Question

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a solid of weight 20 N is dippedd in water. if the weight of the solid in water is 18N what is the relative density of the solid?

Answer 1

$$\begin{gathered} \mathbf{Relative}\mathbf{}\mathbf{density}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solid}\mathbf{:} \\ \mathrm{RD}=\frac{\mathrm{Weight}\mathrm{of}\mathrm{solid}\mathrm{in}\mathrm{air}}{\mathrm{Loss}\mathrm{of}\mathrm{weight}\mathrm{in}\mathrm{water}} \\ =\frac{20}{20-18} \\ =10 \end{gathered}$$