A solid spherical ball having density d and volume v floats on the interface of two immiscible liquids. The density of the liquid in the upper portion is d_(1) and that of the liquid in the lower portion is d_(2). What parts of the ball will remain in the liquids in the upper and lower portions respectively, if d_(1)ltdltd_(2)?

A solid spherical ball having density d and volume v floats on the int

1.	A solid spherical ball having density d and volume v floats on the interface of two immiscible liquids. The density of the liquid in the upper portion is d_(1) and that of the liquid in the lower portion is d_(2). What parts of the ball will remain in the liquids in the upper and lower portions respectively, if d_(1)ltdltd_(2)?
Answer» Solution :SUPPOSE a volume x of the ball remains in the liquid in the UPPER PORTION. So, a volume (V -x) remains in the liquid in the lower portion. According to the principle of floatation, weight of the body = weight of displaced liquid or, `vd=xd_(1)+(v-x)d_(2)or,x(d_(1)-d_(2))=v(d-d_(2))` or, `x/v=(d-d_(2))/(d_(1)-d_(2))=(d_(2)-d)/(d_(2)-d_(1))` `therefore` The PART of the ball that remains in the liquid in the upper portion = `(d_(2)-d)/(d_(2)-d_(1))` , and the part of the ball remains in the liquid in the lower portion = `(1-(d_(2)-d)/(d_(2)-d_(1)))=(d-d_(1))/(d_(2)-d_(1))`.