SSC JE Mechanical Engineering 27 Sep 2019 Shift 1 Paper

Torsion Equation:
$\frac{\mathit{T}}{\mathit{J}}=\frac{\mathrm{\tau }}{\mathit{r}}=\frac{\mathit{G}\mathrm{\theta }}{1}$
Where,$\mathit{T}=$ torque applied on the shaft in $\mathit{N}-\mathit{m}$
$\mathit{J}={\mathit{I}}_{\mathit{X}\mathit{x}}+{\mathit{I}}_{\mathit{y}\mathit{y}}=\frac{\mathrm{\pi }{\mathit{d}}^4}{32}=$ polar moment of inertia ${\mathit{m}}^4$
$\mathrm{\tau }=$ shear stress in $\mathit{M}\mathit{P}\mathit{a}$,
r = radius of the shaft in $\mathit{m}$,
$\mathit{G}=$ shear modulus or modulus of rigidity in MPa
$\mathrm{\theta }=$ angle of twist in radian and
$\mathit{l}=$ length of the shaft in $\mathit{m}.$
${\mathrm{\tau }}_{\max }$ occurs at ${\mathit{r}}_{\max }$ i.e at $\mathit{R}$.
$\frac{\mathit{T}}{\mathit{J}}=\frac{\mathrm{\tau }}{\mathit{r}}\Rightarrow {\mathrm{\tau }}_{\max }=\frac{\mathit{T}}{\mathit{J}}$ ${\mathit{r}}_{\max }=\frac{\mathit{T}}{\frac{\mathrm{\pi }{\mathit{d}}^4}{32}}.\frac{\mathit{d}}{2}=\frac{16\mathit{T}}{\mathrm{\pi }{\mathit{d}}^3}$