# Physics 9702 Doubts | Help Page 142

__Question 700: [Electromagnetism]__
A stiff straight copper wire XY is
held fixed in uniform magnetic field of flux density 2.6 × 10

^{−3}T, as shown in Fig.1.
Wire XY has length 4.7 cm and makes
an angle of 34° with the magnetic field.

**(a)**Calculate force on the wire due to a constant current of 5.4 A in the wire.

**(b)**Current in the wire is now changed to an alternating current of r.m.s. value 1.7 A. Determine total variation in the force on the wire due to the alternating current.

**Reference:**

*Past Exam Paper – November 2014 Paper 41 & 42 Q6*

__Solution 700:__**(a)**Force F = BIL sinθ = (2.6 × 10

^{–3}) × 5.4 × (4.7 × 10

^{–2}) × sin 34° = 3.69 × 10

^{–4}N

**(b)**

Peak current = 1.7 × √2 = 2.4 A

Maximum force = (2.6 × 10

^{–3}) × 2.4 × (4.7 × 10^{–2}) × sin 34° = 1.64 × 10^{–4}N
{Force is a vector, it has
both a magnitude and a direction. Since an a.c. is applied, the force will be
maximum in one direction, then reduces to zero and then has a maximum value in
the opposite direction. The magnitude of the force from the zero to the maximum
value is as calculated above. But the total variation in the force is from the
maximum value in one direction to the maximum value in the opposite direction. So,
we take twice the value above.}

Total variation in force = 2 × (1.64
× 10

^{–4}) = 3.3 × 10^{–4}N

__Question 701: [Work, Energy and Power]__
Steel sphere is dropped vertically
onto a horizontal metal plate. Sphere hits the plate with a speed u, leaves it
at a speed v, and rebounds vertically to half of its original height.

Which expression gives the value of
v / u ?

A 1 / 2

^{2 }B 1 / 2 C 1 / √2 D 1 – 1/√2**Reference:**

*Past Exam Paper – June 2011 Paper 11 Q14*

__Solution 701:__**Answer: C.**

Let the distance fallen by the steel
sphere be s.

For the fall,

Kinetic energy just before hitting
the plate = Potential energy at maximum height

½ mu

^{2}= mgs
Speed u = √(2gs) = √2 (√gs)

For the sphere after rebound,

Kinetic energy just after hitting
the plate = Potential energy at maximum height

½ mv

^{2}= mg (s/2)
Speed v = √(gs)

Ratio = v / u = 1 / √2

__Question 702: [Direct Sensing]__
Student designs an electronic sensor
to monitor whether temperature in a refrigerator is above or below a particular
value. The circuit is shown in Fig.1.

**(a)**Name components used in the output device.

**(b)**An operational amplifier (op-amp) is used as the processing unit. Describe function of this processing unit.

**(c)**State function of

(i) resistors C and D,

(ii) resistor B.

**(d)**Output device of the circuit in Fig.1 is changed so that the new output device may be used to switch on a high-voltage circuit.

(i) State component that is used in
the new output device.

(ii) Draw on Fig.2 to show how
component in (i), together with a diode, are connected so that the high voltage
may be switched on when the output of the op-amp is negative.

**Reference:**

*Past Exam Paper – November 2012 Paper 43 Q9*

__Solution 702:__**(a)**Light-emitting diode

**(b)**The op-amp gives a high or a low output / +5 V or –5 V output that is dependent on which of the inputs is at a higher potential.

**(c)**

(i) This provides a reference/constant
potential

(ii) It determines the temperature
of the ‘switch-over’

**(d)**

(i) A relay

(ii) The relay should be connected
correctly for op-amp output and high-voltage circuit and the diode with correct
polarity in output from op-amp

{Current flows from a
point of higher potential to a point of lower potential. Since the output of
the op-amp is negative, current flows from earth to the op-amp. So, the flow is
upwards. Thus, the diode should point upwards.}

__Question 703: [Vectors]__**(a)**Explain the differences between the quantities

*distance*and

*displacement*.

**(b)**State Newton’s first law.

**(c)**Two tugs pull a tanker at constant velocity in direction XY, as represented in Fig.1.

Tug 1 pulls tanker with force T

_{1}at 25.0° to XY. Tug 2 pulls tanker with force of T_{2}at 15.0° to XY. The resultant force R due to the two tugs is 25.0 × 10^{3}N in direction XY.
(i) By reference to the forces
acting on the tanker, explain how tanker may be described as being in
equilibrium

(ii)

1. Complete Fig.2 to draw a vector
triangle for forces R, T

_{1}and T_{2}
2. Use vector triangle in Fig.2 to
determine magnitude of T

_{1}and of T_{2}**Reference:**

*Past Exam Paper – June 2012 Paper 23 Q1*

__Solution 703:__**(a)**

Displacement is a vector while
distance is a scalar.

