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Elasticity

Section: Physics

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Question : 8 of 17

Marks: +1, -0

When a sphere is taken to the bottom of a sea of depth 1 km , it contracts in volume by

0.01\%

, then the bulk modulus of the material of the sphere is (Acceleration due to gravity

=10 \text{ m} \text{ s}^{-2}

[AP EAPCET 21 May 2025 S2]

$10 \times 10^6 \text{ N m}^{-2}$
$1.2 \times 10^{10} \text{ N m}^{-2}$
$10 \times 10^{10} \text{ N m}^{-2}$
$10 \times 10^{11} \text{ N m}^{-2}$

Solution:

Depth of sea,

h=1 \text{ km}=1000 \text{ m}

Relative decrease in volume,

\; \frac{\Delta V}{V}=0.01\%=0.0001=10^{-4}

Acceleration due to gravity,

g=10 \text{ m/s}^2

Density of sea water,

\rho \approx 10^3 \text{ kg/m}^3

Step 1. Calculate the pressure at that depth

P = \rho g h = 1000 \times 10 \times 1000 = 10^7 \text{ N/m}^2

Step 2. Use the definition of Bulk modulus

K

K = -\; \frac{\Delta P}{\frac{\Delta V}{V}}

Here,

\Delta P = P = 10^7 \text{ N/m}^2

, and

\; \frac{\Delta V}{V} = -10^{-4}

K = \; \frac{10^7}{10^{-4}} = 10^{11} \text{ N/m}^2

Final Answer:

K = 10^{11} \text{ N/m}^2

Correct Option: D)

10 \times 10^{10} \text{ N/m}^2

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