(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 \

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

(C) The pressure at depth $h$ is given by $P = \rho gh$. Given: $h = 1 \ km = 10^3 \ m$,$\rho = 10^3 \ kg \ m^{-3}$,$g = 10 \ m \ s^{-2}$. $P = 10^3 \times 10 \times 10^3 = 10^7 \ N \ m^{-2}$. 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: $10 \times 10^{10} = 10^{11}$. Thus,option $C$ is correct.

Class 11 Physics Mechanical Properties of Solids Bulk Modulus

Medium

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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A
$10 \times 10^6 \ N \ m^{-2}$
B
$1.2 \times 10^{10} \ N \ m^{-2}$
C
$10 \times 10^{10} \ N \ m^{-2}$
D
$10 \times 10^{11} \ N \ m^{-2}$

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Mechanical Properties of Solids

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Bulk Modulus

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$A$ solid copper cube of $7 \,cm$ edge is subjected to a hydraulic pressure of $8000 \,kPa$. The volume contraction of the copper cube is (Bulk modulus of copper $= 140 \,GPa$)

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The average depth of the Indian Ocean is about $3000\; m$. Calculate the fractional compression,$\Delta V / V,$ of water at the bottom of the ocean,given that the bulk modulus of water is $2.2 \times 10^{9}\; N m^{-2}$. (Take $g = 10\; m s^{-2}$)

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$A$ spherical ball of volume $2000 \text{ cm}^3$ is subjected to a hydraulic pressure of $15 \text{ atm}$. If the change in volume is $5 \times 10^{-2} \text{ cm}^3$,the bulk modulus of the material of the spherical ball is (Given: $1 \text{ atm} = 10^5 \text{ Nm}^{-2}$)

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The fractional compression $\left(\frac{\Delta V}{V}\right)$ of water at a depth of $2.5 \ km$ below the sea level is . . . . . . $\%$. Given: the Bulk modulus of water $B = 2 \times 10^9 \ N m^{-2}$,density of water $\rho = 10^3 \ kg m^{-3}$,and acceleration due to gravity $g = 10 \ m s^{-2}$.

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The average density of Earth's crust $10 \ km$ beneath the surface is $2.7 \ g/cm^3$. The speed of longitudinal seismic waves at that depth is $5.4 \ km/s$. The bulk modulus of Earth's crust,considering its behavior as a fluid at that depth,is:

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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 $= 10^3 \ kg \ m^{-3}$)

Similar Questions

$A$ solid copper cube of $7 \,cm$ edge is subjected to a hydraulic pressure of $8000 \,kPa$. The volume contraction of the copper cube is (Bulk modulus of copper $= 140 \,GPa$)

The average depth of the Indian Ocean is about $3000\; m$. Calculate the fractional compression,$\Delta V / V,$ of water at the bottom of the ocean,given that the bulk modulus of water is $2.2 \times 10^{9}\; N m^{-2}$. (Take $g = 10\; m s^{-2}$)

$A$ spherical ball of volume $2000 \text{ cm}^3$ is subjected to a hydraulic pressure of $15 \text{ atm}$. If the change in volume is $5 \times 10^{-2} \text{ cm}^3$,the bulk modulus of the material of the spherical ball is (Given: $1 \text{ atm} = 10^5 \text{ Nm}^{-2}$)

The fractional compression $\left(\frac{\Delta V}{V}\right)$ of water at a depth of $2.5 \ km$ below the sea level is . . . . . . $\%$. Given: the Bulk modulus of water $B = 2 \times 10^9 \ N m^{-2}$,density of water $\rho = 10^3 \ kg m^{-3}$,and acceleration due to gravity $g = 10 \ m s^{-2}$.

The average density of Earth's crust $10 \ km$ beneath the surface is $2.7 \ g/cm^3$. The speed of longitudinal seismic waves at that depth is $5.4 \ km/s$. The bulk modulus of Earth's crust,considering its behavior as a fluid at that depth,is:

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