Physics 9702 Doubts | Help Page 150
Question 739: [Current
of Electricity > Resistance]
A pencil is used to draw line of
length 30 cm and width 1.2 mm. Resistivity of the material in the pencil is 2.0
× 10–5 Ω m and the resistance of the line is 40 kΩ.
What is the thickness of the line?
A 1.25 × 10–10 m
B 1.25 × 10–8 m
C 1.25 × 10–7 m
D 1.25 × 10–5 m
Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q32
Solution 739:
Answer: C.
Resistance R = ρL / A
where ρ is the resistivity of the
material, L is the length and A is the cross-sectional A
The line drawn can be considered to
be a cuboid (in 3-dimension).
Let the thickness of the line be y.
Cross-sectional A = y × (1.2×10-3)
R = ρL
/ A
A = ρL / R
y × (1.2×10-3) = ρL / R
Thickness y = (2.0
× 10–5) (0.3) / (40000
× 1.2×10-3) = 1.25×10-7 m
Question 740: [Kinematics]
Diagram shows a velocity-time graph
for a mass moving up and down on the end of a spring.
Which point represents velocity of
the mass when at the lowest point of its motion?
Reference: Past Exam Paper – November 2014 Paper 13 Q7
Solution 740:
Answer: D.
At the lowest and highest point in
the motion, the mass is momentarily at rest. So, at these points, the velocity
should be zero. [A and C are incorrect]
Additionally at the extremes of the
motion, the velocity changes direction.
If the mass was moving upwards
towards the highest point, at the highest point, the velocity would changes
from being previously positive to negative. That is, the mass will then be
moving downwards afterwards. So, at the highest point, the velocity changes
from positive to negative. This is point B in the graph.
If the mass was moving downwards
towards the lowest point, at the lowest point, the velocity would changes from
being previously negative to positive. That is, the mass is about to move upwards
afterwards. So, at the lowest point, the velocity changes from negative to
positive. This is point D in the graph.
Question 741: [Measurement
> C.R.O]
Diagram shows a trace of a wave on a
cathode-ray oscilloscope.
The vertical and horizontal gridlines
have spacing of 1.0 cm. Voltage scaling is 4 V cm–1 and time scaling
is 5 ms cm–1.
What are the amplitude and period of
the wave?
amplitude
/ V period / ms
A 1.5
4
B 5.0
10
C 6.0
20
D 12.0
20
Reference: Past Exam Paper – June 2011 Paper 11 Q4
Solution 741:
Answer: C.
The amplitude is given by the
vertical spacing from the equilibrium position and the period is given by the
horizontal spacing.
Note that the equilibrium position
is found at the middle of the 3rd square along the vertical.
Amplitude = 1.5cm = 1.5 (4) = 6V
Period = 4cm = 4 (5) = 20ms
Question 742:
[Pressure]
A child drinks liquid of density ρ
through a vertical straw.
Atmospheric pressure is p0
and the child is capable of lowering the pressure at top of the straw by 10%.
The acceleration of free fall is g.
What is maximum length of straw that
would enable the child to drink the liquid?
A p0 / 10ρg B 9p0 / 10ρg C p0
/ ρg D 10p0 / ρg
Reference: Past Exam Paper – June 2003 Paper 1 Q20
Solution 742:
Answer: A.
Imagine a beaker containing the liquid
and a vertical straw (not an inclined straw).
Atmospheric pressure p0
acts on the surface of the liquid.
Now, the child is sucking at top of
the straw with a pressure of 0.9p0 (since the child is capable of
lowering the pressure at the top of the straw by 10%). As the child sucks up
the liquid rises up the straw of length h.
The difference in pressure at the
surface of the liquid in the beaker (= p0) and the pressure at the
top of the straw (= 0.9p0) is represented by the pressure due to the
column of liquid in the straw (= hρg).
p0 – 0.9p0 = hρg
hρg = 0.1p0 = p0
/ 10
Length of straw, h = p0 /
10ρg
Question 743: [Waves
> Intensity]
Intensity I of a sound at a point P
is inversely proportional to the square of the distance x of P from the source
of the sound. That is
Air molecules at P, a distance r
from S, oscillate with amplitude 8.0 μm.
