Monday, December 29, 2014

Physics 9702 Doubts | Help Page 41

  • Physics 9702 Doubts | Help Page 41

Question 251: [Waves]
(a) State what is meant by progressive wave.

(b) Variation with distance x along progressive wave of a quantity y, at a particular time, is shown.
(i) State what quantity y could represent.
(ii) Distinguish between quantity y for
1. transverse wave
2. longitudinal wave

(c) Wave nature of light may be demonstrated using phenomena of diffraction and interference.
Outline how diffraction and how interference may be demonstrated using light.
In each case, draw a fully labelled diagram of apparatus that is used and describe what is observed.

Reference: Past Exam Paper – November 2009 Paper 21 Q5

Solution 251:
(a) A progressive wave is one in which there is transfer / propagation of energy as a result of oscillations / vibrations

(i) Displacement / velocity / acceleration (of the particles in the waves)

1. The displacement etc. is normal to the direction of energy transfer / travel of wave / propagation of wave

2. The displacement etc. is along / in the same direction of energy transfer / travel of wave / propagation of wave


Suitable object (laser and slit), means of observation (screen)
EITHER Laser OR Lamp and aperture OR distant source
Light region is observed where darkness is expected (observation: light spreads into the geometric shadow region)


Suitable object ((coherent) monochromatic light, double slit), means of observation and illumination (screen)
Light and dark fringes are observed
Appropriate reference to a dimension for diffraction or for interference (for interference, the dimensions are as follows: Slit separation 0.5mm, distance of slits from screen ≈ 2.5m)

Question 252: [Measurement Techniques]
Two species of ant have the same shape but all the linear dimensions of the giant ant are X times those of the normal ant. Assume that the weight each ant can lift depends only on the cross-sectional area of its muscles.
If the ‘relative strength’ of an ant is defined as the weight it can lift divided by its own weight, what is the value of the ratio of relative strength of the giant ant to relative strength of the normal ant?
A 1 / X3                       B 1 / X                        C 1                  D X                 E X2

Reference: Past Exam Paper – N78 / II / 7

Solution 252:
Answer: B.
In this question, the values are relative to each other. Calculations are performed by proportions.

First, let calculate the relative weight of each ant.
Weight of ant = mg

Mass m = Density ρ x Volume
Since the ants are of the same species, their densities are the same. So, the mass is proportional to the volume.

Let the linear dimensions of the normal ant = L
Volume of normal ant = L3 = V (let this volume be V)

Linear dimensions of the giant ant = XL
Volume of giant ant = (XL)3 = X3L3 = X3V

Let mass of normal ant = m    (Weight of normal ant = mg)
Mass of giant ant = X3m         (Weight of giant ant = X3mg)

Now, let’s calculate the relative weight that each ant can lift.
The weight each ant can lift depends only on the cross-sectional area of its muscles.

Cross-sectional area of normal ant = L2. Let the weight that can be lifted by this cross-sectional area be W.

Cross-sectional area of giant ant = (XL)2 = X2L2. Weight that can be lifted by this cross-sectional area is X2W.

Relative strength of giant ant = X2W / (X3mg) = W / Xmg
Relative strength of normal ant = W / mg

Ratio = (W / Xmg) / (W / mg) = (Wmg) / (XmgW) = 1 / X  

Question 253: [Physical Quantities & Units > Units]
The behaviour of many real gases deviates from pVm = RT but can be represented quite closely over certain ranges of temperature and pressure by an equation of the form
(p + a/Vm2) (Vm – b) = RT
in which the values of a and b are characteristic of the particular gas.
What are the units of a and b?
            a                                              b
A Pa m-6 mol2                                      m3 mol-1
B Pa m6 mol-2                                      m-3 mol
C Pa m-6 mol2                                      m-3 mol-1
D Pa m6 mol-2                                      m3 mol-1
E none because they are dimensionless constant

Reference: Past Exam Paper – J84 / II / 27

Solution 253:
Answer: D.
Units of RT: [JK-1mol-1] [K] = J mol-1 = Nm mol-1 

For the equation to be homogeneous, when expanded, the units on the left-hand side of the equation should be equal to the units on the right-hand side of the equation.

