Physics 9702 Doubts | Help Page 153
Question 757: [Waves
> Diffraction]
A parallel beam of white light
passes through diffraction grating. Orange light of wavelength 600 nm in the
fourth order diffraction maximum coincides with blue light in the fifth order diffraction
maximum.
What is wavelength of the blue light?
A 450 nm B 480 nm C
500 nm D 750 nm
Reference: Past Exam Paper – June 2014 Paper 11 Q26
Solution 757:
Answer: B.
Diffraction grating: d sinθ = nλ
For the 4th order
diffraction maximum of the orange light,
dsinθ = 4(600)
For the 5th order
diffraction maximum of the blue light,
dsinθ
= 5λ
Since the same diffraction grating is
considered, d is the same for both and since the maximum coincides, sinθ is the same
for both.
So, 5λ = 4(600)
Wavelength of blue light, λ = 4(600)
/ 5 = 480nm
Question 758: [Kinematics]
Science museum designs an experiment
to show the fall of a feather in a vertical glass vacuum tube.
Time of fall from rest is to be
close to 0.5 s.
What length of tube is required?
A 1.3 m B 2.5 m C
5.0 m D 10.0 m
Reference: Past Exam Paper – November 2012 Paper 12 Q8
Solution 758:
Answer: A.
Let the length of tube be s.
Equation for uniformly accelerated
motion: s = ut + ½ at2
The feather falls from rest. So, u =
0.
Time of fall, t = 0.5s (about
0.5s)
Acceleration a = acceleration of
free fall = 9.81ms-2
Length s = ut + ½ at2 = 0
+ ½ (9.81)(0.5)2 = 1.226m
Question 759: [Current
of Electricity > Resistance]
A metal wire of length 0.50 m has
resistance of 12 Ω.
What is resistance of a wire of
length 2.0 m and made of the same material, but with half the diameter?
A 12 Ω B 48 Ω C
96 Ω D 192 Ω
Reference: Past Exam Paper – November 2014 Paper 13 Q33
Solution 759:
Answer: D.
Resistance R of a wire = ρL / A
Resistivity of wire, ρ = RA / L
Since the wires are made of the same
material in both cases, the resistivity ρ is the same.
Let the diameter of the first wire
be d.
Cross-sectional area, A = πd2
/ 4
For the 1st wire: ρ = RA / L = 12 (πd2) / 4(0.5) = 6πd2
For the 2nd wire:
Diameter = d / 2
Cross-sectional area = π(0.5d / 2)2 = πd2 / 16
Resistivity ρ = RA / L = R (πd2
/ 16) / 2.0 = R (πd2) / 32
Equating the values of ρ,
R (πd2) / 32 = 6πd2
Resistance R = 6 × 32 = 192 Ω
Question 760: [Matter
> Hooke’s law]
To determine mass of food in a pan,
a scale is used that has high sensitivity for small masses but low sensitivity
for large masses.
To do this, two springs are used,
each with different spring constant k. One of the springs has a low spring
constant and the other has a high spring constant.
Which arrangement of springs would
be suitable?
Reference: Past Exam Paper – November 2010 Paper 12
Q22
Solution 760:
Answer: A.
We need a scale that has high
sensitivity for small masses but low sensitivity for large masses.
For choice B, the sensitivity will
be the same for both large and small masses.
For choice D, 1 spring will extend
more than the other, giving incorrect measurement values.
Hooke’s law: F = kx
So, extension x = F / k
Force F = Weight = mg
For small m (F is also small), we
need a high value for extension x. So, the spring constant k should also be
small (x = F / k). Similarly, for large m and thus large F, we want a small
value for x, so spring constant k should be also be large.
So, it is incorrect to assume that a
high spring constant is what is required for a large deflection. Here the
thinner spring stretches most when the loading is light but it can go no
further than the rigid box shown. Sensitivity falls thereafter.
Choice C would have high sensitivity
for large m as well since extension x = F / k and here the spring constant k
used is small.
For example, a circular scale on a
commercial balance with similar difference of sensitivity will perhaps have
scale occupying 10 cm for first 100 g, followed by 40 cm for total mass of 4 kg
Question 761: [Current
of Electricity > Potential divider]
In the potentiometer circuit shown,
reading on the ammeter is zero.
The light-dependent resistor (LDR)
is then covered up and ammeter gives a non-zero reading.
Which change could return the
ammeter reading to zero?
A Decrease the supply voltage.
B Increase the supply voltage.
C Move the sliding contact to the
left.
D Move the sliding contact to the
right.
Reference: Past Exam Paper – November 2014 Paper 11
& 12 Q36
Solution 761:
Answer: C.
For the ammeter to read zero, the
potential difference across it should be zero. That is, the terminals of the
ammeter are at the same potential.
The resistance of an LDR decreases when
the light intensity on it increases.
As the LDR is covered up, the light
intensity decreases. This causes its resistance to increase. Considering the
resistor and the LDR in the middle section, an increase in the resistance of
the LDR causes the p.d. across it to increase (from Ohm’s law: V = IR). So, the
terminal connected between the LDR and the resistor is now at a higher
potential than before.
To return the ammeter reading to
zero, the sliding contact should be moved in such a way that it is at a higher
potential (equal to the new potential of the other terminal of the ammeter).
The right end of the uniform metal
wire is connected to the negative terminal of the supply and so, is at a
potential of zero. The left end of the wire is connected to the positive
terminal of the supply and is at the maximum potential (equal to the e.m.f. of
the supply). Thus, as we move from right to left, the potential increases. So,
to increase the potential of the sliding contact, we need to move it to the
left.
if the tube is in vacuum we can we take g as 9.81?
ReplyDeleteVacuum means the absence of things like air (and thus, no air resistance)
DeleteThe gravitational field is still present.
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ReplyDeleteAlhamdullilah
DeleteYes I love this blog..clears my concepts up..Alhamdulilah!!!
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ReplyDeleteQ 760) Do mind explaining a bit more as to why option A is correct and not C?
ReplyDeleteHigh sensitivity means that when a mass is inserted, the scale shows a greater deflection (or here, the system expands by a significant amount).
DeleteWe want a high sensitivity for small masses, i.e. when small masses are added the spring extends by a large amount.
From Hooke’s law, for a longer extension the spring constant should be low (since x = F/k). Thus, by inserting the low k spring into the box, it would extend by large amounts, as the masses are added, until it reaches the rigid box where it can no longer extend.
At this point, the spring inside the boss can no longer extend and the mass it supports is no longer small. Only the lower spring would. However, since this spring has a high k, it would extend by small amount at these larger values of masses.
In the case of C, large masses would also result in high sensitivity since the spring with low k is outside the box and would extend by large amount.
can you please explain the concept of upthrust.
ReplyDeletea brief explanation is available at
Deletehttp://physics-ref.blogspot.com/2014/11/types-of-forces-gravitational-electric.html