When a sphere is taken to the bottom of a sea | vedclass

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(C) The pressure at depth $h$ is given by $P = \rho gh$. Given: $h = 1 \ km = 10^3 \ m$,$\rho = 10^3 \ kg \ m^{-3}$,$g = 10 \ m \ s^{-2}$. $P = 10^3 \

Vedclass · Accepted Answer

$10 	imes 10^{10} \ N \ m^{-2}$

Vedclass · Answer

(C) The pressure at depth $h$ is given by $P = ho gh$. Given: $h = 1 \ km = 10^3 \ m$,$ho = 10^3 \ kg \ m^{-3}$,$g = 10 \ m \ s^{-2}$. $P = 10^3 	imes 10 	imes 10^3 = 10^7 \ N \ m^{-2}$. The fractional change in volume is $\frac{\Delta V}{V} = 0.01 \% = \frac{0.01}{100} = 10^{-4}$. The bulk modulus $B$ is defined as $B = \frac{P}{\Delta V / V}$. $B = \frac{10^7}{10^{-4}} = 10^{11} \ N \ m^{-2}$. Wait,checking the options: $10 	imes 10^{10} = 10^{11}$. Thus,option $C$ is correct.

When a sphere is taken to the bottom of a sea of depth $1 \ km$,it contracts in volume by $0.01 \%$. The bulk modulus of the material of the sphere is (Acceleration due to gravity $= 10 \ m \ s^{-2}$,density of sea water $= 10^3 \ kg \ m^{-3}$)

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