when i checked solution, i saw some different formula for J otherthan that is familier. so pls explain in detail

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For permissible shear stress $\text{'}{f}_s\text{'}$ the torque transmitted by a thin tube of mean diameter $D$ and wall thickness $\text{'}t\text{'}$ is :

$($b$)\{\frac{D}{2}\}t{f}_s$

$($c$)\{(D^2)\}t{f}_s$

$($d$)\{\frac{D^2{t}^2}{4}\}{f}_s$