(p + a/Vm2) (Vm – b) = RT
pVm – pb + (a/Vm) – (ab/Vm2) = RT

Each of the 4 terms on the left-hand side should have the same units as RT {Nm mol-1}

(Pressure = Force / Area)
Units of pressure p: Pa = Nm-2

Consider the term pb.
In terms of units,
[Nm-2] [b] = [Nm mol-1]
Units of b = [b]: [Nm mol-1] / [Nm-2] = [Nm mol-1] [N-1m2] = m3 mol-1

Comparing pb with pVm, b should have the same units as Vm, the molar volume {Units = m3 mol-1}

[So, B, C and E are incorrect]

Vm is the molar volume. That is, volume per mole.
Units of Vm: m3 mol-1

Consider the term a/Vm,
In terms of units,
[a] / [m3 mol-1] = [Nm mol-1]
Units of a = [a]: [Nm mol-1] [m3 mol-1] = N m4 mol-2

Since the units of pressure p: Pa = Nm-2, multiply by Pa and divide by Nm-2 such that the overall units is unchanged (this does not affect the overall units).
Units of a = [a]: N m4 mol-2 = [N m4 mol-2] [Pa] / [Nm-2] = Pa m6 mol-2

Question 254: [Forces > Equilibrium of forces]
(a) Explain what is meant by
(i) the moment of a force
(ii) the torque of a couple                                [4]

(b) A desk lamp is illustrated.

The lamp must be constructed so that it does not topple over when fully extended as shown. The base of the lamp is circular and has a radius of 10cm. Other dimensions are shown on the figure. The total weight of the light and shade is 6.0N and each of the two uniform arms has weight 2.0N.

(i) On Fig, draw an arrow to represent the weight of the base.
(ii) The lamp will rotate about a point if the base is not heavy enough. On Fig, mark this point and label it P.
(iii) Calculate the following moments about P.
1. moment of first arm
2. moment of second arm
3. moment of light bulb and shade
(iv) Use the principle of moments to calculate the minimum weight of base required to prevent toppling.                                                  [7]

Reference: Past Exam Paper – J99 / II / 3

Solution 254:
(i) The moment of a force is the product of the force and the perpendicular distance of its line of action to the pivot.

(ii) The torque of a couple is the product of one of the forces of the couple and perpendicular distance between the lines of action of the forces.

(i) & (ii)

The weights of each arm act at the middle of the arms (half the lengths of the arms). Point P is at a distance of 0.10m from the centre of the base.
Distance of weight of 1st arm from point P = 0.15 – 0.10 = 0.05m
Moment of 1st arm = (2N) (0.05m) = 0.10Nm

Distance of weight of 2nd arm from point P = (0.30 – 0.10) + 0.15 = 0.35m
{(0.30 – 0.10) is the remaining length of the 1st arm from point P and 0.15m is the centre of the 2nd arm where the weight acts.}
Moment of 2nd arm = (2N) (0.35m) = 0.70Nm

Distance of light bulb and shade from point P = (0.30 – 0.10) + 0.30 = 0.50m
Moment of light bulb and lamp = (6N) (0.50m) = 3.00Nm

The principle of moments states that, for equilibrium, the sum of clockwise moments about a point should be equal to the sum of anticlockwise moments.

The moments calculated above (moments of 1st arm, 2nd arm and light bulb and shade) are all clockwise moments about point P.

The weight of the base, which acts at its centre (a distance of 0.10m to the left of point P) produces an anticlockwise moment.
Let the weight of the base be W.
Moment of weight W of base = W (0.10) = 0.10W

For equilibrium,
0.10W = 0.10 + 0.70 + 3.00
Weight W = 3.80 / 0.1 = 38N
So, the minimum weight of the base is 38N.

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