Question

Young’s modulus of rubber is ${{10}^{4}}N{{m}^{-2}}$ and area of cross section is 2 $c{{m}^{2}}$. If force of $2\times {{10}^{5}}$ dyn is applied along its length, then its initial length l becomes
A. 3l
B. 4l
C. 2l
D. None of these

Answer 1

Hint: Young’s modulus (Y) is the ratio of longitudinal stress to the longitudinal strain, i.e. $\text{Y=}\dfrac{\text{stress}}{\text{strain}}$. Calculate the stress in the rubber and its Young’s modulus is given. Then you can find the value strain from the formula. Hence, you can calculate the final length of the rubber.

Formula used:

$\text{Y=}\dfrac{\text{stress}}{\text{strain}}$
$\sigma =\dfrac{F}{A}$
$1N={{10}^{5}}dyn$
$strain=\dfrac{l'-l}{l}$

Complete step by step answer:
Young’s modulus (Y) is the ratio of longitudinal stress to the longitudinal strain.

Hence, $\text{Y=}\dfrac{\text{stress}}{\text{strain}}$.

Longitudinal stress is defined as perpendicular force applied to the surface per unit area of the surface.

Let the stress in the body be $\sigma$, force applied be F and surface area be A.

Hence, $\sigma =\dfrac{F}{A}$.

When a force is applied on a body, the dimensions of the body change. The ratio of the change in any dimension of the body to the original dimension is called strain. The strain along the direction of perpendicular force is called longitudinal strain.

Consider a cylindrical tube. Suppose a force is applied on the both ends of the cylinder such that the force is perpendicular to the cross sectional surface of the cylinder.

Here, the force (F) applied on the rubber is equal to $2\times {{10}^{5}}$ dyn. This force is applied on an area of 2 $c{{m}^{2}}$. Therefore, the stress ($\sigma$) in the rubber is $\sigma =\dfrac{F}{A}=\dfrac{2\times {{10}^{5}}}{2}={{10}^{5}}dync{{m}^{-2}}$.

It is given that the value of young’s modulus for the rubber is ${{10}^{4}}N{{m}^{-2}}$.

We know that $1N={{10}^{5}}dyn$ and 1m = 100cm.

Therefore, $Y={{10}^{4}}N{{m}^{-2}}={{10}^{4}}\times {{10}^{5}}dyn\times {{\left( 100cm \right)}^{-2}}={{10}^{5}}dync{{m}^{-2}}$.

Here, the strain will be in the length of the rubber. Looking at the given options, we understand that the length increases when the force is applied. This means that the force is a pulling force.

We know that $\text{Y=}\dfrac{\text{stress}}{\text{strain}}\Rightarrow strain = \dfrac{stress}{Y} = \dfrac{{{10}^{5}}}{{{10}^{5}}} = 1$.

It is given that the initial length of the rubber is l. Let the increased length be l’.

Therefore, change in length is l’-l.
Hence, the strain in the rubber will be $\dfrac{l'-l}{l}$.
And we just now found that strain is equal to 1.
Therefore, $\dfrac{l'-l}{l}=1\Rightarrow l'-l=l\Rightarrow l'=2l$.

Hence, the correct option is C.

Note: Strain can be positive as well as negative. The value of strain depends on the nature of applied force. If it is a pulling force, the body will elongate and if it is a compressing force, the body will shrink or compress. The change in the length (longitudinal strain) of the body is always in the direction of applied force.