Point Q is situated a distance 2r
from S.
What is the amplitude of oscillation
of air molecules at Q?
A 1.4 μm B 2.0 μm C
2.8 μm D 4.0 μm
Reference: Past Exam Paper – June 2008 Paper 1 Q26
Solution 743:
Answer: D.
Intensity I is proportional to the
square of the amplitude (A2).
Now, the intensity I is also
inversely proportional to the square of the distance x form the source (1/x2).
So, doubling the distance x causes
the intensity to be (I / 4).
Recall: Intensity I is proportional
to the square of the amplitude (A2).
Intensity I is proportional to 8.02
Intensity I/4 is proportional to
(8.02 / 4)
So, at distance = 2r,
(Amplitude)2 = (8.02
/ 4)
Amplitude = √(8.02
/ 4) = 8.0 / 2 = 4.0μm
OR
Since I α A2 and I α 1 / x2,
It can be concluded that A2
α 1 / x2
So, amplitude A α 1 / x
Thus, if distance x is doubled,
amplitude A is halved.
OR
At distance = r, Intensity IP
α 82
At distance = 2r, Intensity IQ
= IP / 4 (as calculated above)
At distance = 2r, Intensity (IP
/ 4) α AQ2
OR
Since we are dealing with the same
quantities in both cases, we can take the ratio of intensities and amplitudes
at P and at Q.
IP / IQ = AP2
/ AQ2
IP / (IP/4) =
82 / AQ2
4 = 82 / AQ2
Amplitude AQ = √(8.02
/ 4) = 8.0 / 2 = 4.0μm
Question 744: [Electric
field]
Two vertical conducting plates X and
Y are positioned so that they are separated by distance of 6.0 mm in air. A 60
V d.c. supply is connected as shown.
What is electric field strength at
E, a point midway between the plates?
A 1.0 × 104 V m–1 towards
X
B 1.0 × 104 V m–1 towards
Y
C 2.0 × 104 V m–1 towards
X
D 2.0 × 104 V m–1 towards
Y
Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q31
Solution 744:
Answer: B.
The direction
of the electric field is from positive to negative. Since plate X is connected
to the positive terminal of the supply, it is also positive. So, the direction
of the electric field is from plate X to plate Y. [A
and C are incorrect]
Electric field
strength E = V / d = 60 / 0.006 = 1 × 104 Vm-1
The electric field strength between 2
parallel plates is constant anywhere between them.
11/M/J/11 Q.27,32
ReplyDeleteFor 11/M/J/11 Q.27, see solution 753 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-152.html
9702/23/o/N/12 QUESTION 2 AND 3
ReplyDeleteFor Q3, check solution 1003 at
Deletehttp://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-208.html
For Q2, check solution 587 at
http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-115.html
This is amazing!!! Whats your website??
ReplyDeleteI need help in maths. Can u give me any reference??
ReplyDeleteCan I also get reference for chemistry??
ReplyDeleteFor chemistry:
Deletehttp://chemistry-ref.blogspot.com/
but it is not updated as the physics one.
We do not have one for maths. But if anyone is willing to contribute to build up for these subjects, you are welcomed.
I would Love to contribute to help with producing these references for pure maths. how can i help??
DeleteYou are welcomed.
DeleteFor more details, go to
http://physics-ref.blogspot.com/2016/01/participating-to-physics-reference-blog.html
You are fantastic
ReplyDeleteI have a question, in question no. 740, what does point B mean?
ReplyDeleteat point B, the graph touches the time axis, i.e. the velocity is zero. it is momentarily at rest.
Deletehowever, the acceleration (given by gradient) is not zero. it is negative, indicating that the motion would be downwards the next instant.