Question

A mild steel torsion bar is used for measurement of a torque of $200Nm$. If the shear modulus of mild steel is $100GN/m^2$ and $50mm$ diameter then the angle of twist in $\left(\mu rad\right)$ is

• A. $40.53$
• B. $81.53$
• C. $32.53$
• D. $20.53$

Answer 1

The correct option is B $81.53$

$\text{For torsion bar we know}$

$16T=\pi G{d}^3\theta$

$\theta =\frac{16T}{\pi G{d}^3}$

$=\frac{16\times 200}{3.14\times 100\times {10}^9\times (50\times {10}^{-3}{)}^3}$

$=81.53\times {10}^{-6}rad$

$\theta =81.53\left(\mu rad\right)$