KEAM (Engineering) 2007 Questions on Elasticity

You will often find questions involving elastic potential energy in entrance examinations for admission to professional and other degree courses. Here is a question which appeared in Kerala Government Engineering Entrance 2007 test paper:

A wire of natural length L_, Young’s modulus Y and area of cross section A is extended by x. Then the energy stored in the wire is given by

(a) (YA/2L)x² (b) (YA/3L)x² (c) (YL/2A)x² (d) (YA/2L²)x² (e) (A/2YL)x²

The elastic potential energy per unit volume is ½ stress×strain. Since the Young’s modulus Y = stress/strain, we can modify the above expression as (½)Y×strain². The elastic potential energy stored in the entire wire is (½)Y×strain² × volume of the wire = (½)Y×strain² × AL = (½)Y×(x/L)² × AL = (YA/2L)x².

Here is another question which appeared in Kerala Government Engineering Entrane 2007 test paper:

The length of a rubber cord is l₁ metre when the tension is 4N and l₂ metre when the tension is 6N. The length when the tension is 9N is

(a) (2.5 l₂ – 1.5 l₁) m (b) (6 l₂ – 1.5 l₁) m (c) (3 l₁ – 2 l₂) m

(d) (3.5 l₂ – 2.5 l₁) m (e) (2.5 l₂ + 1.5 l₁) m

You can use Hooke’s law to solve this question. Since the increase in length is directly proportional to the force (tension) applied, in accordance with Hooke’s law, we have

l₂ – l₁ = K(6 – 4) where K is the constant of proportionality.

[Note that the increase in length from l₁ to l₂ is produced by the increase in tension from 4N to 6 N].

If the length of the rubber cord is l₃ when the tension is 9 N, we have

l₃ – l₁ = K (9 – 4)

Dividing the first equation by the second, we obtain

(l₂ – l₁)/(l₃ – l₁) = 2/5, from which l₃ = (2.5 l₂ – 1.5 l₁) metre.

You will find more questions on elasticity on clicking on the label ‘elasticity’ below this post or on the on the side of this page.