9702/Oct Nov/13/2013/Q22

A lift is supported by two steel cables each of length 20 m.

Each of the cables consists of 100 parallel steel wires, each wire of cross-sectional area 3.2×10^–6 m². The Young modulus of steel is 2.1×10¹¹Nm^–2.

Which distance does the lift move downward when a man of mass 70 kg steps into it?
A 0.010 mm

B 0.020 mm

C 0.10 mm

D 0.20 mm

Solution:

Answer: C

Hooke’s law: F = ke

The lift is supported by 2 steel cables, each of which consists of 100 parallel steel wires. So, there is a total of 200 parallel steel wires in parallel.

For parallel spring,
Effective spring constant, k_eff = k₁ + k₂ + k₃ + … 

First, we need to find the spring constant, k for 1 steel wire.

For 1 wire,
Young modulus, E = stress / strain = (F/A) / (e/L) = FL / Ae
Hooke’s law: F = ke
Young modulus, E = (ke)L / Ae = kL / A
Spring constant, k = EA / L = (2.1 × 10¹¹) (3.2 × 10^–6) / 20 = 33600 Nm^-1 

Effective spring constant, k_eff = 200k
Hooke’s law: F = k_eff e
Mass of person = 70kg. Weight = mg = 700N           (take g = 10 ms^-2)
Extension, e = F / k_eff = 700 / (200 × 33600) = 0.00010m = 0.10 mm

Reference: PYQ - Oct/Nov 2013 Paper 13 Q22

9702/May Jun/12/2013/Q23

The diagram shows a large crane on a construction site lifting a cube-shaped load.

A model is made of the crane, its load and the cable supporting the load.

The material used for each part of the model is the same as that in the full-size crane, cable and
load. The model is one tenth full-size in all linear dimensions.

A             10⁰

B             10¹

C             10²

D             10³

Solution:
Answer: B

Stress = force per unit area = F / A

Let the force (weight) and the (cross-sectional) area in the full-size crane be F and A respectively.

Stress in cable of full-size crane = F / A

As mentioned in the question, the model is one-tenth full-size in all linear dimensions.

For the model,

The load, which has a cube-shaped load, is in 3-dimension. So, for each of the 3 dimensions, the length of the model is reduced by a factor of 1/10. Therefore, the volume of the load is reduced by a factor of (1/10)³ = 1 / 1000.

Weight = mg and Mass = density x volume. Since the same material is used, the density is the same. So, the mass is proportional to the volume. A reduction in the volume causes the mass to be reduced by the same factor. The weight, which depends on the mass (g is constant), is also reduced by the same factor.

Force (weight) in model = F / 1000

Similarly, the cross-sectional area (which depends on (diameter)²) will be reduced by a factor of (1/10)² = 1 / 100.

(Cross-sectional) Area in model = A / 100

Stress in cable of model crane = (F/1000) / (A/100) = 0.1 (F/A)

Ratio = (F/A) / 0.1(F/A) = 10  

Reference: PYQ - May/Jun 2013 Paper 12 Q23

9702/May Jun/13/2015/Q21

A W-shaped tube contains two amounts of mercury, each open to the atmosphere. Air at
pressure P is trapped in between them. The diagram shows two vertical distances x and y.

Atmospheric pressure is equal to the pressure that would be exerted by a column of mercury of
height 760 mm. The pressure P is expressed in this way.

Which values of x, y and P are possible?

Solution:

Answer: B.

This is a relatively complex situation.

Atmosphere pressure P_A = column of mercury of height 760mm

Compare the left part of the diagram with the middle (left) part.

Since the position of the mercury column is lower at the middle, the pressure P is greater than the atmospheric pressure acting at the left ‘open’ tube.

P – P_A = x

P – 760 = x                              (1)

Now, compare the right part of the diagram with the middle (right) part.

P – P_A = 50 – y

P – 760 = 50 – y

P + y = 810                             (2)

The values form the table should satisfy both equation (1) and (2).

Choice A:

Put x = 20 and P = 780 in eqn (1): 780 – 760 = 20    [correct]

Put y = 20 and P = 780 in eqn (2): 780 + 20 = 810    [incorrect]

Choice B:

Put x = 20 and P = 780 in eqn (1): 780 – 760 = 20    [correct]

Put y = 30 and P = 780 in eqn (2): 780 + 30 = 810    [correct]

Choice C:

Put x = 30 and P = 810 in eqn (2): 810 – 760 = 30    [incorrect]

Choice D:

Put x = 30 and P = 790 in eqn (1): 790 – 760 = 30    [correct]

Put y = 30 and P = 790 in eqn (2): 790 + 30 = 810    [incorrect]

Reference: PYQ - May/Jun 2015 Paper 13 Q21

9702/May Jun/13/2015/Q23

The graph shows the non-linear force-extension curve for a wire made from a new composite
material.

What could be the value of the strain energy stored in the wire when it is stretched elastically to
point P?

A 0.09 J
B 0.10 J
C 0.11 J
D 0.20 J

Solution:

Answer: C.

The strain energy is given by the area under the force-extension curve.

Consider a straight line joining point (0, 0) and point P (2.0, 100). This line would be below the curve shown in the question.

For the linear force-extension curve (drawn joining point (0, 0) to point P),

Strain energy = ½ Fx = ½ (100)(2x10^-3) = 0.1J.

But the shape of graph implies that the area of under the curve shown in the question should be greater than the area calculated. [A and B are incorrect]

However, the answer should be about 10% greater than this straight line graph value to account for the difference. Choice D (0.20 J) is twice the value calculated. This is too big. [D is incorrect]

Reference: PYQ - May/Jun 2015 Paper 13 Q23