Physics 9702 Doubts | Help Page 229
Question 1082: [Waves
> Double slit]
(a) State one difference and one similarity between longitudinal and
transverse waves.
(b) A laser is placed in front of two slits as shown in Fig. 6.1.
The laser emits light of wavelength
6.3 × 10–7 m.
The distance from the slits to the
screen is 2.5 m. The separation of the slits is 0.35 mm.
An interference pattern of maxima
and minima is observed on the screen.
(i) Explain why an interference
pattern is observed on the screen.
(ii) Calculate the distance between
adjacent maxima.
(c) State and explain the effect, if any, on the distance between
adjacent maxima when the laser is replaced by another laser emitting
ultra-violet radiation.
Reference: Past Exam Paper – November 2014 Paper 22 Q6
Solution 1082:
(a)
Difference:
The vibration / oscillation (of
particles) / displacement of the particles is parallel to the energy transfer /
wavefronts in longitudinal waves and perpendicular for transverse waves
OR Transverse waves can be
polarised, longitudinal waves cannot be polarized
Similarity: both transfer / propagate energy
(b)
(i) Waves from the slits are
coherent / have a constant phase relationship
The waves overlap (at screen) with a
phase difference or have a path difference
EITHER Maxima are obtained where the
phase difference is an integer ×360° (or ×2Ï€ rad) OR the path difference is an integer
×λ
OR equivalent explanation of minima
e.g. (n+½)×360°
(ii)
{λ = wavelength, D =
distance from slits to screen, a = separation of slits}
Maxima spacing = λD / a
Maxima spacing = (6.3×10–7
× 2.5) / 0.35 × 10–3
Maxima spacing = 4.5 × 10–3
m
(c) The (ultra-violet radiation has) a shorter wavelength,
hence there will be a smaller separation / distance
Question 1083: [Current
of Electricity > Capacitors]
Some capacitors are marked ‘48 μF,
safe working voltage 25 V’.
Show how a number of these
capacitors may be connected to provide a capacitor of capacitance
(a) 48 μF, safe working voltage 50 V,
(b) 72 μF, safe working voltage 25 V.
Reference: Past Exam Paper – November 2002 Paper 4 Q5
Solution 1083:
(a)
Two capacitors in series OR any circuit such that V ≤ 25V across any C
{Facts:
For capacitors in series:
1/C = 1/C1 + 1/C2 + …
For capacitors in
parallel: C = C1 + C2 + …
From Kirchhoff’s laws, the
sum of p.d. across any branch in a parallel combination is the same for any all
branches.
To have a voltage of 50V
from these capacitors, we can connect 2 capacitors in series. The sum of p.d.
is then (25 + 25 =) 50 V. However, this would produce an overall capacitance of
24 μF. To increase the capacitance, we connect the 2 capacitors in series above
in parallel with another pair of capacitors in series. Since this is a parallel
connection, the voltage of 50 V across the branch is still the same, but the
overall capacitance is now 48 μF }
{The overall capacitance
needs to be 72 μF. This can be obtained by connecting ‘2 capacitance in series’
with another capacitor in parallel.
For the 2 capacitors in
series 1/C = 1/48 + 1/48 giving C = 24 μF
Overall capacitance of
parallel connection = 24 + 48 = 72 μF
One of the branches
contain a single capacitor with a safe working voltage of 25V. Since the sum of
p.d. should be the same in all branches, the p.d. across EACH capacitor in the
other branch would be (25 / 2 =) 12.5 V. The safe working voltage is 25V: this
represents the maximum voltage that can be used. Any lower voltage can also be
used though.}
Question 1084: [Kinematics > Projectile motion]
Water emerges horizontally from a
hose pipe with a velocity of 4.0 ms-1 as shown in the figure. The
pipe is pointed at P on a vertical surface 2.0 m from the pipe. If the water
strikes at S, calculate PS.
Reference: ???
Solution 1084:
This is a projectile motion problem.
The motion can be broken down into a horizontal and a vertical motion.
The vertical distance PS needs to be
found. Time, t is the quantity which is common for both the vertical and
horizontal motion. [For the amount of time the water
would have travelled a distance of 2.0m horizontally, it would also have
dropped a distance PS vertically]
For the horizontal motion,
Speed, u = 4.0 ms-1
Distance to travel, s = 2.0 m
No acceleration (The horizontal motion is not affected by the acceleration of
free fall. Thus, the horizontal motion occurs at a constant speed.)
Speed u = s / t giving time t = s / u = 2.0 / 4.0 =
0.5 s
For the vertical motion,
Distance dropped, s = PS = ???
Acceleration (due to gravity), a =
9.81 ms-2
Initial speed = 0 ms-1 (since the water emerges horizontally, it has no vertical
component initially)
Time, t = 0.5s
Equation for uniformly accelerated
motion: s = ut + ½at2
PS = 0 + 0.5(9.81)(0.5)2 = 1.72625
= 1.73 m
No comments:
Post a Comment
If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation