Page 107 - Physics

Sinking and fl oating

Example 5.1 Factors aff ecti ng upthrust

In an experiment to study the behaviour According to Archimedes’ principle, an
of the weight of an object a student fi nds object submerged in a fl uid experiences
that the weight of the object is 4.9 N an upthrust that is equal to the weight of
in air and its weight when completely the fl uid displaced. That is,
submerged in water is reduced to 3.1 N. Upthrust (U) = Weight of displaced fl uid (w ).
f
Calculate the upthrust.
But,
Solution Weight of the fl uid displaced
Weight of the object in air = 4.9 N w ( ) = mass m ( ) × acceleration due to gravity g ( )
weight of the object in water = 3.1 N f f
Mass of the fl uid (m ) = density of the fl uid
upthrust = weight of an object in air − ρ ( ) × volume of the displaced fluid V ( )
f
weight of an object in water f f
= 4.9 N − 3.1 N m = ρ V f
f
f
= 1.8 N In Chapter Four, you learned that the
volume of the displaced fl uid is equivalent
Therefore, upthrust acting on the object
is 1.8 N. to the volume of the immersed object.

That is,
Example 5.2 Volume of the fluid displaced (V ) =
f
A body immersed in water displaces volume of object (V )
1.1 N. If its weight while in water was o
3.3 N, fi nd its weight in air. ( ) = mass m f × acceleration due to gravity g ( )
( )
w
f

Solution Since, V = V
o
f
Weight of the displaced water = 1.1 N m = ρ V
f f 0
Weight of the body in water = 3.3 N
U = m g
upthrust = weight of an object in air − f
weight of an object in water Thus, U = ρ V g
0
f
Upthrust = 1.1 N That is,
upthrust = densityof the fluid×
Weight of the body in air = Apparent
loss in weight + Apparent weight volumeof theobject×g
= 3.3 N + 1.1 N = 4.4 N Hence, upthrust is affected by:

3.3 N + 1.1 N = 4.4 N (i) the volume of the immersed object;
Therefore, weight of the body in air (ii) the density of the fl uid in which the
is 4.4 N. object is immersed; and
(iii) the acceleration due to gravity.

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Physics Form 1 Final.indd 101 16/10/2024 20:56