Question

If a rubber ball is taken at the depth of 6000 m in a ocean. If the density of the water is 10³ kg/m³. g= 10 m/s² and then the bulk modulus of rubber is 2 x 10⁹N/m². The percentage change in its radius is will be:​

Answer 1

Answer:

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

Hence the ans is D