Physics 9702 Doubts | Help Page 149
Question 733: [Current
of Electricity]
Six resistors, each of resistance R,
are connected as shown.
Combined resistance is 66 kΩ.
What is the value of R ?
A 11 kΩ B 18 kΩ C
22 kΩ D 36 kΩ
Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q37
Solution 733:
Answer: D.
For the parallel combination of 2
resistors,
Combined resistance = [1/R + 1/R]-1
= R/2
For the parallel combination of 3
resistors,
Combined resistance = [1/R + 1/R +
1/R]-1 = R/3
Total combined resistance = R + R/2
+ R/3 = 11R / 6
11R / 6 = 66
Resistance R = 6 × 66 / 11 = 36 kΩ
Question 734: [Waves
> Diffraction]
Diffraction can be observed when a wave
passes an obstruction. Diffraction effect is greatest when the wavelength and
the obstruction are similar in size.
For waves travelling through air,
what is the combination of wave and obstruction that could best demonstrate
diffraction?
A microwaves passing a steel post
B radio waves passing a copper wire
C sound waves passing a human hair
D visible light waves passing a gate
post
Reference: Past Exam Paper – November 2012 Paper 13 Q28
Solution 734:
Answer: A.
In this question, we need to
estimate the sizes involved.
Below is the approximate range of wavelengths
for the waves:
Microwave: 1mm – 1m
Radio wave: 1m – 100 000km
Visible light: 390nm – 750nm
Sound waves: 17mm – 17m (in audible
frequencies)
A steel post is quite similar in
size as the wavelength of microwaves. [A is correct]
As for the others, there is a huge
mismatch in size between the wavelength and the obstacle.
Question 735: [Waves]
Which statement about sound waves in
air at constant temperature is correct?
A Amplitude is inversely
proportional to velocity.
B Frequency is inversely
proportional to wavelength.
C Velocity is proportional to
wavelength.
D Wavelength is proportional to amplitude.
Reference: Past Exam Paper – June 2011 Paper 11 Q23
Solution 735:
Answer: B.
Speed v = fλ
Frequency f = v / λ (The frequency is inversely proportional
to wavelength.)
At constant temperature, the
velocity of the sound waves is constant, even if the wavelength changes. In
such a case, the frequency also changes and the speed is still constant. Thus, as the wavelength increases, the speed
does not increase or decrease – it remains constant. So, the velocity is NOT
proportional to wavelength at constant temperature. [C
is incorrect]
Choices A and C are also incorrect /
irrelevant.
Question 736: [Waves
> Intensity]
Sound wave X has intensity 1012
times greater than that of sound wave Y.
By how much is amplitude of X
greater than the amplitude of Y?
A 106 times
B 3.16 × 106 times
C 5 × 1011 times
D 1012 times
Reference: Past Exam Paper – June 2007 Paper 1 Q22
Solution 736:
Answer: A.
Intensity is proportional to
(amplitude A)2.
The intensity of sound wave X is 1012
times greater than that of sound wave Y.
Let the amplitude of sound wave X by
AX and that of sound wave Y be AY.
AX2 is 1012
times greater than AY2
AX is √(1012) = 106 times greater than AY
Question 737: [Matter
> Deformation]
Suspended copper wire is gradually
loaded until it is stretched just beyond the elastic limit, and it is then
gradually unloaded.
Which graph (with arrows indicating
the sequence) best illustrates variation of the tensile stress with
longitudinal strain?
Reference: Past Exam Paper – June 2003 Paper 1 Q19
Solution 737:
Answer: B.
Stress = Force / Area
Strain = Extension / Original Length
As the load is increased (stress
increases), the extension of the wire increases (strain increases).
