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Question: A ball falling in a lake of depth \(200\,m\) shows \(0.1\% \) decrease in its volume at the bottom. ...

A ball falling in a lake of depth 200m200\,m shows 0.1%0.1\% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball:
A. 19.6×108Nm219.6 \times {10^8}N{m^{ - 2}}
B. 19.6×1010Nm219.6 \times {10^{ - 10}}N{m^{ - 2}}
C. 19.6×1010Nm219.6 \times {10^{10}}N{m^{ - 2}}
D. 19.6×108Nm219.6 \times {10^{ - 8}}N{m^{ - 2}}

Explanation

Solution

The bulk modulus is one of the measures of the mechanical properties of the solids. Also, the bulk modulus is defined as the ratio of the pressure and the strain. Therefore, we will use the formula of bulk modulus of elasticity to calculate the bulk modulus of the ball.

Formula used:
The formula of the bulk modulus of elasticity is given by
B=ΔPΔVVB = \dfrac{{\Delta P}}{{\dfrac{{\Delta V}}{V}}}
Here, BB is the bulk modulus, ΔP\Delta P is the change of the pressure or the force applied per unit area on the material, ΔV\Delta V is the change in the volume of the material due to the compression, and VV is the initial volume of the material in the units.

Complete step by step answer:
In this example, a ball falls in a lake that is of depth 200m200\,m. After falling off the ball, there is a 0.1%0.1\% decrease in the volume at the bottom.
Now, the formula of the bulk modulus is given by
B=PΔVVB = \dfrac{P}{{\dfrac{{\Delta V}}{V}}}
Also, the value of the pressure is given by
P=ρghP = \rho gh
PP is the pressure, ρ\rho is the density of the liquid, gg is the acceleration due to gravity, and hh is the height of the liquid.
Also, the decrease in the volume of the lake is given by
ΔVV=0.1%\dfrac{{\Delta V}}{V} = 0.1\%
ΔVV=0.1100\Rightarrow \,\dfrac{{\Delta V}}{V} = \dfrac{{0.1}}{{100}}
Therefore, putting this value in the formula of the bulk modulus as shown below
B=ρgh0.1%B = \dfrac{{\rho gh}}{{0.1\% }}
B=1000×9.8×2000.1100\Rightarrow \,B = \dfrac{{1000 \times 9.8 \times 200}}{{\dfrac{{0.1}}{{100}}}}
B=1000×9.8×200×1000.1\Rightarrow \,B = \dfrac{{1000 \times 9.8 \times 200 \times 100}}{{0.1}}
B=19.6×1070.1\Rightarrow \,B = \dfrac{{19.6 \times {{10}^7}}}{{0.1}}
B=196×107\Rightarrow \,B = 196 \times {10^7}
B=19.6×108Nm2\therefore\,B = 19.6 \times {10^8}N{m^{ - 2}}
Hence, the bulk modulus of the material of the ball is 19.6×108Nm219.6 \times {10^8}N{m^{ - 2}} .

Hence, option A is the correct option.

Note: The bulk modulus of elasticity is one of the measures of the mechanical properties of solids. Other elastic modules include Young’s modulus and Shear modulus. In any case, the bulk elastic properties of a material are used to determine how much it will compress under a given amount of external pressure.