Example 9.5 The average depth of IndianOcean is about 3000 m. Calculat

1.	Example 9.5 The average depth of IndianOcean is about 3000 m. Calculate thefractional compression, Δν/ν, of water atthe bottom of the ocean, given that the bulkmodulus of water is 2.2 × 109 N m-2. (Takeg = 10 m s-2)
Answer» Thank u 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

Example 9.5 The average depth of IndianOcean is about 3000 m. Calculate thefractional compression, Δν/ν, of water atthe bottom of the ocean, given that the bulkmodulus of water is 2.2 × 109 N m-2. (Takeg = 10 m s-2)

Answer»

Thank u

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

Discussion

No Comment Found

Related InterviewSolutions

7. A body has a weight 50 N in air. Afterimmersing completely in liquid its weight is20 N. Then, the buoyant force on the bodywill be
Name the particulate which causes the disease mentioned below:(a)minamata (b)silicosis.
- Where does a body weight more at the surface of the carth or in a mine?
DelyAbody79entire chainthe ch.57. A body of mass 50gm collides elastically with anotherbody of mass 30 gm at rest. Then the percentage lossof the velocity of the colliding body during collision is1) 25% 2) 75% 3) 50% 4) 67%3) 120380.beide
What will be the acceleration of a body of mass5 kg if a force of 200 N is applied on it?
Rig. 5.168.13 Two bodles of masses 10 kg and 20 kg respectively kept on a smooth, hortzontalsurface are tied to the ends of a light string a horizontal force F 600 Nsapplied to lu A (t) B along the direction of string. What is the tension in thestring in each case
culate the gravitational force befween two protons kept at a distance67 x 10-27 kg. (1 femtometre = 10-15 m)
15. A uniform chain of length 2 m is kept on atable such that a length of 60 cm hangs freelyfrom the edge of the table. The total mass ofthe chain is 4 kg. What is the work done inpulling the entire chain onto the table ?(g = 10 m/s)1) 7.2J273.63) 120J4) 1200
Two block of mass 3 kg and 6 kg respectively are placed on a smooth 1.0 m/shorizontal surface. They are connected by a light spring of force 3 kgconstant k 200 N/m. Initially the spring is unstretched. The indicatedzmvelocities are imparted to the blocks. The maximum extension of thespring will be:A30 cm(C) 20 cm2.0 mls(B) 25 cm(D) 15 cm
88. A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the accelerationof the stone be 9.8 m/s directed towards the ground, calculate the initial velocity of the stone with which itis thrown upwards.