If a solid circular shaft of steel 2 cm in diameter is subjected to a permissible shear stress of 10 kN/cm^{2}, then the value of the twisting moment (Tr) will be)

This question was previously asked in

UPRVUNL JE ME 2016 Official Paper Shift 1

- 10π kN-cm
- 20π kN-cm
- 15π kN-cm
- 5π kN-cm

Option 4 : 5π kN-cm

**Concept:**

From the torsional equation of the shaft

\(\mathop \tau \nolimits_{\max } = \frac{{16T}}{{\pi \mathop D\nolimits^3 }}\)

\(T = \frac{{\mathop \tau \nolimits_{\max } \pi \mathop D\nolimits^3 }}{{16}}\)

__Calculation:__

**Given:**

D = 2 cm,\(\mathop \tau \nolimits_{\max } = 10 ~{kN}/{cm^2}\)

\(T = \frac{10 ~\times ~\pi ~\times ~8}{16}=5\pi\)kN-cm

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