Question

a 20kg load is suspended by a wire of cross section 0.4 mm² The stress produced in Nm-2 is​

Answer 1

Given :

Mass = 20 kg

Area of cross section = 0.4 mm² = 0.4 × 10¯⁶m²

To Find :

The stress produced

Solution :

Stress is the external force per unit area of the body,

$\bf Stress = \dfrac{Froce }{Area}$

by substituting the values in the formula,

$\dashrightarrow \sf Stress = \dfrac{Froce }{Area}$

$\dashrightarrow \sf Stress = \dfrac{mg }{0.4 \times {10}^{ - 6} }$

$\dashrightarrow \sf Stress = \dfrac{20 \times 10 }{0.4 \times {10}^{ - 6} }$

$\dashrightarrow \sf Stress = \dfrac{200 }{0.4 \times {10}^{ - 6} }$

$\dashrightarrow \sf Stress = 50 \times {10}^{7}$

$\dashrightarrow \sf Stress = 5 \times {10}^{8} \: Nm^{-2}$

Thus, the stress produced by the body is 5 × 10⁸ N/m².