Question

A brass rod cross-sectional area $1c{m^2}$ and length $0.2m$ is compressed lengthwise by a weight of $5kg$ . If Young’s modulus of elasticity of brass is $1 \times {10^{11}}N{m^{ - {2^{}}}}$ and $g = 10m{s^{ - {2^{}}}}$ , then increase in the energy of the rod will be
a. ${10^{ - 5}}$ joules
b. $2.5 \times {10^{ - 5}}$ joule
c. $5 \times {10^{ - 5}}$ joule
d. $2.5 \times {10^{ - 4}}$ joule

Answer 1

Hint: Young’s modulus is the property of every material. It is defined as the material to withstand the compression or the elongation with respect to its length. In other words how the material is easily bent or stretched.
The Young’s Modulus Is temperature and pressure dependent. The unit of Young’s modulus is $N{m^{ - 2}}$. It is also known as Modulus of Elasticity. It is denoted as E or Y.
Stress: The stress is defined mechanically as restoring force applied per unit area of the material.
Strain: It is defined as the amount of deformation of a solid due to stress. It is the ratio of the change in size, shape to original size or shape. Strain may be classified as Normal strain and Shear strain.

Formula Used:
We will be using the relation of increase in energy $U = \dfrac{1}{2}F\Delta l$.

Complete step by step answer:
Modulus of elasticity provides the mechanical behavior of any material in response to the induced stress due to loading. It is given by,
$U = \dfrac{1}{2}F\Delta l$

Young’s Modulus $Y = \dfrac{{stress}}{{strain}}$
${\text{Stress = }}\dfrac{{{\text{Force}}}}{{{\text{Cross - sectional area}}}}$
${\text{Strain = }}\dfrac{{{\text{change in length}}}}{{{\text{Original length}}}}$
$Y = \dfrac{{F \times L}}{{A \times \Delta l}}$
Where, $F$-force exerted by the object under tension.
$A$-Cross sectional area
$\Delta l$- Change in length
$L$- Original length

Rearranging the above equations, We get
$\Delta l = \dfrac{{F \times L}}{{Y \times A}}$
Substitute the given values in the above equation,
$U = \dfrac{1}{2} \times \dfrac{{{F^2} \times L}}{{Y \times A}}$
$ \Rightarrow \dfrac{1}{2} \times \dfrac{{{{50}^2} \times 0.2}}{{{{10}^{11}} \times {{10}^{ - 4}}}}$
$U = 2.5 \times {10^{ - 5}}J$
The increase in energy of the brass rod is $2.5 \times {10^{ - 5}}$ joule.

Hence, the correct answer is option (B).

Note: Young’s Modulus is a measure of stress- strain relationship of an object. The Physicist Thomas Young describes the elastic properties of a solid undergoing tension or compression. Young’s Modulus, Bulk modulus, Rigidity modulus, Poisson’s ratio are the types of elasticity.