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Upthrust Problems

The document contains 5 physics problems involving concepts like density, buoyancy, upthrust, and relative density. It provides the questions, given information, and step-by-step solutions for calculating various values related to submerged objects. For example, in one problem it calculates the upthrust and weight of an object in liquid using the density of the liquid and volume of the object. In another, it uses the law of floatation to find the volume of a block above and below the surface of a liquid.

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6K views3 pages

Upthrust Problems

The document contains 5 physics problems involving concepts like density, buoyancy, upthrust, and relative density. It provides the questions, given information, and step-by-step solutions for calculating various values related to submerged objects. For example, in one problem it calculates the upthrust and weight of an object in liquid using the density of the liquid and volume of the object. In another, it uses the law of floatation to find the volume of a block above and below the surface of a liquid.

Uploaded by

srinu247
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Q1: A body of volume 100cm 3 weighs 5kgf in air.

It is completely immersed in a liquid of


density 1.8 x 103 kg m-3. Find: (a) the upthrust due to liquid and (b) the weight of the body
in liquid.

Solution:

Given

Volume of the body, V = 100cm3 = 100 x 10-6 m3

Weight of the body in air, W = 5kgf

Density of the liquid, ρ = 1.8 x 103 kg m-3

(a) Upthrust due to the liquid = V ρ g


= 100 x 10-6 x 1.8 x 103 x g
= 0.18kgf. [Since 1kgf = g N]
(b) Weight of the body in the liquid = Weight of the body in air – upthrust
= 5kgf – 0.18kgf
= 4.82kgf.

Q2: A block of mass 10kg and volume 0.1m3 floats in a liquid of density 150 kg/m3.
Calculate:
a) Volume of the block above the surface of the liquid.
b) Density of the block.
Solution:

Given

Mass of the block, m = 10kg

Volume of the block, V = 0.1 m3

Density of the liquid, ρl = 150 kg m-3

Let Vl be the volume of the block immersed in the liquid.

m 10
Density of the block, ρ = = = 100 kg m-3
V 0.1

By Law of floatation:

Weight of the block = weight of the liquid displaced by immersed part of the block
V ρ g = V1 ρl g
V ρ = V1 ρl
Vρ
V1 =
ρl
0.1 x 10 0 10
= == = 0.067 m3
150 150
Volume of the block above the surface of liquid = V - V1
= 0.1 – 0.067
= 0.033 m3

Q3: A solid of R.D 4 is found to weigh 0.200kgf in air. Find its apparent weight in water.
Solution:
Given
Relative density of a solid = R.D = 4
Weight of solid in air, W = 0.200kgf
Let Weight of solid in water be W1
We know that,
Weight of the solid ∈air
R.D =
weight of the solid ∈air−weight of the solid ∈water

Also, weight of the solid ∈air−weight of the solid ∈water = Upthrust

Weight of the solid ∈air


R.D =
upthrust

0.200
R.D =
upthrust

0.200
Upthrust = = 0.05kgf
4

Apparent weight of solid in water = weight of solid in air – upthrust

= 0.200-0.05 = 0.15kgf

Q4: A solid lighter than water is found to weigh 7.5gf in air. When tied to a sinker the
combination is found to weigh 62.5gf. if sinker alone weighs 72.5gf in water. Find the R.D of the
solid.

Solution:

Given
Weight of the solid in air = 7.5gf

Weight of the sinker in water = 72.5gf

Weight of the solid in water + Weight of the sinker in water = 62.5gf

Weight of the solid in air + Weight of the sinker in water = 7.5 + 72.5 = 80gf

Weight of solid in air - Weight of the solid in water = 80 – 62.5 = 17.5gf

Weight of the solid ∈air


R.D =
weight of the solid ∈air−weight of the solid ∈water
7.5
=
17.5

= 0.428

Q5: A block of wood is floating on water with its dimensions 50cm x 50cm x50cm inside water.
Calculate buoyant force acting on the block. Take g= 9.8 ms-2.

Solution:

Given

Volume of the block, V = 50cm x 50cm x50cm

= 125000 cm-3

= 125000 x 10-6 m3

= 0.125m3

g = 9.8 ms-2

Density of water, ρ = 1000 kgm-3

Buoyant force = V ρ g

= 0.125 x 1000 x 9.8

= 1225N

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