91Ó°ÊÓ

i) Volume and density of liquid 1 is $V_1=0.50\mathrm{L},{\mathrm{ÒÏ}}_1=2.6\mathrm{g}/{\mathrm{cm}}^3$

ii) Volume and density of liquid 2 is $V_2=0.50\mathrm{L},{\mathrm{ÒÏ}}_2=1.0\mathrm{g}/{\mathrm{cm}}^3$

iii) Volume and density of liquid 3 is$V_3=0.40\mathrm{L},{\mathrm{ÒÏ}}_3=0.80\mathrm{g}/{\mathrm{cm}}^3$

We can use the formula for force (weight) in terms of density and volume to find the forces on three liquids separately. Then adding them, we can get the force on the bottom of the container due to three liquids.

Formula:

Weight of a body in terms of force,

$$\begin{gathered} W=mg \\ =\mathrm{ÒÏ}Vg, \end{gathered}$$

where, (i)

Force on the bottom of the container due to liquids:

We know that$1\mathrm{L}=1000{\mathrm{cm}}^3$

For first liquid, using equation (i)

role="math" localid="1657193879159" $$\begin{gathered} W_1=2.6×\left(0.50×1000\right)×980 \\ =1.27×{10}^6\mathrm{g}·\mathrm{cm}/{\mathrm{s}}^2 \\ =1.27×{10}^6\mathrm{dyne} \\ =12.7\mathrm{N}\left(âˆ\mu 1\mathrm{N}={10}^5\mathrm{dyne}\right) \end{gathered}$$

Similarly, we can find the weight of second liquid using equation (i)

role="math" localid="1657194157414" $$\begin{gathered} W_2=0.25×1.0×1000×980 \\ =2.45×{10}^5\mathrm{g}·\mathrm{cm}/{\mathrm{s}}^2 \\ =2.45×{10}^5\mathrm{dyne} \\ =2.45\mathrm{N}\left(âˆ\mu 1\mathrm{N}={10}^5\mathrm{dyne}\right) \end{gathered}$$

For third liquid, using equation (i) we have

$$\begin{gathered} W_3=0.40×0.80×1000×980 \\ =3.14×{10}^5\mathrm{g}·\mathrm{cm}/{\mathrm{s}}^2 \\ =3.14×{10}^5\mathrm{dyne} \\ =3.14\mathrm{N}\left(âˆ\mu 1\mathrm{N}={10}^5\mathrm{dyne}\right) \end{gathered}$$

Total force on the bottom of the cylinder is

$$\begin{gathered} F=W_1+W_2+W_3 \\ =12.7+2.45+3.14 \\ =18.29\mathrm{N} \\ ≈18.3\mathrm{N} \end{gathered}$$

Therefore, the force on the bottom of the container due to three liquids is$18.3\mathrm{N}$