In a Searle's experiment, the load is 5.000 ± 0.005 kg, the length is 2.00 ± 0.01 m, the diameter…

Step 1. The modulus. The numerator is $4MgL = 4 \times 5.000\ \text{kg}$ $\times 9.8\ \text{m/s}^2$ $\times 2.00\ \text{m}$ $= 392\ \text{N m}$. The denominator is $\pi d^2 \Delta L = \pi \times (0.60 \times 10^{-3}\ \text{m})^2$ $\times 1.75 \times 10^{-3}\ \text{m}$ $= 1.98 \times 10^{-9}\ \text{m}^3$. So $Y = \dfrac{392\ \text{N m}}{1.98 \times 10^{-9}\ \text{m}^3}$ $= 1.98 \times 10^{11}\ \text{N/m}^2$.

Step 2. The four terms. $\dfrac{\Delta M}{M} = \dfrac{0.005}{5.000} = 0.10\%$. $\dfrac{\Delta L}{L} = \dfrac{0.01}{2.00} = 0.50\%$. $2\dfrac{\Delta d}{d} = 2 \times \dfrac{0.01}{0.60} = 3.33\%$. $\dfrac{\Delta(\Delta L)}{\Delta L} = \dfrac{0.01}{1.75} = 0.57\%$.

Step 3. The sum. $0.10 + 0.50 + 3.33 + 0.57$ $= 4.5\%$.