Question

when a rubber ball is taken to the bottom of a Sea of depth 1400m its volume decreases by 2 percentage the bulk modulus of rubber ball is ?

Answer 1

Given,

g of water = 10 m/s^2

h = 1400m

Change in volume, i.e, ΔV/V = 2% = 2/100

rho of water = 1000 kg/m^3  (standard parameter)

We know that,

Bulk Modulus = PV/∆V

Now, P=rho *g * h = 1000 * 10 * 1400 = 14*10^6 kg/ms^2

∴ Bulk modulus = (14*10^6)*(100/2) = 7*10^8 kgm^2/s^2

Answer 2

volume decreased by 2% .

Let initial volume is V

then, % change in volume = ∆V/V × 100 = 2%

so, ∆V/V = 2/100 = 0.02

now, pressure in 1400 depth from sea level , P

= $P_0+\rho gh$

= 1.05 × 10^5 N/m² + 1000 × 10 × 1400

= 1.05 × 10^5 + 140 × 10^5

= 141.05 × 10^5 N/m²

so, Bulk modulus , B = P/(∆V/V)

= 141.05 × 10^5/0.02 N/m²

≈ 70.5 × 10^7 N/m²

≈ 7 × 10^8 N/m²

hence, option (1) is correct.