### Neet Solved Paper 2015 Question 16

##### Question: The approximate depth of an ocean is 2 700 m. The compressibility of water is $ 45.4\times {10^{-11}}P{a^{-1}} $ and density of water is $ 10^{3}kg/m^{3} $ . What fractional compression of water will be obtained at the bottom of the ocean?

#### Options:

A) $ 0.8\times {10^{-2}} $

B) $ 1.0\times {10^{-2}} $

C) $ 1.2\times {10^{-2}} $

D) $ 1.4\times {10^{-2}} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

Given d = 2700m $ \rho =10^{3}kg/m^{3} $

Compressibility $ =45.4\times {10^{-11}} $ per pascal

The pressure at the bottom of ocean is given by

$ p=\rho gd $ $ =10^{3}\times 10\times 2700=27\times 10^{6},Pa $

So, fractional compression = compressibility $ \times $ pressure

$ =45.4\times {10^{-11}}\times 27\times 10^{6}=1.2\times {10^{-2}} $