A piece of wood with a density of 766 kg/m 3 is tied with a string to the bottom

Question

A piece of wood with a density of 766 kg/m3 is tied with a string to the bottom of a water-filled flask. The wood is completely immersed, and the tension in the string is 0.79 N.

(a) What is the volume of the wood?
m3
(b) If the string breaks and the wood floats on the surface, does the water level in the flask rise, drop, or stay the same?
---Select--- rise? drop? or stay the same ?

(c) Assuming the flask is cylindrical with a cross sectional area of 65 cm2, find the change in water level after the string breaks.
cm

Explanation / Answer

The upward thurst or buoyant force ( Fb ) exerted by water due to displace by wood is given by:

Fb = V pwater g                        ( V = volume of wood, pwater = density of water , g = acceleration due to gravity)

From fig,    Fb   = T + mblock * g                     ( T = tension in string)

Fb = T + V * p'block *g                                   ( mass = density*volume,              p'block = density of block)

V*pwater *g = T + V*p'block *g

V = T / ( pwater - p'block ) g                         ( pwater = 1000 kg/m3 )

by pytting all the values:

V ( volume of block) = 3.44*10-4 m3

( b) When the string break water level will drop. Because when the block is floating , then amount of water displace will be reduced by the volume of wood above the surface.

( C) Volume of wood block = 344 cm3

volume of water the wood will displace when floating = p'wood * Vwood / pwater                 ( By Vol. = mass / density)

= 344* 766/ 1000                         ( pwater = 1g/cc)

= 263.5 cm3

Now volume below the water surface is decreased by= 344- 263 = 80.4 cm3

The level of flask will fall by = 80.4 / 65             ( by Vol. = area * Height,      H = Vol. / area)

                                         = 1.23 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.