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Question 314: [Physics of Materials ~ Matter > Solids]
A steel of length l and diameter d is found to have a strain of 8.0x10^-4 when supporting a certain load. If this wire is replaced by a second steel wire of the same length but of diameter d/2, the strain will be
A 3.2x10^-3                   B 1.6x10^-3                   C 8.0x10^-4       D 4.0x10^-4       E 2.0x10^-4

Reference: Past Exam Paper – J76 / II / 40 & J85 / I / 30



Solution 314:
Answer: A.
Since a steel wire is used in both cases, the Young modulus of both wires is the same.
Young modulus, E = stress / strain = Force / (Area x strain)

Cross-sectional area of a wire = πr² = πd²/4

Young modulus, E = F / [(πd²/4) x strain] = 4F / [(πd²) x strain]

For the 1^st case:
Young modulus, E = (4F/π) [1 / (8.0x10^-4) d²]

For the 2^nd case: [diameter is now d/2]
Young modulus, E = (4F/π) [1 / (strain) (d/2)²] = (4F/π) [4 / (strain) d²]

Since the Young modulus of the steel wire is the same,
(4F/π) [1 / (8.0x10^-4) d²] = (4F/π) [4 / (strain) d²]
Strain = 4 (8.0x10^-4) = 3.2x10^-3 










Question 315: [A2 Practical > Analysis, Conclusions and Evaluation]

Student is investigating how period T of a simple pendulum depends on its length l, as shown.

Time t for 10 oscillations is recorded for a pendulum of length l. Period T of the pendulum is determined. Procedure is then repeated for different lengths.

It is suggested that T and l are related by the equation

T = al^b

where a and b are constants.

(a) A graph is plotted of lg T on y-axis and lg l on x-axis. Determine expressions for gradient and y-intercept in terms of a and b.

(b) Values of l and t are given in Fig.

Reference: Past Exam Paper – November 2010 Paper 51 & 52 Q2(a), (b)





Solution 315:

(a)

T = al^b

lg T = lg (al^b) = lg a + b lg l

Comparing with y = mx + c, [lg T = y, lg l = x. So, m = b and c = lg a]

Gradient = b

Y-intercept = lg a

(b)

Period T = t / 10

ΔT / T = Δt / t

Since period T = t / 10, ΔT should be Δt / 10 [for the above equation to hold]

Consider t = 19.6 ± 0.2 s

Period T = (19.6 ± 0.2) / 10 = 1.96 ± 0.02 s

lg T = lg (1.96) = 0.2923

{In logarithmic quantities the number of significant figures is determined by the number of decimal places, thus the number of decimal places in a logarithmic quantity should be the same or one more than the number of significant figures in the raw data.}

Uncertainty in lg T = lg (1.98) – lg (1.96) = 0.004

{The uncertainty in a logarithmic quantity is obtained by taking the log of the maximum value – the log of the exact value OR log (exact value) – log (minimum value)}

So, lg T = 0.2923 ± 0.004

The others are calculated in the same way.

For lg (T / s), 3 decimal places are accepted.

Question 316: [Forces > Hooke’s law]
A force of 10N acting on a certain spring gives an extension of 40mm. Two such springs are connected end to end and this double-length spring is extended by 40mm.
Assuming that the springs conform to Hooke’s law, what is the strain energy?
A 0.05J            B 0.10J            C 0.20J            D 0.40J            E 0.80J

Reference: Past Exam Paper – J88 / I / 30



Solution 316:
Answer: B.

Hooke’s law: F = ke

where F is the load, k the spring constant and e is the extension / compression.

From the force of 10N and the extension of 40mm,

Spring constant, k = F / e = 10 / 0.04 = 250 Nm^-1

For springs in series and in parallel, the following formulae for the ‘effective’ spring constant apply:

Springs in parallel:    k_eff = k₁ + k₂ + ….

Springs in series:       1/k_eff = 1/k₁ + 1/k₂ + ….

Since the 2 springs are in series,
1 / k_eff = (1/250) + (1/250)
Effective spring constant, k_eff = [(1/250) + (1/250)]^-1 = 125 Nm^-1 
{A force-extension graph of a system of springs obeying Hooke’s law is a straight line with positive gradient. The area under the force-extension graph gives the strain energy.}
Strain energy = ½ Fe = ½ ke² = 0.5 (125) (0.04)² = 0.10 J









Question 317: [Physical Quantities]
The symbol g represents acceleration of free fall.
Which of these statements is correct?
A g is gravity.
B g is reduced by air resistance.
C g is the ratio weight / mass.
D g is the weight of an object.

Reference: Past Exam Paper – November 2007 Paper 1 Q7



Solution 317:
Answer: C.
As stated in the question, the symbol g represents the acceleration of free fall.

Gravity is a force. Force and acceleration are different quantities. Similarly, the weight of an object has unit ‘Newton’, which is the unit of force, not acceleration.

The acceleration of free fall is constant for a specific body (planet) – it cannot be reduced by air resistance. The resultant force on a body may be reduced by air resistance, but not the acceleration of free fall.

Weight = mg. So, g = weight / mass. This is the correct choice.

More information about gravity can be found at Types of Forces - Gravitational, Electric,Frictional, Viscous drag and Upthrust (http://physics-ref.blogspot.com/2014/11/types-of-forces-gravitational-electric.html)