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