Answer to Question #211626 in Optics for Tawanda

Question

A rectangular block of low density material weighs 25.0 N. A thin string is attached to the center

of the horizontal bottom of the block and tied to the bottom of a beaker partially filled with water.

When 20% of the block’s volume is submerged, the tension in the string is 15.0 N.

(i) Calculate the buoyant force acting on the block.(ii)The string breaks when the tension reaches 65 N, and at this moment, 25%

of the block’s volume is still below the water line. Determine the fraction

of the block’s volume that is below the top surface of the oil.

[3,5]

Answer 1

$F_0=T-F_T=65-15=50~N,$

$F_B=P+F_T=25+15=40~N,$

$\frac{F_0}{F_B}=\frac{\rho_0\Delta Vg}{\rho_w \frac V4 g},\implies$

$\frac {\Delta V}{V}=\frac{4\rho _w F_0}{\rho _0 F_B}=0.625=62.5\%.$