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Question

The compressibility of water is 4 x 10$$^{-5}$$ per unit atmosphere pressure. The decrease in volume of 100cm$$^{3}$$ water under a pressure of 100atm will be


A
0.4 cm3
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B
4 x 105cm3
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C
0.025 cm3
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D
0.04cm3
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Solution

The correct option is A 0.4 cm$$^{3}$$
1 atm pressure $$= 10^{5} N/m^{2}$$
$$C=\dfrac{1}{B}=4\times \dfrac{10^{-5}}{10^{5}} m^{2}/N$$
$$=4\times 10^{-10} m^{2}/N$$
$$P=100  atm  , V=100  cm^{3}$$
$$\dfrac{10^{10}}{4}=\dfrac{-100\times 10^{5}}{\dfrac{\Delta V}{100\times 10^{-6}}}$$
$$\dfrac{10^{10}}{4}=\dfrac{-10^{3}}{\Delta V}$$
$$\Delta V=-4\times 10^{-7} m^{3}$$
$$=-0.4 cm^{3}$$
-ve sign indicates that volume is decreased.

Physics

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