Physics 9702 Doubts | Help Page 64
Question 357: [Current
of Electricity > Internal resistance]
Diagram shows two circuits. In these
circuits, only internal resistances differ.
Which line in table is correct?
Reference: Past Exam Paper – November 2003 Paper 1 Q34
Solution 357:
Answer: B.
The current in circuit X is greater
than that in circuit Y since the total resistance in circuit Y is greater than
that in X.
Current in circuit X = 1.5 / (3.0 +
0.5) = 0.43A
Current in circuit Y = 1.5 / (3.0 +
2.0) = 0.30A
From Ohm’s law, potential difference
across the 3.0Ω resistor = IR. So, this would be
greater in circuit X. [C and D are incorrect]
p.d. across
3.0Ω resistor in circuit X = 0.43 (3) = 1.29V
p.d. across
3.0Ω resistor in circuit X = 0.30 (3) = 0.9 V
Power dissipated in 3.0Ω resistor = I2R . Again this would
be greater in circuit X. [A is incorrect]
Power dissipated in 3.0Ω resistor in circuit X = (0.43)2 (3)
= 0.55W
Power dissipated in 3.0Ω resistor in circuit Y = (0.30)2 (3)
= 0.27W
Question 358: [Radioactive
decay]
Radiation from a radioactive source
enters an evacuated region in which there is a uniform magnetic field
perpendicular to the plane of the diagram. This region is divided into two by a
sheet of aluminium about 1mm thick. The curved, horizontal path followed by the
radiation is shown in the diagram below.
Which of the following correctly
describes the type of radiation and its point of entry?
Type
of radiation Point of
entry
A alpha X
B alpha Y
C beta X
D beta Y
E gamma X
Reference: Past Exam Paper – N87 / I / 27
Solution 358:
Answer: D.
From the diagram, the type of
radiation emitted passes through the aluminium and is deviated in the field.
The range of α-particles in air is about 5cm. They are
easily stopped by a piece of paper. (So, they cannot pass through a sheet of
aluminium.) [A and B are incorrect]
γ-rays are not charged, so they are not
deflected by magnetic or electric fields. [E is
incorrect]
From Flemming’s left hand rule [Thumb: Force, Forefinger: field, Middle finger: current (this
is opposite to the direction of flow of negative charge (β-particles here))] if
the radiation enters at Y (that is current is to the right) and the field is
into the paper, then the force on the particle is upwards. The other choice
does not satisfy Flemming’s left hand rule. [D is
incorrect]
Question 359: [Nuclear
Physics]
In repeating Rutherford’s α-particle scattering experiment, a student
used the apparatus shown, in a vacuum, to determine n the number of α-particles
incident per unit time on a detector held at various angular positions θ.
Which graph best represents the
variation of n with θ?
Reference: Past Exam Paper – N95 / I / 28
Solution 359:
Answer: C.
From Rutherford’s experiment, it was
found that most of the α-particles
were not deflected or only deflected through small angles. Thus, the α-particles
behaved as though the gold foil was not there. (So, most of the α-particles
should be detected when θ = 0o.) [A is
incorrect]
It was also seen that less than 1%
of the α-particles were deflected through very large angles of more than 90o.
These α-particles seemed to have struck something massive but very few of the α-particles
were deflected through nearly 180o (i.e. back towards where it came
from). [B and D are incorrect]
Question 360: [Quantum
Physics]
Some data for work function energy Φ
and threshold frequency f0 of some metal surfaces are given in Fig.
(a)
(i) State what is meant by threshold
frequency.
(ii) Calculate threshold frequency
for platinum.
(b) Electromagnetic radiation having continuous spectrum of
wavelengths between 300 nm and 600 nm is incident, in turn, on each of the
metals listed in Fig.
Determine which metals, if any, will
give rise to emission of electrons.
(c) When light of particular intensity and frequency is incident on a
metal surface, electrons are emitted.
State and explain effect, if any, on
the rate of emission of electrons from this surface for light of the same
intensity and higher frequency.
Reference: Past Exam Paper – June 2013 Paper 41 & 43 Q7
Solution 360:
(a)
(i) The threshold frequency is the
lowest frequency of e.m. radiation giving rise to emission of electrons (from
the surface).
(ii)
Energy E = hf
Threshold frequency f0 =
(9.0x10-19) / (6.63x10-34) = 1.4x1015Hz
(b)
EITHER {Here,
we calculate the frequency corresponding to a wavelength of 300nm and compared
this to the threshold frequencies of the metals.}
a wavelength of 300nm corresponds to
a frequency of 10x1015Hz (and 600nm ≡ 5.0x1014Hz)
OR {Here, we
calculate the energy of a 300nm photon and compare it to the work function energies
of the metals.}
a wavelength of 300nm corresponds to
an energy of 6.6x10-19J (and 600nm ≡ 3.3x10-19J)
OR {Here, we
calculate the threshold wavelength of the metals and compare them to the
wavelength of 300nm.}
zinc λo
= 340nm, platinum λo = 220nm (and sodium λo = 520nm)
So, there is emission from sodium and
zinc.
(c)
Each photon has larger energy {since E = hf}. So, there must be fewer photons per
unit time {in order to maintain the same intensity}.
Thus, there are fewer electrons emitted per unit time.
Question 361: [Pressure]
A pipe is closed at one end and
contains gas, trapped by column of water.
Atmospheric pressure is 1.0 × 105
Pa. Density of water is 1000 kg m–3.
What is the pressure of the gas?
(Use g = 10 m s–2.)
A 0.3 × 105 Pa
B 0.5 × 105 Pa
C 1.5 × 105 Pa
D 1.7 × 105 Pa
Reference: Past Exam Paper – June 2012 Paper 12 Q23
Solution 361:
Answer: C.
The gas inside the tube exerts a
pressure on the liquid (at the level of the dotted line).
The atmosphere also exerts a pressure
on the liquid from outside and the vertical length of the column of the liquid
which is above the level of the dotted line also exerts a pressure. These 2 pressures act in a similar way (they tend to cause
the liquid to go inside the tube) while the pressure due to the gas acts in an
opposite way (it tends to cause the liquid to go outside the tube).
The liquid is in equilibrium (it is
not moving). So, the pressure at the level of the dotted line is due to the gas
should be equal to the sum of pressures due to the column of liquid above that
level and due to the atmosphere.
Pressure of gas = Pressure of column
of liquid + Atmospheric pressure
Pressure of column of liquid = hρg =
5 x 1000 x 10 = 50000Pa = 0.5x105Pa
Pressure of gas = (0.5x105Pa)
+ (1.0x105Pa) = 1.5x105Pa
For solution 357, can the formula for power dissipated be P=VI?
ReplyDeleteyes, but you would have to use the correct value of V which is the p.d. across that component. it's better to use the value of R though as it is provided in the question
Deleteon 2018 p11 q 13
ReplyDeletejune or november?
DeleteKindly solve this Q w18_qp_11 Q13
ReplyDeletego to
Deletehttp://physics-ref.blogspot.com/2020/08/a-vertical-tube-closed-at-one-end-is.html