Question

If the density of the material increases, the value of Young’s modulus :
A) increases.
B) decreases.
C) first increases, then decreases.
D) first decreases, then increases.

Answer 1

Hint
Use the formula ${{\;\gamma = }}\dfrac{{{\text{stress}}}}{{{\text{strain}}}}$ and evaluate the expression of young’s modulus in terms of density by using the relation \[{\text{density = }}\dfrac{{{\text{mass}}}}{{{\text{volume}}}}\] and hence evaluate the answer. For creating the term volume in the expression of Young’s modulus, multiply numerator and divide denominator by L.

Complete step by step answer
Given that the density of the material is increasing.
We know that ${{\;\gamma = }}\dfrac{{{\text{stress}}}}{{{\text{strain}}}}$
Putting the required values we have,
$\Rightarrow \;\gamma = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}\; $
$\Rightarrow \;\gamma = \dfrac{{FL}}{{\Delta LA}} $
Multiplying numerator and denominator by L we have,
$ \Rightarrow \;\gamma = \dfrac{{F{L^2}}}{{\Delta L(AL)}}$
We know,
${\text{density = }}\dfrac{{{\text{mass}}}}{{{\text{volume}}}} $
$\Rightarrow d = \dfrac{m}{v} $
$\Rightarrow v = \dfrac{m}{d} $
Putting the volume in above equation we have,
$\Rightarrow \;\gamma = \dfrac{{F{L^2}d}}{{\Delta L \times m}}$
As it is clear from the above formula that Young’s modulus is directly proportional to density, hence on increasing density, the value of Young’s modulus increases.
Hence the correct answer is option (A) ( increases).

Note
Alternatively we know that $Stress = \;\gamma \times strain$. And $\;\gamma $ is the property of material to oppose the deformation or in other words resist the deformation. So, in a body when density increases, it means stiffness increases and thus the body is more rigid and if the body is more rigid it cannot be deformed.