If weight of body in air is $w_1=8.00\pm 0.05N$ and weigh in water is $w_2=6.00\pm 0.05N$, then relative density $\rho=\frac{w_1}{w_1-w_2}$ with max permissible error is?
I solved by doing the following, although it’s probably wrong $$\frac{\Delta \rho}{\rho}=\frac{\Delta w_1}{w_1}+\frac{\Delta(w_1-w_2)}{w_1-w_2}$$
Plugging in the values we end up with $$=\frac{0.05}{8}+\frac{0.1}{2}$$ I took $\Delta (w_1-w_2)$ and $0.05+0.05=0.1$ and $w_1-w_2=8-6=2$ Calculating I get the final value as $\frac{0.45}{8}$. The options are not matching. The correct answer is $4.00\pm 5.62$%