Question

If the ratio of the density of ice to that of wood is $\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)$and the relative density of ice is$0.9$, then the density of wood is _____$\mathrm{kg}{\mathrm{m}}^{-3}$(Take the density of water=${10}^3\mathrm{kg}{\mathrm{m}}^{-3}).$)

• A. $0.8$
• B. $800$
• C. $8\times {10}^3$
• D. $80$

Answer 1

The correct option is B

$800$

Step 1: Given data:

$$\begin{gathered} \frac{\mathrm{Density}\mathrm{of}\mathrm{ice}\left({\mathrm{\rho }}_{\mathrm{i}}\right)}{\mathrm{Density}\mathrm{of}\mathrm{wood}\left({\mathrm{\rho }}_{\mathrm{w}}\right)}=\frac{9}{8}...........\left(\mathrm{i}\right) \\ \left({\mathrm{\rho }}_{\mathrm{w}}\right)\mathrm{Density}\mathrm{of}\mathrm{water}={10}^3\mathrm{kg}{\mathrm{m}}^{-3} \\ \mathrm{Relative}\mathrm{density}\mathrm{of}\mathrm{ice}=0.9 \end{gathered}$$

Step 2: On applying the relative density formula, we get:

$$\begin{gathered} \because \mathrm{Relative}\mathrm{Density}=\frac{{\mathrm{\rho }}_{\mathrm{ice}}}{{\mathrm{\rho }}_{\mathrm{water}}} \\ 0.9=\frac{{\mathrm{\rho }}_{\mathrm{ice}}}{1000} \\ \because {\mathrm{\rho }}_{\mathrm{ice}}=0.9\times 1000 \\ =900\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3$}\right. \end{gathered}$$

By putting the density of ice into ${\mathrm{eq}}^{\mathrm{n}}\left(\mathrm{i}\right)\mathrm{we}\mathrm{get}$

$$\begin{gathered} \frac{{\mathrm{\rho }}_{\mathrm{ice}}}{{\mathrm{\rho }}_{\mathrm{wood}}}=\frac{9}{8} \\ \frac{900}{{\mathrm{\rho }}_{\mathrm{wood}}}=\frac{9}{8} \\ {\mathrm{\rho }}_{\mathrm{wood}}=\frac{900\times 8}{9} \\ =800\left(\raisebox{1ex}{$\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{m}}^3$}\right.\right) \end{gathered}$$

Hence the correct option is B, The density of the wood$=800kg{m}^{-3}$.