Two spheres weighing 10 KN and 25 KN are each 1.6 m in diameter. They are connected by a short rope and placed in water. Find the tension in the connecting rope. Find also what portion of the lighter sphere will protrude above the water surface.

SOLUTION:

Here,

Considering the lower sphere,

Buoyant force = net downward force

or,`4/3pi**0.8^3**9.81**1=25-T`

or,`T=3.958 KN`

Similarly, Considering the upper sphere,

Buoyant force = net downward force

or,`(V_(i\n)*rho_w*g)/1000=10+T`

or,`V_(i\n)**9.81=10+3.958`

`V_(i\n)=1.4228 m^3`

Thus the volume above water is,

`=V-V_(i\n)`

`=4/3 pi**0.8^3-1.4228`

`=0.722 m^3`

Now, the percentage of volume above water is,

`=0.722/2.145 **100%`

`=33.6.... Show More

SOLUTION:

Here,

Considering the lower sphere,

Buoyant force = net downward force

or,`4/3pi**0.8^3**9.81**1=25-T`

or,`T=3.958 KN`

Similarly, Considering the upper sphere,

Buoyant force = net downward force

or,`(V_(i\n)*rho_w*g)/1000=10+T`

or,`V_(i\n)**9.81=10+3.958`

`V_(i\n)=1.4228 m^3`

Thus the volume above water is,

`=V-V_(i\n)`

`=4/3 pi**0.8^3-1.4228`

`=0.722 m^3`

Now, the percentage of volume above water is,

`=0.722/2.145 **100%`

`=33.66%`