Question

oose the correct answer:
The depth at which density of water becomes 1%
greater than that at the surface, (compressibility of
water is 50 * 10-11/Pa) is
(1) 1000 m
(2) 500 m
(3) 2000 m
(4) 4000 m
​

Answer 1

answer : option (3) 2000m

explanation : compressibility is just inverse of bulk modulus .

so, compressibility = (∆V/V)/P ......(1)

we know, density = mass / volume

or, d = m / V

if mass is constant then, density is directly proportional to volume.

so, % change in volume = % change in density

so, (∆V/V) × 100 = 1

or, ∆V/V = 0.01

and P = $P_0+\rho gh$

where $P_0$ is atmospheric pressure and $\rho$ density of water.

so, P = 10^5 N/m² + 1000 kg/m³ × 10m/s² × h

now, from equation (1),

50 × 10^-11 /Pa = 0.01/( 10^5 + 10⁴ × h)

or, 5× 10^-10 (10^5 + 10⁴ × h) = 0.01 = 10^-2

or, 10^5 + 10⁴ × h = 0.2 × 10^8

or, 10⁴ × h = (200 - 1) × 10^5

or, 10⁴ × h = 199 × 10^5

or, h = 1990 m ≈ 2000 m

hence, answer should be 2000m.

Answer 2

Answer:

2000m

Explanation:

∆density / density = ∆P / B

0.01 = density × g × h × K

h = 0.01 / density of water ×g×K

h= 0.01 / 10^3 × 10 × 50× 10^-11

h = 0.2 × 10^4

h = 2000m