Question

state hooks law calculate the frictional compressive delta v per v of water at the bottom of the ocean having depth 3000 m the bulk modulus of water is 2.2 ×10 race to power 9 nm-2 take g = 10 ms-2​

Answer 1

Explanation:

We know one thing 

We know one thing P = P₀ + ρgh 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² Again, we have to use formula, 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² Again, we have to use formula, B = P/{-∆V/V} 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² Again, we have to use formula, B = P/{-∆V/V} Here, B is bulk modulus and { -∆V/V} is the fractional compression 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² Again, we have to use formula, B = P/{-∆V/V} Here, B is bulk modulus and { -∆V/V} is the fractional compression So, -∆V/V = P/B 

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² Again, we have to use formula, B = P/{-∆V/V} Here, B is bulk modulus and { -∆V/V} is the fractional compression So, -∆V/V = P/B Put , P = 3.01 × 10⁷ N/m² and B= 2.2 × 10⁹ N/m²

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² Again, we have to use formula, B = P/{-∆V/V} Here, B is bulk modulus and { -∆V/V} is the fractional compression So, -∆V/V = P/B Put , P = 3.01 × 10⁷ N/m² and B= 2.2 × 10⁹ N/m²∴ fractional compression = 3.01 × 10⁷/2.2 × 10⁹ = 1.368 × 10⁻²

We know one thing P = P₀ + ρgh Where P₀ is the atmospheric pressure , g is acceleration due to gravity, h is the height from the Earth surface and ρ is density of water Here, P₀ = 10⁵ N/m² , g = 10m/s² , h = 3000m and ρ = 10³ Kg/m³ Now, P = 10⁵ + 10³ × 10 × 3000 = 3.01 × 10⁷ N/m² Again, we have to use formula, B = P/{-∆V/V} Here, B is bulk modulus and { -∆V/V} is the fractional compression So, -∆V/V = P/B Put , P = 3.01 × 10⁷ N/m² and B= 2.2 × 10⁹ N/m²∴ fractional compression = 3.01 × 10⁷/2.2 × 10⁹ = 1.368 × 10⁻²Hence the ans is D