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}} $
