Question

A 6.4 kg solid sphere made of metal whose density is 3400 kg/m^3 is held under an unknown liquid by a cord. The tension in the cord is 38 N. (a)What is the density of the liquid?

(b)When the cord is cut, what percentage of the volume of the sphere will be submerged in the liquid?

Answer 1

Weight of sphere is equal to 64N and tension in rope is equal to 38 N that means total up thrust by the liquid on the sphere is equal to 64 + 38 = 102N

Now, up thrust is given by the weight of total liquid displaced by the sphere.

102 N= volume of sphere × density of liquid × g

102 N =(mass of sphere/density of sphere )× density of liquid

102 = 6.4 × density of liquid × 10/3400

Density of liquid = 5593.54 kg/m³.

B.) When the rope is cut then for equilibrium case weight of sphere equals to weight of liquid diapldisp by immersed Part of sphere

64 N = V(immersed) × 5593.54 ×10 (let sphere is immersed to x percent then v(immersed) is equal to x.vol of sphere/100)

So 64 = x × 6.4 × 5593.54 × 10 /( 3400 × 100)

X = 60.78 ans.