A load of 50 N is suspended from a wire that has an area of cross-section of 1 mm² - AQA - A-Level Physics - Question 19 - 2019 - Paper 1

The stress ($\sigma$) in a material is defined as the load (force) applied per unit area. The formula for stress is given by:

$\sigma = \frac{F}{A}$

Where:

• $F$ is the force (in Newtons)
• $A$ is the area (in square meters)

Given:

• Load, $F = 50 \, \text{N}$
• Cross-sectional area, $A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2$ (Note: conversion from mm² to m²)

Now, substituting the values:

$\sigma = \frac{50}{1 \times 10^{-6}} = 50 \times 10^{6} \, \text{Pa} = 5 \times 10^{7} \, \text{Pa}$