Mechanical Properties of Solids
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Question : 13
Total: 21
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10 3 k g m – 3 45.8 × 10 – 11 P a – 1 .
Solution:
Here, P = 80.0 atm .
= 80.0 × 1.013 × 10 5 P a
Compressibility,
= 45.8 × 10 − 11 P a − 1
Density of water at surface,ρ = 1.03 × 10 3 k g m − 3
Let ρ′ be the density of water at the given depth. If V and V′ are volume of certain mass M of ocean water at surface and at a given depth, then
V =
V " =
∴ Δ V = V − V " = M (
−
)
∴
= M (
−
) × ρ M = 1 −
= 1 −
. . . ( i )
As, Bulk modulusB =
=
∴
= ( 80.0 × 1.013 × 10 5 ) × 45.8 × 10 − 11 = 3.712 × 10 − 3
Putting this value in (i) we get
1 −
= 3.712 × 10 − 3
orρ ′ =
= 1.034 × 10 3 k g m − 3
Compressibility,
Density of water at surface,
Let ρ′ be the density of water at the given depth. If V and V′ are volume of certain mass M of ocean water at surface and at a given depth, then
As, Bulk modulus
Putting this value in (i) we get
or
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