From Hooke’s law, the extension is
proportional to the force (strain is proportional to stress) up to the limit of
proportionality, which is before the elastic limit. So, up to this point, the
graph is a straight line. [A is incorrect]
The loading part (when the load is
being increased) is shown by the upward arrow. From Hooke’s law, when the load
is zero, the extension is also zero. So, the loading part should pass through
the origin. [C is incorrect]
Beyond the limit of proportionality,
the strain is no longer proportional to the stress. Since the wire has been stretched
just beyond the elastic limit, the loading part (upwards arrow) should be
curved at some point.
So, the deformation is no longer
elastic – that is when the load is removed, the wire does not regain its
original length. However, as the load is decreased, some of the extension also
decreases. However, this part of the graph is a straight line, not curved. [D is incorrect]
Question 738: [Dynamics
> Newton’s 3rd law]
A ball falls vertically and bounces
on the ground.
The following statements are about
forces acting while the ball is in contact with the ground.
Which statement is correct?
A The force that the ball exerts on
the ground is always equal to the weight of the ball.
B The force that the ball exerts on
the ground is always equal in magnitude and opposite in direction to the force
the ground exerts on the ball.
C The force that the ball exerts on
the ground is always less than the weight of the ball.
D The weight of the ball is always
equal in magnitude and opposite in direction to the force that the ground
exerts on the ball.
Reference: Past Exam Paper – June 2004 Paper 1 Q10
Solution 738:
Answer: B.
Force is a vector – it has both a magnitude
and a direction.
Newton’s 3rd law states
that when 2 bodies are in contact, the force they exert on which other is equal
and opposite to each other. [B is correct]
The time of contact of the ball with
the ground can be described as follows. The ball just hits the ground, then it
is in contact with the ground and finally, the ball is just about to leave the
ground.
During these processes, the forces
change accordingly.
As the ball just hits the ground,
the force it exerts on the ground at that instant is equal to the weight. [C is incorrect] After this instant, the ball (which
was falling down) comes to a stop momentarily. Thus, the force on the ball is
reduced to zero. (If the force on the ball was ALWAYS equal to the weight, the
ball would just keep on accelerating downwards) [A
is incorrect]
The weight of the ball always acts
towards the ground (downward). As stated above, the force on the ball changes
during the time of contact. As the ball is just about to leave the ground to
move up, the resultant force on the ball has increased and is now upwards. But
the weight is always downwards. [D is incorrect]
11/M/J/11 Q.25,27,32
ReplyDeleteFor 11/M/J/11 Q.25, see solution 129 at
Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-24.html
M/J/08 Q.26 paper12
ReplyDeleteCheck solution 743 at
Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-150.html
Oct/Nov 2014 P41 Q5 b i & ii part
ReplyDeleteOct/Nov 2014 P41 Q5 b (i) & (ii) part
ReplyDeleteCheck solution 619 at
Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-122.html
Salam! in q 738, when the ball is about to move upwards, the force exerted by the ground on it is upward n greater than its weight which is why it accelerates upwards ryt?
ReplyDeletefor option A, this is what i think, let me know if this is wrong.
1. In absence of air resistance if the ball fallas and hits the ground, the force it exerts on the ground will be equal to its weight. (and if not, how can the ball not exert W because that is the force with which it fell, right?)
2. but if air resistance is appreciable, then the ball would exert force on the ground equal to the resultant of its W and resistive force.
Is this why A and D are both wrong?
Actually in this question, we should not be thinking of the forces in terms of weight and air resistance.
DeleteThis question is mostly on Newton's laws of motion. From the third law itself, choice B is correct.
Now, from Newton's 2nd law of motion, the force is equal to the rate of change of momentum. Remember, the ball is falling - so it has some velocity. When it hits the ground, its speed is momentarily zero. So, there is a change in velocity.
Momentum = mv. Since velocity changes, momentum also changes, thus giving rise to a force (Newton's 2nd law).
Since the way the velocity changes is different at different stages, the FORCE we are talking about differs from the weight.
Sir, have you solved this question before?
ReplyDelete9702/13/M/J/17 question 28
Thank you.
go to
Deletehttp://physics-ref.blogspot.com/2020/04/monochromatic-light-of-wavelength-is.html