Question

Young’s modulus of elasticity for A and B are $Y_A=1.1\times {10}^{11}Pa$ and $Y_B=2.0\times {10}^{11}Pa$ which is more elastic and why?

Answer 1

Hint: In order to solve this question, we should know about Young’s modulus of elasticity. Young modulus of elasticity is simply the ratio of pressure acting on a body called stress and strain which is the ratio of change in length to original length, here we will use the stress and strain concept to know which body is more elastic.

Formula used:
If s denoted for the stress acting on a body and x denote for strain produced in the body due to this stress, then Young’s modulus of elasticity Y can be written as
$Y={\displaystyle \frac{s}{x}}.$

Complete step by step answer:
According to the question, we have given the Young’s modulus of elasticity for A and B are $Y_A=1.1\times {10}^{11}Pa$ and $Y_B=2.0\times {10}^{11}Pa$ The body will be more elastic to which we have to apply larger amount of stress s to produce same amount of strain x.
Let $s_A$ and $s_B$ be the amount of stress applied on body A and B to produce same amount of strain x then using formula, $Y={\displaystyle \frac{s}{x}}.$ we get,
$Y_A={\displaystyle \frac{s_A}{x}}$ and $Y_B={\displaystyle \frac{s_B}{x}}$
on equating both equations we get,
${\displaystyle \frac{s_A}{Y_A}}={\displaystyle \frac{s_B}{Y_B}}$
or
${\displaystyle \frac{Y_B}{Y_A}}={\displaystyle \frac{s_B}{s_A}}$ and since,
$Y_A=1.1\times {10}^{11}Pa$ and $Y_B=2.0\times {10}^{11}Pa$ which shows that,
$Y_B>Y_A$
so,
$s_B>s_A$
So, stress in body B is larger than in body A.
Hence, body B is more elastic due to its high value of stress as well as young’s modulus of elasticity.

Note: It should be remembered that, young’s modulus of elasticity is directly proportional to the stress applied on the body so, larger the value of young’s modulus of elasticity, larger the stress and results in the body will be more elastic.