Question

The Young’s modulus of the material of a wire is $6 \times {10^{12}}N{m^{ - 2}}$ and there is no transverse in it, then its modulus of rigidity will be:
A) $3 \times {10^{12}}N{m^{ - 2}}$
B) $2 \times {10^{12}}N{m^{ - 2}}$
C) ${10^{12}}N{m^{ - 2}}$
D) $\text{None of the above}$

Answer 1

Hint: First, consider the relation between young’s modulus and modulus of rigidity of the material of the wire, i.e., $Y = 2\eta (1 + \sigma )$ . As no transverse is there in the wire, we can take $\sigma = 0$ . Now, from the above relation, calculate the modulus of rigidity $\eta$ .

Complete step by step solution:
From the relation between young’s modulus and rigidity modulus, we know that, $Y = 2\eta (1 + \sigma )$ ;
Where
$Y =$ Young’s modulus of the material of the wire,
$\eta =$ modulus of rigidity of the material of the wire,
$\sigma =$ transverse strain of the wire
By the given problem, there is no transverse strain in the wire, so $\sigma = 0$ .
So, we have $Y = 2\eta$ from the above relation.
Here, young’s modulus of the material of the wire is $Y = 6 \times {10^{12}}N{m^{ - 2}}$
Therefore, the value of modulus of rigidity of the material of the wire will be,
$\Rightarrow \eta = \dfrac{Y}{2}$
$\Rightarrow \dfrac{{6 \times {{10}^{12}}}}{2}N{m^{ - 2}}$
$\Rightarrow 3 \times {10^{12}}N{m^{ - 2}}$

The correct solution is (A), $3 \times {10^{12}}N{m^{ - 2}}.$

Additional information:
Young’s modulus is the ratio of longitudinal (tensile or compressive) stress to the longitudinal (tensile or compressive) strain, within elastic limit, with no force applied to prevent the associated lateral change in the dimension.
$Y = \dfrac{{F/A}}{{\Delta l/l}}$
Rigidity modulus or shear modulus is the ratio of shearing stress (tangential stress) to the shearing strain (angle of shear), within the elastic limit.
$\eta = \dfrac{{F/A}}{{\tan \theta }}$
When $\theta$ is very small we have $\tan \theta \approx \theta$-
$\eta = \dfrac{{F/A}}{\theta }$
The relation between the two elastic constants is $Y = 2\eta (1 + \sigma )$ .

Note: Here, transverse strain $(\sigma )$ is the ratio of the change in diameter of a circular bar of a material to its diameter because of deformation in the longitudinal direction. It is also known as lateral strain. This quantity is dimensionless because of being a ratio between two quantities of the same dimension.