Physics 9702 Doubts | Help Page 135
Question 671: [Radioactivity]
(a) Describe structure of an atom of the nuclide 23592U.
(b) Deflection of α-particles by a thin metal foil is investigated
with the arrangement shown in Fig.1. All the apparatus is enclosed in a vacuum.
The detector of α-particles, D, is
moved around path labelled WXY.
(i) Explain why apparatus is
enclosed in a vacuum.
(ii) State and explain readings
detected by D when it is moved along WXY.
(c) A beam of α-particles produces a current of 1.5 pA. Calculate
number of α-particles per second passing a point in the beam.
Reference: Past Exam Paper – November 2012 Paper 21 Q6
Solution 671:
(a) An atom of the nuclide 23592U has 92 protons
in the nucleus and 92 electrons around nucleus. It also has 143 neutrons (in
the nucleus).
(b)
(i) The α-particle travels short
distance in air.
(ii) There is a very small
proportion in the backwards direction / at large angles. The majority pass
through the metal foil with no /small deflections.
EITHER most of the mass is in a very
small volume (nucleus) and is charged
OR most of atom is empty space
(c)
Current I = Q / t
{The current is due to a
number n of α-particles. An α-particle is a helium nucleus
and thus has a charge of +2e.
Charge q of 1 α-particle = 2 × (1.6 × 10–19) C
Total charge Q = nq
Current I = (nq) / t = (n/t)
q
(n/t): number of
α-particles per second passing a point in the beam
(n/t) = I / q}
n / t = (1.5 × 10–12) /(
2 × 1.6 × 10–19) = 4.7 × 106 s–1
Question 672: [Work,
Energy and Power]
A man has mass of 80 kg. He ties
himself to one end of a rope which passes over a single fixed pulley. He pulls
on other end of the rope to lift himself up at an average speed of 50 cm s–1.
What is average useful power at
which he is working?
A 40 W B 0.39 kW C
4.0 kW D 39 kW
Reference: Past Exam Paper – June 2011 Paper 13 Q17
Solution 672:
Answer: B.
Average speed = 50 cm s–1
= 0.5 m s–1
Since we are considering an average
speed, the acceleration on the system is zero. So, the resultant force on the rope
is zero.
The weight of the man (= mg) acts downwards.
So, the man should pull on the rope with an equal and opposite force, for the
resultant to be zero.
Power = Fv = (80×9.81)
× 0.5 = 392 W = 0.39 kW
Question 673: [Work,
Energy and Power]
(a) State what is meant by work done.
(b) Trolley of mass 400 g is moving at a constant velocity of 2.5 m s–1
to the right as shown in Fig.1.
Show that kinetic energy of the
trolley is 1.3 J.
(c) Trolley in (b) moves to point P as shown in Fig.2.
At point P speed of the trolley is
2.5 m s–1.
A variable force F acts to the left
on trolley as it moves between points P and Q.
Variation of F with displacement x
from P is shown in Fig.3.
Trolley comes to rest at point Q.
(i) Calculate distance PQ.
(ii) On Fig.4, sketch variation with
x of velocity v for the trolley moving between P and Q.
Reference: Past Exam Paper – November 2013 Paper 22 Q3
Solution 673:
(a) Work done on a body is the product of the force applied on the body
multiplied by the distance moved (displacement) by the body in the direction of
the force.
(b)
Mass = 400g = 0.4 kg
Kinetic energy = ½ mv2 =
½ × 0.4
×
(2.5)2 = 1.25J ≈ 1.3J
(c)
(i)
Acceleration is not
constant. Equation for uniformly accelerated motion cannot be used.
Let x be the distance between P and
Q.
(from previous part) Work done = 1.3 J (or 1.25J)
Also, Work done = area under graph =
½ Fx
1.3 = (14x) / 2
So, x = 0.19m
Or, if Work done is taken to be 1.25J, x = 0.179m = 0.18m
(ii)
{From Fig.3, the force
increases with distance, x. So, a straight line (which represents a constant
acceleration) is incorrect.
A connection should be
made between the increase in the opposing force, F with the distance. This is
shown by an increase in gradient in the in variation of velocity with distance.
That is, the speed decreases by a greater value over the same amount of
distance as x increases.
Note that in a graph, a
vertical straight line corresponds to a gradient = infinity while a horizontal
straight line corresponds to a gradient = 0. Recall that gradient = ∆y/∆x.}
Take the origin to be at P.
