Tuesday, February 10, 2015

Physics 9702 Doubts | Help Page 64

  • 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





2 comments:

  1. For solution 357, can the formula for power dissipated be P=VI?

    ReplyDelete
    Replies
    1. yes, 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

      Delete

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