle of thickness 3.4 cm, as shown in Fig.1.

The ultrasound wave is reflected at
a muscle-bone boundary. Intensity of the ultrasound received back at the
transducer is I

_{R}.
Some data for bone and muscle are
given in Fig.2.

specific
acoustic impedance / kg m

^{−2}s^{−1}linear absorption coefficient / m^{−1}
bone 6.4 × 10

^{6}130
muscle 1.7 × 10

^{6}23
(i) Intensity reflection coefficient
Î± for two media having specific acoustic impedances Z

_{1}and Z_{2}is given by
Î± = (Z

_{1}− Z_{2})^{2}/ (Z_{1}+ Z_{2})^{2}.
Calculate fraction of the ultrasound
intensity that is reflected at the muscle-bone boundary.

(ii) Calculate fraction of the
ultrasound intensity that is transmitted through a thickness of 3.4 cm of
muscle.

(iii) Use answers in (i) and (ii) to
determine the ratio I

_{R}/ I**Reference:**

*Past Exam Paper – November 2014 Paper 43 Q11*

__Solution 615:__**(a)**The specific acoustic impedance of a medium is the product of the density of the medium and the speed of the wave in the medium

**(b)**

(i) Intensity reflection coefficient
Î± = (6.4 – 1.7)

^{2}/ (6.4 + 1.7)^{2}= 0.34
(ii) I / I

_{0}= e^{–Î¼x}= exp (–23 × 3.4 × 10^{–2}) = 0.46
(iii)

{A fraction of the
intensity is lost on transmission and a fraction is lost on reflection. There
are two transmission losses and one reflection loss (the ultrasound is
transmitted, reflected at the boundary, and finally transmitted back), and each
loss is cumulative, so it is necessary to multiply the fractions together.

So, we consider I / I

_{0}twice – that is, (I / I_{0})^{2}and Î± once. We multiply them all.}
Ratio of I

_{R}/ I (= [(I / I_{0}]^{2}× Î±) = (0.46)^{2}× 0.34 = 0.072

__Question 616: [Simple Harmonic motion]__**(a)**Define

*simple harmonic motion*.

**(b)**A horizontal plate is vibrating vertically, as shown in Fig.1.

Plate undergoes simple harmonic
motion with a frequency of 4.5 Hz and amplitude 3.0 mm.

A metal cube of mass 5.8 g rests on
the plate.

Calculate, for the cube, the energy
of oscillation.

**(c)**Amplitude of oscillation of the plate in (b) is gradually increased. The frequency remains constant.

At one particular amplitude, the
cube just loses contact momentarily with the plate.

(i) State position of the plate in
its oscillation at the point when the cube loses contact.

(ii) Calculate this amplitude of
oscillation.

**Reference:**

*Past Exam Paper – November 2011 Paper 43 Q3*

__Solution 616:__**(a)**In simple harmonic motion, the acceleration is proportional to the displacement / distance from fixed point and is in opposite directions / directed towards fixed point.

**(b)**

{The cube is a temporarily
at rest when the displacement is maximum (amplitude)}

Energy of oscillations at rest
position = ½ mÏ‰

^{2}x_{0}^{2}and Ï‰ = 2Ï€f
Energy of oscillations = ½ × (5.8 ×
10

^{–3}) × (2Ï€ × 4.5)^{2}× (3.0 × 10^{–3})^{2}= 2.1 × 10^{–5}J**(c)**

(i)

{The reaction force
between the cube and the plate is a minimum when the plate is at its maximum
displacement above its mean position. This will, therefore, be the position of
the plate when the cube just loses contact.}

The cube loses contact at the
maximum displacement above the rest position.

(ii)

{As stated previously, at
the maximum displacement, the plate is temporarily at rest. The plate would be
at its maximum displacement when its acceleration is 9.81 m s

^{–2}. This is because the acceleration of free fall, g, is always present. The plate has reached its maximum displacement above the mean position, so it won’t be moving any more up. Actually, it is temporarily at rest because the downward acceleration due to gravity has reduced its upwards velocity to zero.}
Acceleration = (–)Ï‰

^{2}x_{0}and acceleration = 9.81 or g
9.81 = (2Ï€ × 4.5)

^{2}× x_{0}
Amplitude x

_{0}= 1.2 × 10^{–2}m
for solution 612, how did you get the power off the cell as EI, did you use the formula P=VI? And if you did, why can't we use the same formula to find the power of the resistor R but instead its I^2R

ReplyDeletePower = VI

DeleteFor the cell, the p.d. V across its terminal is equal to the e.m.f. E of the cell.

For a resistor, V is the p.d. across the resistor and I is the current through the resistor. From Ohm's law: V = IR. So, we can replace V = IR in the equation P = VI. This gives P = I^2 R

This symbol for different significance for the different components we are calculating

The symbol HAS different ...

DeleteThanks a lot man

ReplyDeletesolution of 616 c 2 ,, even if the plate doesnt go up any further how do I know that is the point when the cube loses contact and not any earlier?

ReplyDelete