Tuesday, May 26, 2015

Physics 9702 Doubts | Help Page 153

  • 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.


  1. if the tube is in vacuum we can we take g as 9.81?

    1. Vacuum means the absence of things like air (and thus, no air resistance)

      The gravitational field is still present.

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  5. Q 760) Do mind explaining a bit more as to why option A is correct and not C?

    1. High sensitivity means that when a mass is inserted, the scale shows a greater deflection (or here, the system expands by a significant amount).

      We 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.


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