Have you ever thought why a tree does not grow beyond a limiting height? It is the breaking stress of the tree that limits its height. Here is a question that high lights this point:
A tree has a breaking stress of 4.5×105 Nm–2. If the density of the tree is 900 kg m–3, the maximum height up to which it can grow is (assuming that the tree is cylindrical in shape)
(a) 100 m (b) 75 m (c) 50 m (d) 30 m (e) 25 m
The weight of the tree is AHρg where A is the area of cross section, H is the height, ρ is the density and ‘g’ is the acceleration due to gravity.
Maximum stress will be developed at the base of the tree and will be equal to AHρg/A = Hρg.
Since the breaking stress is 4.5×105 Nm–2, we can write
Hρg = 4.5×105 Nm–2 so that H = 4.5×105 Nm–2/ρg = 4.5×105 Nm–2/9000 nearly.
This works out to 50 m.
Now, consider the following MCQ involving compressibility:
A solid sphere made of copper having compressibility K is placed in an air chamber and the pressure of air is reduced by P. The fractional change in the radius of the sphere will be
(a) PK (b) 3PK (c) P/3K (d) K/3P (e) PK/3
Compressibility (K) is the reciprocal of bulk modulus (B) and we have
B = –P/(dV/V) where dV is the change in volume V due to a pressure change P. The negative sign indicates that a decrease in pressure will produce a increase in volume.
In terms of compressibility, the above equation can be written as
B = 1/K = –P/(dV/V) so that the fractional change in volume is given by
dV/V = PK.
But V = (4/3) πR3 where R is the radius of the sphere so that dV/V = 3 dR/R. The fractional change in radius is therefore given by dR/R = (dV/V)/3 = PK/3.
You may find all posts on elasticity on this site by clicking on the label ‘elasticity’ below this post or on the side of this page.
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