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Question: The bulk modulus of water if its volume changes from 100 litre to 99.5 litre under a pressure of 100...

The bulk modulus of water if its volume changes from 100 litre to 99.5 litre under a pressure of 100 atm is

(Take 1 atm=105 N m−2)= 10^{5}\text{ N }\text{m}^{- 2})

A

2×107 N m−22 \times \text{1}\text{0}^{7}\text{ N }\text{m}^{- 2}

B

2×108 N m−22 \times \text{1}\text{0}^{8}\text{ N }\text{m}^{- 2}

C

2×109 N m−22 \times \text{1}\text{0}^{9}\text{ N }\text{m}^{- 2}

D

2×1010 N m−22 \times \text{1}\text{0}^{\text{10}}\text{ N }\text{m}^{- 2}

Answer

2×109 N m−22 \times \text{1}\text{0}^{9}\text{ N }\text{m}^{- 2}

Explanation

Solution

: According to definition of bulk modulus,

B=−PΔV/VB = - \frac{P}{\Delta V/V}

Here, p =100 atm =100×105Nm−2=107Nm−2100 \times 10^{5}Nm^{- 2} = 10^{7}Nm^{- 2}

ΔV=(99.5−100)litre=−0.5litre\Delta V = (99.5 - 100)litre = - 0.5litre

V = 100 litre

∴B=−107Nm−2(−0.5litre/100litre)=2×109Nm−2\therefore B = - \frac{10^{7}Nm^{- 2}}{( - 0.5litre/100litre)} = 2 \times 10^{9}Nm^{- 2}