The displacement is the straight
line between two points / distance is the sum of lengths moved / example
showing difference

**(b)**Newton’s first law states that a body continues at rest or at constant velocity unless acted on by a

__resultant__(external) force.

**(c)**

(i) The sum of T

_{1}and T_{2}equals the frictional force. These two forces are in opposite directions.
(ii)

1. A scale vector triangle should be
drawn with correct orientation / vector triangle with correct orientation both
with arrows. The scale should be given or a mathematical analysis for the
tensions

{

__Adding vectors graphically:__
Adding vectors graphically
is done by

1. firstly drawing the 1

^{st}vector (can be any of the vectors present),
2. then drawing the next
vector, with its

**tail at the head of the 1**.^{st}vector
3. The resultant vector
formed is a straight line which starts from the

**tail of the first vector and ends at the head of the last vector**.
(So, the tail of the
resultant vector starts at the tail of the 1

^{st}vector drawn and its head is at the head of the last vector drawn)
Consider the diagram below
as an example: }

{First, for T

_{2}, lightly draw a line (long enough) at an angle of 15^{o}to the horizontal (at the right end – by using a protractor). This can be either above or below the horizontal resultant line.
Then for T

_{1}, again lightly draw a line (long enough) at an angle of 25^{o}{from the rules of angles} to the horizontal (at the left end of the horizontal resultant line). This is drawn on the same side (either above or below the horizontal line) as T_{1}was drawn.
The point of intersection
indicates the end of the arrow for T

_{2}and the start of the arrow for T_{1}. Draw arrows on each of them and indicate clearly the directions of the vectors and their angles. An angle of about 140^{o}should be formed between T_{1}and T_{2}.
Erase any unnecessary
lines beyond the point of intersection.

[Alternatively, T

_{1}could be drawn first. …]^{3}N force is of length of about 12.3cm. So, 1.0cm represents about 2.0325N. The lengths of the vectors for T

_{2}and T

_{1}should be ABOUT 8.07cm and 4.97cm respectively. By using proportion, the answers of part (ii) can be obtained.}

2.

Magnitude of T

_{1}= 10.1 x 10^{3}(± 0.5 x 10^{3}) N
Magnitude of T

_{2}= 16.4 x 10^{3}(± 0.5 x 10^{3}) N

__Question 704: [Current of Electricity > Charge]__
Current in the circuit shown is 4.8
A.

What is the direction of flow and
rate of flow of electrons through the resistor R?

direction
of flow rate of flow

A X
to Y 3.0
× 10

^{19}s^{–1}
B X
to Y 6.0
× 10

^{18}s^{–1}
C Y
to X 3.0
× 10

^{19}s^{–1}
D Y
to X 6.0
× 10

^{18}s^{–1}**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q31 & Paper 13 Q32*

__Solution 704:__**Answer: C.**

Electrons flow from the negative
terminal of the cell (the shorter vertical arrow for the cell symbol). So, the
direction of flow of electrons is from Y and X. [A and B are incorrect]

Charge Q = It

Charge Q = ne

where n is the number of electrons
and e is the charge of an electron

ne = It

Rate of flow of electrons = n / t

Rate of flow of electrons = I / e = 4.8
/ (1.6×10

^{-19}) = 3.0 × 10^{19}s^{–1}
12/O/N/10 Q.26,33

ReplyDelete11/M/J/11 Q.9,10

For 12/O/N/10 Q.26, go to

Deletehttp://physics-ref.blogspot.com/2014/11/9702-november-2010-paper-12-worked.html

Plz explain ON 09 question 8 part c (ii)?

ReplyDeletewhich paper? which variant?

Deletecan u solve q4 from w13 paper 12?

ReplyDeleteGo to

Deletehttp://physics-ref.blogspot.com/2014/06/9702-november-2013-paper-11-12-worked.html

please explain october/november 2012, variant-11, Q.21 (paper -1)?

ReplyDeleteCheck solution 803 at

Deletehttp://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-161.html

on page 142 of your blog on question 700

ReplyDeletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-142.html

link^

i do not understand why you have multiplied 1.7 with underoot 2 (2^0.5)

1.7 is the rms current. To find the peak current, we need to multiply by root 2.

Delete