Answer:
The water level would rise by [tex]5\; \rm cm[/tex].
Step-by-step explanation:
Assume that the initial water level is [tex]h\; \rm cm[/tex].
Before the cube was placed in the tank, the volume of the water was the same as the volume of a rectangular prism with a length of [tex]160\; \rm cm[/tex], a height of [tex]80\; \rm cm[/tex], and a height of [tex]h\; \rm cm[/tex]:
[tex]\begin{aligned} & 160\; {\rm cm} \times 80\; {\rm cm} \times h\; {\rm cm} \\ =\; & (80 \times 160) \, h \; \rm cm^{3} \end{aligned}[/tex].
Let the rise in the water level in the tank be [tex]x\; \rm cm[/tex]. The new water level would be [tex](x + h)\; \rm cm[/tex].
The volume of the water and the submerged cube, combined, would be the same as that of a rectangular prism with a length of [tex]160\; \rm cm[/tex], a width of [tex]\rm 80\; \rm cm[/tex], and a height of [tex](x + h)\; \rm cm[/tex]:
[tex]\begin{aligned} & 160\; {\rm cm} \times 80\; {\rm cm} \times (x + h)\; {\rm cm} \\ =\; & (80 \times 160) \, (x + h) \; \rm cm^{3} \end{aligned}[/tex].
In other words:
- Volume of water in this tank: [tex](80 \times 160)\, h\; \rm cm^{3}[/tex], whereas
- Volume of water in this tank, plus the volume of the submerged cube: [tex](80 \times 160) \, (x + h) \; \rm cm^{3}[/tex].
Therefore, the volume of the submerged cube could be expressed as:
[tex]\begin{aligned} & (80 \times 160) \, (x + h) \; {\rm cm^{3}} - (80 \times 160) \, h \; \rm cm^{3} \\ =\; &(80 \times 160)\, x\; {\rm cm^{3}}\end{aligned}[/tex].
On the other hand, the volume of the cube could be expressed as:
[tex](40\; \rm cm)^{3} = 64000\; \rm cm^{3}[/tex].
Equate these two expressions for the volume of the cube and solve for [tex]x[/tex], the rise in the water level in the tank:
[tex](80 \times 160)\, x = 64000[/tex].
[tex]x = 5[/tex].
In other words, the rise in the water level in this tank would be [tex]5\; \rm cm[/tex].
