Physics 9702 Doubts | Help Page 141
Question 695: [Sensing
devices]
(a) State name of an electrical sensing device that
will respond to changes in
(i) length
(ii) pressure
(b) Relay is sometimes used as the output of a sensing
circuit.
Output of particular sensing circuit
is either +2V or -2V.
On Fig.1, draw symbols for relay and
any other necessary component so that the external circuit is switched on only
when the output from the sensing circuit is +2V.
Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q10
Solution 695:
(a)
(i) Strain gauge
(ii) Piezo-electric / quartz crystal
/ transducer
(b) For the circuit, a coil of relay is connected
between the sensing circuit output and earth and the switch is across the
terminals of the external circuit.
A diode is connected in series with
the coil with a correct polarity for the diode (downwards). A second diode with
correct polarity is connected (upwards and parallel to coil of relay).
{Note that the following
explanations have been taken directly from the application booklet.
The diode D1
conducts only when the output is positive with respect to earth and thus the relay
coil is energized only when the output is positive. When the current in the relay
coil is switched off, a back e.m.f. is generated in the coil that could damage
the sensing circuit. A diode D2 is connected across the coil to
protect the sensing circuit from this back e.m.f.}
Question 696: [Current
of Electricity]
(a) A wire has length 100 cm and diameter 0.38 mm. Metal of the wire
has resistivity 4.5 × 10–7 Ω m.
Show that resistance of the wire is
4.0 Ω.
(b) The ends B and D of the wire in (a) are connected to a cell X, as
shown in Fig.1.
Cell X has electromotive force
(e.m.f.) 2.0 V and internal resistance 1.0 Ω.
A cell Y of e.m.f. 1.5 V and
internal resistance 0.50 Ω is connected to the wire at points B and C, as shown
in Fig.1.
Point C is distance l from
point B. The current in cell Y is zero. Calculate
(i) current in cell X,
(ii) potential difference (p.d.)
across the wire BD,
(iii) distance l.
(c) Connection at C is moved so that l is increased. Explain
why the e.m.f. of cell Y is less than its terminal p.d.
Reference: Past Exam Paper – November 2014 Paper 23 Q6
Solution 696:
Go toA wire has length 100 cm and diameter 0.38 mm. The metal of the wire has resistivity 4.5 × 10–7 Ω m.
Question 697: [Measurement
> Prefixes]
A
signal has frequency of 2.0 MHz.
What
is period of the signal?
A 2
μs B 5 μs C 200 ns D 500 ns
Reference: Past Exam Paper – November 2010 Paper 11 Q1 & Paper 13 Q4
Solution 697:
Answer: D.
Frequency f = 2.0 MHz = 2.0×106 Hz
Period = 1 / f = 1 / (2.0×106) = 0.5×10-6 = 500×10-9
= 500ns
Question 698: [Kinematics
+ Dynamics]
A helicopter has a cable hanging
from it towards the sea below, as shown in Fig.1.
A man of mass 80 kg rescues child of
mass 50.5 kg. The two are attached to the cable and are lifted from sea to the
helicopter. Lifting process consists of an initial uniform acceleration
followed by a period of constant velocity and then completed by a final uniform
deceleration.
(a) Calculate combined weight of the man and child.
(b) Calculate tension in the cable during
(i) initial acceleration of 0.570 m
s–2,
(ii) period of constant velocity of
2.00 m s–1.
(c) During final deceleration the tension in the cable is 1240 N.
Calculate this deceleration.
(d) (i) Calculate time over which the man and child are
1. moving with uniform acceleration,
2. moving with uniform deceleration.
(ii) Time over which the man and
child are moving with constant velocity is 20 s. On Fig.2, sketch a graph to
show variation with time of the velocity of the man and child for the complete
lifting process.
Reference: Past Exam Paper – June 2011 Paper 23 Q3
Solution 698:
(a) {Combined mass = 80 + 50.5 = 130.5kg}
Combined weight = m × g = 130.5 ×
9.81 = 1280 N
(b)
(i)
Resultant force F = ma
{Tension is upwards while
the weight acts downwards. The resultant force / acceleration is upwards.}
T – 1280 = 130.5 × 0.57
Tension T = 1280 + 74.4 = 1350 N
(ii) Tension T = 1280 N
{For constant velocity, the resultant force is zero. Thus, tension =
weight.}
(c)
{Resultant force F = ma
For a deceleration, the
direction of the resultant force should be opposite to the direction of motion.
Motion is upwards, so the resultant force is downwards.
Tension – Weight =
Resultant force = ma (a is negative
here)}
1240 – 1280 = 130.5 × a
a = (–) 0.31 m s–2
(d) (i)
1.
