An object whose density is ρ₀ falls from a certain height into a liquid whose density is ρ₁. The potential energy U of the object is plotted along the vertical axis and the position of the object (its altitude) is plotted along the horizontal axis. Vertically upward direction is taken as positive and origin is taken at the interface of the liquid. If ρ₁/ρ₀ = N/(3 − √3), find the value of N.
Physics · Mixed Chapter
JEE Mains

Question

An object whose density is ρ₀ falls from a certain height into a liquid whose density is ρ₁. The potential energy U of the object is plotted along the vertical axis and the position of the object (its altitude) is plotted along the horizontal axis. Vertically upward direction is taken as positive and origin is taken at the interface of the liquid. If ρ₁/ρ₀ = N/(3 − √3), find the value of N.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Above the liquid, the slope dU/dx=mg; the graph makes 45° with the x-axis, so its slope magnitude is 1 in the plotted scale. Step 2: Below the interface, the effective vertical force is (ρ₀−ρ₁)Vg and the slope is negative, with magnitude tan30°=1/√3. Step 3: Therefore (ρ₁−ρ₀)/ρ₀=1/√3, hence ρ₁/ρ₀=1+1/√3. Step 4: Equating 1+1/√3=N/(3−√3) yields N=2.