Test: Elastic Moduli - JEE MCQ

Young's Modulus of elasticity =stress/strain
Y= [F/a/△l/l] ​ or Y= [Fl​/a△l]
or △l= [Fl/aY] ​= Fl​/ πr²Y
In the given problem, △l∝ l​/ r²
When both l and r are doubled, △l is halved.

Stress = Force per area and strain is unit less.
Therefore in CGS units dyne per cm square.
Hence A is correct.

We know that magnitude of bulk modulus
K=[P/(dV/V)]
Now, percentage change in volume is. 10^-3 %
Therefore, (dV/V)x100=10^-3
So, dV/V=10^-5
Hence, k is,
K=40atm/10^-5=40x1.01x10⁵ /10^-5 =4.04x10¹¹ N/m² 

Same elastic properties in all direction is called the homogenity of a material.

As the temperature increases, the inter-atomic distance also increases. So, it results in an increase in the area (stress = force /area). As the area increases stress decreases, this results in a decrease in Young's modulus.

Explanation:

When a deforming force is applied to a body in a direction parallel to its surface, it produces a shear strain. Shear strain is the deformation that occurs when one layer of a material slides past another layer. This type of deformation produces shear stress, which is the force that is acting parallel to the surface of the material. Shear stress is calculated as the force per unit area and is expressed in units of pressure, such as pascals (Pa) or pounds per square inch (psi).

Modulus of rigidity is defined as the ratio of Shear stress to shear strain.

E=stress/strain
Strain x E= Stress

F =AYΔl/L
Δl = FL/AY