At P, v = 2.5ms-1, x =
0m. And at Q, v = 0ms-1, x = 0.19m.
Since force increases with distance,
the curve will have an increasing gradient.
Question 674: [Direct Sensing]
An
electronic sensor may be represented by the block diagram of Fig.1.
(a)
State function of the processing unit.
(b) Student designs a sensing unit for temperature
change. A 4 V supply, a fixed resistor of resistance 2.5 kΩ and thermistor are
available. Thermistor has resistance 3.0 kΩ at 6 °C and resistance 1.8 kΩ at 20
°C.
Complete circuit diagram of Fig.2 to
show how the resistor and the thermistor are connected to provide an output
that is greater than 2 V at 6 °C and less than 2 V at 20 °C. Mark clearly the
output VOUT.
(c) Suggest two uses of relay as part of an output
device.
Reference: Past Exam Paper – November 2013 Paper 41 & 42 Q9
Solution 674:
(a)
The processing unit operates on/takes signal
from a sensing device so that it gives an output voltage.
(b)
The
thermistor and resistor should be in series between +4V line and earth.
Vout
shown clearly across thermistor
{The
lower line (Earth) is at 0V and the upper line is at +4V. As we go from the
upper line towards the Earth symbol, the potential should decrease.
The resistance
of the thermistor 1.8 kΩ at 20 °C and 3.0 kΩ at 6 °C. The resistor has a fixed resistance
of 2.5 kΩ. By connecting the 2 resistors in series, a potential divider circuit
is formed.
The
circuit is to be drawn such that at 20 °C, the output voltage is less than 2V. At
20 °C, the temperature of the thermistor is less than at 6 °C, so the p.d.
across it will be less. Actually, it would be [1.8 / (1.8+2.5)] × 4 = 1.67.
This is only a difference in potential (p.d.).
At
this temperature, if the thermistor is connected to the upper line of +4V, then
the upper terminal of the thermistor would be at +4V and its lower terminal (which
is connected to the fixed resistor) would be at a potential of 4 – 1.67 = 2.33V.
This is greater than 2V, so the output cannot be drawn from the upper line (of
+4V) to the lower terminal of the thermistor. Also, the upper terminal of the
fixed resistor is at 2.33V and its lower terminal (connected to Earth symbol)
is at 0V. So, the output cannot be drawn between these 2 points too.
The
correct way is to connect the thermistor to Earth. Since the p.d. across it at
the temperature of 20 °C is 1.67V, the upper terminal of the thermistor would
be at 1.67V and its lower terminal at 0V. So, the output can be drawn across
these 2 points.}
(c) Choose any 2:
Remote
sensing
Switching
large current by means of a small current
Isolating
circuit from high voltage
Switching
high voltage by means of a small voltage/current
Here is the remaining question for now:
ReplyDelete21/M/J/13 Q.6(c)(d)
Check solution 676 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-136.html
Sir, why aren't you posting any worked solutions for Paper 5 especially the diagram in question 1.
ReplyDeleteI don't plan to solve the practical papers right now. Maybe later (a lot of time later, actually). Drawing on the computer takes too much time, ...
DeleteInstead of drawing on the computer, you can draw on a piece of paper and take a picture of to upload it on this website or on your Facebook page. I really need an urgent help in paper 5 since I am giving my A Level in the October/November 2015 session.
ReplyDeleteI'll think about it. But I don't think I do this right now. Anyway, use the solutions for the other papers in the meantime.
DeleteBut check again, if I ever if the questions, you could read them.
Don't worry this idea had been tried before by so many people and the written working solutions were clear enough to be read. Just don't picture the whole A4 page in one snapshot. For one snapshot it should not be taken for more than half an A4 sheet. It has got to work very well with you. Go to this page to see an example: https://www.facebook.com/alvlmaths?fref=ts
ReplyDeleteThat's not really the issue. Originally, I intended to delay the paper 5 for later. + For other reasons, ...
ReplyDeleteCan you please tell me by which month this year do you plan to start with paper 5 ?
ReplyDeleteIt depends. I don't know when I'll start the paper 5. It may be this year, or it may be next year or even later.
DeleteI don't recommend you to depend on this blog for papers right now.
Isn't that Decreasing gradient in solution 673?
ReplyDeleteno, it is increasing but has a negative sign.
DeleteA tangent at Q would be steeper than one at P. thus, the gradient is greater at Q.
n Q 672, why average speed is considered as Zero acceleration? Aint Constant / terminal speed only means Zero acceleration. ?
ReplyDelete