Question

In order to twist one end of a wire, 2m long and 4 mm in diameter, though 45⁰, the torque required is -(h = 5 × 10¹⁰ Nm^–2)

• A. 0.49 Nm
• B. 3.49 Nm
• C. 49 Nm
• D. 4.9 × 10¹⁰ Nm

Answer 1

Answer: (A)

0.49 Nm

Answer 2

$\tau = \frac{22 \times 5 \times 10^{10}(2 \times 10^{- 3})^{4} \times 22 \times 45}{7 \times 7 \times 180 \times 2 \times 2}Nm$ = 0.49Nm