Sunday, June 7, 2015

Physics 9702 Doubts | Help Page 165

  • Physics 9702 Doubts | Help Page 165

Question 825: [Electric field]
An electron enters a region of space where there is uniform electric field E as shown.

Initially, electron is moving parallel to, and in the direction of, the electric field.
What is the subsequent path and change of speed of the electron?
path of electron           speed of electron
A         linear                           decreases
B         linear                           increases
C         parabolic                      decreases
D         parabolic                      increases

Reference: Past Exam Paper – June 2014 Paper 13 Q31

Solution 825:
Answer: A.
Electric field direction is drawn from positive to negative – that is, it shows the direction of the electric force on a positive charge. So, the force on the electron (which is negatively charged) is opposite to the direction of its motion. This causes deceleration (reducing the speed) since the force is opposite to the direction of motion.

The path of the electron will be linear (since the direction of the force is (anti-) parallel and opposite to the direction of motion). Additionally, the field is uniform.

Question 826: [Waves > Interference > Double slits]
Light of wavelength 600 nm is incident on a pair of slits. Fringes with spacing of 4.0 mm are formed on a screen.
What will be the fringe spacing when the wavelength of the light is changed to 400 nm and separation of the slits is doubled?
A 1.3 mm                    B 3.0 mm                    C 5.3 mm                    D 12 mm

Reference: Past Exam Paper – June 2013 Paper 11 Q27

Solution 826:
Answer: A.
For double slits: Separation of slits, a = Dλ / w
where D is the distance of the slits from the screen, λ is the wavelength and w is the fringe spacing

When λ = 600nm, w = 4.0mm
a = D (600 / 4.0)

The separation of the slits, a has now been doubled and the wavelength λ is changed to 400nm. D is still kept constant.

Fringe spacing, w = Dλ / a
The slit separation is now 2a (= 2D (600 / 4.0)).
Fringe spacing, w = D (400) / [2D (600 / 4.0)] = (400×4.0) / (2×600) = 1.3mm

Note that the units have not been converted in these equations since they would have cancelled out directly at the end of the calculations.

Question 827: [Electric field]
Two oppositely-charged horizontal metal plates are placed in vacuum. A positively-charged particle starts from rest and moves from one plate to the other plate, as shown.

Which graph shows how kinetic energy EK of the particle varies with the distance x moved from the positive plate?

Reference: Past Exam Paper – June 2014 Paper 12 Q29

Solution 827:
Answer: D.
Electric field strength, E = V / d
The electric field, E is uniform (since the p.d. V and the plate separation d are constant) and so the electrical potential energy decreases uniformly with the distance moved in the electric field.

From the conservation of energy, the kinetic energy of the particle therefore increases uniformly too. [D is correct]

The answer could also be obtained by reasoning that a constant force (F = Eq) gives a constant acceleration (F = Eq = ma), and then using the equation of uniformly accelerated motion v2 = 2ax shows that kinetic energy (which is proportional to v2) is proportional to x.

Question 828: [Waves > Interference > Diffraction grating]
A narrow beam of monochromatic light is incident normally on diffraction grating. Third-order diffracted beams are formed at angles of 45° to the original direction.
What is the highest order of diffracted beam produced by this grating?
A 3rd                           B 4th                           C 5th                           D 6th

Reference: Past Exam Paper – November 2007 Paper 1 Q25

Solution 828:
Answer: B.
For diffraction grating: d sinθ = nλ

When n = 3 (third order), θ = 45°
d sin45° = 3λ
d / λ = 3 / sin45°

For the highest order of diffracted beam, the angle θ should be less (or equal to) 90°. So, consider the angle θ = 90°.
d sin90° = nλ
n = (d / λ) sin90° = [3sin90°] / sin45° = 4.24.
The order n should be an integer which is less than this.
So, highest order n = 4th

Question 829: [Pressure]
A submarine is in equilibrium in a fully submerged position.

What causes upthrust on the submarine?
A The air in the submarine is less dense than sea water.
B The sea water exerts a greater upward force on the submarine than the weight of the steel.
C The submarine displaces its own volume of sea water.
D There is a difference in water pressure acting on the top and on the bottom of the submarine.

Reference: Past Exam Paper – June 2013 Paper 12 Q11

Solution 829:
Answer: D.
Pressure = hρg
The pressure acting at the top and that acting at the bottom of the submarine are different since the depths h are different. The pressure at the bottom acting (upwards) on the surface of the submarine is greater.

Pressure P = Force / Area
The resultant pressure causes an upward force on the submarine. This is the upthrust.

Since the submarine is in equilibrium, the upthrust should be equal to the weight of the submarine. [B is incorrect]

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