Question

a aluminium rod has a breaking strain 0.2 % the minimum cross sectional area of rod in m square in order to support a load of 10 power 4 Newtons is ?youngs modulus is 7 * 10 power 9

Answer 1

Answer on Question#36837 - Physics - Mechanics

Young’s modulus is

$$Y = \frac{F / A}{\text{breaking strain}},$$

where $A$ - the minimum cross sectional area of rod, $F$ - a load.

We have

$$A = \frac{F}{Y * \text{breaking strain}} = \frac{10^4 * 100}{7 * 10^9 * 0.2} = 7.1 * 10^{-4} m^2.$$

Answer: $7.1 * 10^{-4} m^2$.