The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10-11 Pa -1 and density of water is 103 kg / m 3 . What fractional compression of water will be obtained at the bottom of the ocean?

Question

The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10-11 Pa -1 and density of water is 103 kg / m 3 . What fractional compression of water will be obtained at the bottom of the ocean? (A) 1.0 × 10-2 (B) 1.2 × 10-2 (C) 1.4 × 10-2 (D) 1.8 × 10-2

Tardigrade Admin · Accepted Answer

compressibility is given as K =((Δ V / V )/Δ P ) Δ V =K× Δ P × V Substituting values Δ P =ρ gh Pa , K=45 × 10-11 Pa -1 (Δ V / V )=45 × 10-11 × Δ P (Δ V / V )=45 × 10-11 × 103 × 10 × 2700 (Δ V / V )=1.2 × 10-2

Q. The approximate depth of an ocean is $2700 m$ . The compressibility of water is $45.4 \times 1 0^{- 11} P a^{- 1}$ and density of water is $1 0^{3} k g / m 3.$ What fractional compression of water will be obtained at the bottom of the ocean?

$1.0 \times 1 0^{- 2}$

$1.2 \times 1 0^{- 2}$

$1.4 \times 1 0^{- 2}$

$1.8 \times 1 0^{- 2}$

Q. 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?

1.0×10−2

1.2×10−2

1.4×10−2

1.8×10−2

compressibility is given as K=ΔPVΔV​​ ΔV=K×ΔP×V Substituting values ΔP=ρghPa, K=45×10−11Pa−1 VΔV​=45×10−11×ΔP VΔV​=45×10−11×103×10×2700 VΔV​=1.2×10−2

Q. The approximate depth of an ocean is $2700 m$ . The compressibility of water is $45.4 \times 1 0^{- 11} P a^{- 1}$ and density of water is $1 0^{3} k g / m 3.$ What fractional compression of water will be obtained at the bottom of the ocean?

$1.0 \times 1 0^{- 2}$

$1.2 \times 1 0^{- 2}$

$1.4 \times 1 0^{- 2}$

$1.8 \times 1 0^{- 2}$