A low-density block with a weight of 10.0 N is placed in a beaker of water and t

Question

A low-density block with a weight of 10.0 N is placed in a beaker of water and tied to the bottom of the beaker by a vertical string of fixed length. When the block is 25% submerged, the tension in the string is 17.0 N. The string will break if its tension exceeds 60.0 N. As water is steadily added to the beaker, the block becomes more and more submerged.

(a) What percentage of the block is submerged at the instant the string breaks?

(b) After the string breaks and the block comes to a new equilibrium position in the beaker, what fraction of the block's volume is submerged?

Explanation / Answer

tension in the string = buoyancy force

buoyancy force = Fb = d*Vsubmerged*g

(a)

T = 17 N

let V be the volume of the block

17 = 1000*0.25*V*9.8

Volume of the bloc = V = 6.94*10^-3 m^3

(a)

for T = 60

let x % is submerged

60 = 1000*x*6.94*10^-3*9.8

x = 0.882 = 88.2 % <<<_-----answerr

(b)

in equilibrium

Fb = Fg

1000*x*6.94*10^-3*9.8 = 10

x = 14.7 %

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