{Initial acceleration a = 0.570
m s–2. Initial velocity, u = 0. Final velocity, v = 2.00 m s–1.
a = (v – u) / t}
Time t {= (v
– u) / a = 2.00 / 0.570} = 3.5s
2.
{Deceleration a = - 0.31 m
s–2. Initial velocity, u = 2.00ms-1. Final velocity, v= 0.
a = (v – u) / t}
Time t {= (v
– u) / a = -2.00 / -0.31} = 6.5s
Question 699:
[Alternating Current]
An ideal transformer has 5000 turns
on its primary coil. It is to be used to convert mains supply of 230 V r.m.s.
to an alternating voltage having a peak value of 9.0 V.
(a) Calculate number of turns on the secondary coil.
(b) Output from the transformer is to be full-wave rectified. Fig.1
shows part of the rectifier circuit.
On Fig.1, draw
(i) diode symbols to complete
diagram of the rectifier such that terminal A of the resistor R is positive
with respect to terminal B,
(ii) symbol for a capacitor connected
to provide smoothing of potential difference across the resistor R.
(c) Fig.2 shows variation with time t of the smoothed potential
difference V across the resistor R.
(i) State interval of time during
which the capacitor is being charged from the transformer.
(ii) Resistance of the resistor R is
doubled. On Fig.2, sketch variation with time t of the potential difference V
across the resistor.
Reference: Past Exam Paper – June 2007 Paper 4 Q4
Solution 699:
(a)
r.m.s. output = 9/√2 or peak input =
230√2
NS / NP = VS
/ VP
{NS / 5000 =
(9/√2) / 230 or NS /
5000 = 9 / 230√2}
NS = 138 → 140 turns
(b)
(i) The four diodes should be correctly
positioned (regardless of output polarity) giving correct output polarity (all
‘point to left’)
{The diodes should be
drawn such that current always passes point A first, before passing point B.}
(c)
(i) time t1 to time t2
(ii) The sketch has the same peak
values with the ripple reduced and a reasonable shape
11/O/N/10 Q.31
ReplyDelete12/O/N/10 Q.26,33
For 11/O/N/10 Q.31, see solution 704 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-142.html
In solution 699, the answer comes to be different if I use the peak input rather than the rms output, please help.
ReplyDeleteIf you use the peak input, you should use 9V for the output. Both methods are given in red above.
DeleteLook at the division signs correctly.
Also, can you please explain the last part of the same question, why has the ripple reduced? Your help would be much appreciated
ReplyDeleteIntroducing a resistor causes the smoothing effect on the output potential (see your notes for further details). This causes the ripples to be reduced - that is, the potential TENDS towards a straight line (though it is not a straight line).
DeleteSo, increasing the resistance causes the smoothing to be more effective. The potential approaches a straight line even more.
Is there a difference between increasing capacitance and increasing resistance on smoothing? Because I remember doing a past paper question where the only acceptable answer for better smoothing was increasing C. THe mark scheme rejected increasing R.
ReplyDeleteThe effect of resistance is shown above.
DeleteThe bigger the C, the less the ripple will be.
Try to view the question again. Maybe it was a bit different
In Q669 b(i) Can the diode drawn in the bottom right be in the opposite direction?
ReplyDeleteno, else the current would have two possible path to flow
DeleteI still didn't get the last part (c) of question 696. Can you please explain again. I'll be very much grateful.
ReplyDeleteyou need to tell me what exactly you did not get.
Deleteanyway, this part was discounted
In the question 699 part c(ii),
ReplyDeleteas V=IR,when resistance(R) doubles current decreases as they both are inversly propotional.so when current decreases the current flowng through the resisitor decreases and hence the discharging rate of the capacitor should decrease.so is that the graph should be drawn below the original graph?
There is a p.d. across the capacitor (and the same p.d. is across the resistor as they are in parallel) and this allows current to flow through the resistor. As current flows, the p.d. decreases as the capacitor is discharging.
DeleteIf the resistance is increased, the flow of current decreases and so, the capacitor discharges more slowly. Instead of decreasing by a large amount, the p.d. decreases only a little (as shown in the graph).
Awesome!!! Thanks!
DeleteFor solution 699 (c)(i) How the time during which the capacitor is being charged from the transformer is t1 to t2?
ReplyDeleteWhen the capacitor is being charged, the p.d. across it increases. This corresponds to t1 to t2.
Delete695 b.
ReplyDeletenothing is stopping the current to flow to external circuit. we can have both +2 or -2v at terminals of external circuit. please review the design of circuit
We are making sure that the current flows only through the relay by arranging the diodes in this way.
DeleteThe current itself will not flow through the external circuit. It only flows through the relay which closes the switch of the external circuit (which usually requires a high voltage).
for solution 696,(Current of electricity), why is the e.m.f of cell x divided by 5 to find the current?
ReplyDeletewhere did this 5 come from and why?
detailed have been added
Delete