SATHEE: Mechanical Properties of Fluids

Mechanical Properties of Fluids - Result Question 4

4. The approximate depth of an ocean is $2700 m$ . The compressibility of water is $45.4 \times 10^{- 11} P a^{- 1}$ and density of water is $10^{3} k g / m^{3}$ . What fractional compression of water will be obtained at the bottom of the ocean?

(a) $1.0 \times 10^{- 2}$

(b) $1.2 \times 10^{- 2}$

(d) $0.8 \times 10^{- 2}$

[2015]

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Answer:

Correct Answer: 4. (b)

Solution:

(b) Compressibility of water,

$K = 45.4 \times 10^{- 11} P a^{- 1}$

density of water $P = 10^{3} k g / m^{3}$

depth of ocean, $h = 2700 m$

We have to find $\frac{Δ V}{V} =$ ?

As we know, compressibility,

$K = \frac{1}{B} = \frac{(Δ V / V)}{P} (P = ρ g h)$

So, $(Δ V / V) = K ρ g h$

$= 45.4 \times 10^{- 11} \times 10^{3} \times 10 \times 2700 = 1.2258 \times 10^{- 2}$

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Mechanical Properties of Fluids - Result Question 4

4. The approximate depth of an ocean is 2700m. The compressibility of water is 45.4×10−11Pa−1 and density of water is 103kg/m3. What fractional compression of water will be obtained at the bottom of the ocean?

Answer:

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4. The approximate depth of an ocean is $2700 m$ . The compressibility of water is $45.4 \times 10^{- 11} P a^{- 1}$ and density of water is $10^{3} k g / m^{3}$ . What fractional compression of water will be obtained at the bottom of the ocean?