# 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
E

_{K}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 v

^{2}= 2ax shows that kinetic energy (which is proportional to v^{2}) 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 = 4

^{th}

__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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